12: Epilogue
-
- Last updated
- Save as PDF
In this book, I have focused on corpus linguistics as a methodology, more precisely, as an application of a general observational scientific procedure to large samples of linguistic usage. I have refrained from placing this method in a particular theoretical framework for two reasons.
The first reason is that I am not convinced that linguistics should be focusing quite as much on theoretical frameworks, but rather on linguistic description based on data. Edward Sapir famously said that “unfortunately, or luckily, no language is tyrannically consistent. All grammars leak” (Sapir 1921: 39). This is all the more true of formal models, which have a tendency to attempt to achieve tyrannical consistency by pretending those leaks do not exist or, if they do exist, are someone else’s problem. To me, and to many others whose studies I discussed in this book, the ways grammars leak are simply more interesting than the formalisms that help us ignore these leaks.
The second reason is that I believe that corpus linguistics has a place in any theoretical linguistic framework, as long as that framework has some commitment to modeling linguistic reality. Obviously, the precise place, or rather, the distance from the data analyzed using this method and the consequences of this analysis for the model depend on the kind of linguistic reality that is being modeled. If it is language use, as it usually is in historically or sociolinguistically oriented studies, the distance is relatively short, requiring the researcher to discover the systematicity behind the usage patterns observed in the data. If it is the mental representation of language, the length of the distance depends on your assumptions about those representations.
Traditionally, those representations have been argued to be something fundamentally different from linguistic usage. It has been claimed that they are an ephemeral “competence” based on a “universal” grammar. There is disagreement as to the nature of this universal grammar – some claim that is a “mental organ” (Chomsky 1980), some imagine it as an evolved biological instinct (Pinker 1994). But all proponents of a universal grammar are certain that mental representations of language are dependent on and responsible for linguistic usage only in the most indirect ways imaginable, making corpora largely useless for the study of language. As I have argued in Chapters 1 and 2, the only methodological alternative to corpus data that proponents of this view offer – i.e. introspective grammaticality judgments – suffer from all the same problems as corpus data, without offering any of the advantages.
However, more recent models do not draw as strict a line between usage and mental representations. The Usage-Based Model (Langacker 1991) is a model of linguistic knowledge based on the assumption that speakers initially learn language as a set of unanalyzed chunks of various sizes (“established units”), from which they derive linguistic representations of varying degrees of abstractness and complexity based on formal and semantic correspondences across these units (cf. Langacker 1991: 266f). The Emergent Grammar model is based on similar assumptions but eschews abstractness altogether, viewing language as “built up out of combinations of [...] prefabricated parts”, as “a kind of pastiche, pasted together in an improvised way out of ready-made elements” (Hopper 1987: 144).
In these models, the corpus becomes more than just a research tool, it becomes an integral part of a model of linguistic competence (cf. Stefanowitsch 2011). This view is most radically expressed in the notion of “lexical priming” developed in Hoey (2005), in which linguistic competence is seen as a mental concordance over linguistic experience:
The notion of priming as here outlined assumes that the mind has a mental concordance of every word it has encountered, a concordance that has been richly glossed for social, physical, discoursal, generic and interpersonal context. This mental concordance is accessible and can be processed in much the same way that a computer concordance is, so that all kinds of patterns, including collocational patterns, are available for use. It simultaneously serves as a part, at least, of our knowledge base. (Hoey 2005: 11)
Obviously, this mental concordance would not correspond exactly to any concordance derived form an actual linguistic corpus. First, because – as discussed in Chapters 1 and 2 – no linguistic corpus captures the linguistic experience of a given individual speaker or the “average” speaker in a speech community; second, because the concordance that Hoey envisions is not a concordance of linguistic forms but of contextualized linguistic signs – it contains all the semantic and pragmatic information that corpus linguists have to reconstruct laboriously in their analyses. Still, an appropriately annotated concordance from a balanced corpus would be a reasonable operationalization of this mental concordance (cf. also Taylor 2012).
In less radical usage-based models of language, such as Langacker’s, the corpus is not a model of linguistic competence – the latter is seen as a consequence of linguistic input perceived and organized by human minds with a particular structure (such as the capacity for figure-ground categorization). The corpus is, however, a reasonable model (or at least an operationalization) of this linguistic input. Many of the properties of language that guide the storage of units and the abstraction of schemas over these stored units can be derived from corpora – frequencies, associations between units of linguistic structure, distributions of these units across grammatical and textual contexts, the internal variability of these units, etc. (cf. Stefanowitsch & Flach 2016 for discussion).
This view is explicitly taken in language acquisition research conducted within the Usage-Based Model (e.g. Tomasello 2003, cf. also Dabrowska 2001, Diessel 2004), where children’s expanding grammatical abilities, as reflected in their linguistic output, are investigated against the input they get from their caretakers as recorded in large corpora of caretaker-child interactions. The view of the corpus as a model of linguistic input is less explicit in the work of the major theoretical proponents of the Usage-Based Model, who connect the notion of usage to the notion of linguistic corpora only in theory. However, it is a view that offers a tremendous potential to bring together two broad strands of research – cognitive-functional linguistics (including some versions of construction grammar) and corpus linguistics (including attempts to build theoretical models on corpus data, such as pattern grammar (Hunston & Francis 2000) and Lexical Priming (Hoey 2005)). These strands have developed more or less independently and their proponents are sometimes mildly hostile toward each other over small but fundamental differences in perspective (see McEnery & Hardie 2012, Section 8.3 for discussion). If they could overcome these differences, they could complement each other in many ways, cognitive linguistics providing a more explicitly psychological framework than most corpus linguists adopt, and corpus linguistics providing a methodology that cognitive linguists serious about usage urgently need.
Finally, in usage-based models as well as in models of language in general, corpora can be treated as models (or operationalizations) of the typical linguistic output of the members of a speech community, i.e. the language produced based on their internalized linguistic knowledge. This is the least controversial view, and the one that I have essentially adopted throughout this book. Even under this view, corpus data remain one of the best sources of linguistic data we have – one that can only keep growing, providing us with ever deeper insights into the leaky, intricate, ever-changing signature activity of our species.
I hope this book has inspired you and I hope it will help you produce research that inspires all of us.


8.3: Lab Report
Exercise A: Colony Characteristics
|
Name of Bacterium: |
Typically found where? |
Average colony size (mm): |
Colony Shape: |
Margin: |
Surface: |
Pigmentation (or none) |
|---|---|---|---|---|---|---|
|
Micrococcus luteus |
|
|
|
|
|
|
|
Serratia marcescens |
|
|
|
|
|
|
|
Bacillus |
|
|
|
|
|
|
|
Escherichia coli |
|
|
|
|
|
|
Table 8.1 Colony Characteristics
Below, list three ways we benefit from bacteria. You may think of examples from nature, industrial and medical applications, or others. You may reference our lecture material or search the internet (be sure to cite your source).
A.1.
A.2.
A.3.
Exercise B: Investigating Environmental Microbes
B.1. What surfaces will you swab? Examples: you can compare before and after handwashing (don’t dry your hands), or compare two body areas, or environmental samples such as commonly touched areas (phone vs your pencil).
B.2. What is your hypothesis? Get approval before you swab .
B.3.
B.4. After incubation: Sketch and describe your results below.
B.5. Do your results support your hypothesis? Explain:
Exercise C: Viewing Prepared Slide of Gram-Stained "Three Bacterial Types"
|
Type: |
Describe the shape and arrangement of cells: |
Sketch at 1000x (oil) |
Gram +/-: |
|---|---|---|---|
|
bacillus (bacilli) |
|||
|
coccus (cocci) |
|||
|
spirillum (spirilla) |
Table 8.2 Prepared Slide of Gram-Stained "Three Bacterial Types"
Exercise D
|
Bacterium: |
Color of cells (100x lens) |
Gram result (+/-) |
Shape (morphology) |
Sketch in color, 100x lens |
|---|---|---|---|---|
|
Micrococcus luteus |
||||
|
Serratia marcescens |
||||
|
Bacillus |
||||
|
Escherichia coli |


9.5: The p value of a test
-
- Last updated
- Save as PDF
In one sense, our hypothesis test is complete; we’ve constructed a test statistic, figured out its sampling distribution if the null hypothesis is true, and then constructed the critical region for the test. Nevertheless, I’ve actually omitted the most important number of all: the p value . It is to this topic that we now turn. There are two somewhat different ways of interpreting a p value, one proposed by Sir Ronald Fisher and the other by Jerzy Neyman. Both versions are legitimate, though they reflect very different ways of thinking about hypothesis tests. Most introductory textbooks tend to give Fisher’s version only, but I think that’s a bit of a shame. To my mind, Neyman’s version is cleaner, and actually better reflects the logic of the null hypothesis test. You might disagree though, so I’ve included both. I’ll start with Neyman’s version…
softer view of decision making
One problem with the hypothesis testing procedure that I’ve described is that it makes no distinction at all between a result this “barely significant” and those that are “highly significant”. For instance, in my ESP study the data I obtained only just fell inside the critical region - so I did get a significant effect, but was a pretty near thing. In contrast, suppose that I’d run a study in which X=97 out of my N=100 participants got the answer right. This would obviously be significant too, but my a much larger margin; there’s really no ambiguity about this at all. The procedure that I described makes no distinction between the two. If I adopt the standard convention of allowing α=.05 as my acceptable Type I error rate, then both of these are significant results.
This is where the p value comes in handy. To understand how it works, let’s suppose that we ran lots of hypothesis tests on the same data set: but with a different value of α in each case. When we do that for my original ESP data, what we’d get is something like this
| Value of α | Reject the null? |
|---|---|
| 0.05 | Yes |
| 0.04 | Yes |
| 0.03 | Yes |
| 0.02 | No |
| 0.01 | No |
When we test ESP data (X=62 successes out of N=100 observations) using α levels of .03 and above, we’d always find ourselves rejecting the null hypothesis. For α levels of .02 and below, we always end up retaining the null hypothesis. Therefore, somewhere between .02 and .03 there must be a smallest value of α that would allow us to reject the null hypothesis for this data. This is the p value; as it turns out the ESP data has p=.021. In short:
p is defined to be the smallest Type I error rate (α) that you have to be willing to tolerate if you want to reject the null hypothesis.
If it turns out that p describes an error rate that you find intolerable, then you must retain the null. If you’re comfortable with an error rate equal to p, then it’s okay to reject the null hypothesis in favour of your preferred alternative.
In effect, p is a summary of all the possible hypothesis tests that you could have run, taken across all possible α values. And as a consequence it has the effect of “softening” our decision process. For those tests in which p≤α you would have rejected the null hypothesis, whereas for those tests in which p>α you would have retained the null. In my ESP study I obtained X=62, and as a consequence I’ve ended up with p=.021. So the error rate I have to tolerate is 2.1%. In contrast, suppose my experiment had yielded X=97. What happens to my p value now? This time it’s shrunk to p=1.36×10−25, which is a tiny, tiny 163 Type I error rate. For this second case I would be able to reject the null hypothesis with a lot more confidence, because I only have to be “willing” to tolerate a type I error rate of about 1 in 10 trillion trillion in order to justify my decision to reject.
probability of extreme data
The second definition of the p-value comes from Sir Ronald Fisher, and it’s actually this one that you tend to see in most introductory statistics textbooks. Notice how, when I constructed the critical region, it corresponded to the tails (i.e., extreme values) of the sampling distribution? That’s not a coincidence: almost all “good” tests have this characteristic (good in the sense of minimising our type II error rate, β). The reason for that is that a good critical region almost always corresponds to those values of the test statistic that are least likely to be observed if the null hypothesis is true. If this rule is true, then we can define the p-value as the probability that we would have observed a test statistic that is at least as extreme as the one we actually did get. In other words, if the data are extremely implausible according to the null hypothesis, then the null hypothesis is probably wrong.
common mistake
Okay, so you can see that there are two rather different but legitimate ways to interpret the p value, one based on Neyman’s approach to hypothesis testing and the other based on Fisher’s. Unfortunately, there is a third explanation that people sometimes give, especially when they’re first learning statistics, and it is absolutely and completely wrong . This mistaken approach is to refer to the p value as “the probability that the null hypothesis is true”. It’s an intuitively appealing way to think, but it’s wrong in two key respects: (1) null hypothesis testing is a frequentist tool, and the frequentist approach to probability does not allow you to assign probabilities to the null hypothesis… according to this view of probability, the null hypothesis is either true or it is not; it cannot have a “5% chance” of being true. (2) even within the Bayesian approach, which does let you assign probabilities to hypotheses, the p value would not correspond to the probability that the null is true; this interpretation is entirely inconsistent with the mathematics of how the p value is calculated. Put bluntly, despite the intuitive appeal of thinking this way, there is no justification for interpreting a p value this way. Never do it.


2.1: Barriers to Critical Thinking
Preconceived ideas are ideas that decisively influence our thinking, but which we have not critically reflected upon. Not all preconceived ideas are fallacious (based on false assumptions). However, we must constantly consider our preconceived ideas critically to test their validity. In critical reflection we ask about the grounds for holding a belief. Do these grounds withstand rational scrutiny? Is there sufficient evidence to support the belief? Is the belief appropriate in its context? What weight should we give to counterarguments and counterexamples? And so on.
There are two main obstacles to clear thinking: preconceived ideas and fallacies. Preconceived ideas could be broadened to include social conditioning, labeling, and stereotypes. The section below will give you the opportunity to learn to recognize these obstacles to clear thinking.
 They have an accident in which the father is killed and his son badly injured. The boy is taken to the local hospital and requires urgent surgery. He is prepared for the operation and wheeled into the operating room. The surgeon arrives, looks at the boy and says, “I cannot operate on my own son”.
How is this possible?
- The man is the boy’s ________________
- The boy is the surgeon’s ________________
- The surgeon is the boy’s ________________
Ask a few friends, colleagues or family. See how they fare with the same riddle.
Answer: To many people the story presents a riddle. But the answer is obvious. The surgeon is the boy’s mother. The story is a riddle only if we have a preconceived idea which associates being a surgeon with being male. There is no good reason to hold this idea.
Now we will explore how preconceived ideas such as social conditioning, labeling and stereotyping affect our capacity for critical reasoning. The topic of preconceived ideas is an interesting field and it is worth exploring the various forms they take ranging from racial and gender stereotypes to the inability to see ourselves clearly.


2.5: Responsibilities to Stakeholders
-
- Last updated
- Save as PDF
4. How do businesses meet their social responsibilities to various stakeholders?
What makes a company be admired or perceived as socially responsible? Such a company meets its obligations to its stakeholders. Stakeholders are the individuals or groups to whom a business has a responsibility. The stakeholders of a business are its employees, its customers, the general public, and its investors.
Responsibility to Employees
An organization’s first responsibility is to provide a job to employees. Keeping people employed and letting them have time to enjoy the fruits of their labor is the finest thing business can do for society. Beyond this fundamental responsibility, employers must provide a clean, safe working environment that is free from all forms of discrimination. Companies should also strive to provide job security whenever possible.
Enlightened firms are also empowering employees to make decisions on their own and suggest solutions to company problems. Empowerment contributes to an employee’s self-worth, which, in turn, increases productivity and reduces absenteeism.
Each year, in collaboration with Great Place to Work ® , Fortune conducts an extensive employee survey of the best places to work in the United States. For 2017, the top companies included Google, Wegmans Food Markets, Edward Jones, Genentech, Salesforce, Acuity, and Quicken Loans. Some companies offer unusual benefits to their employees. For example, biotech company Genentech offers employee compensation for taking alternative methods of transportation to work at its South San Francisco campus. Employees can earn $12 per day for walking or biking to work, and those who drive a carpool or vanpool can earn $8 and $16, respectively. In addition, the company offers free commuter bus service for all employees via 27 routes around the Bay Area. 9
Responsibility to Customers
To be successful in today’s business environment, a company must satisfy its customers. A firm must deliver what it promises, as well as be honest and forthright in everyday interactions with customers, suppliers, and others. Recent research suggests that many consumers, particularly millennials, prefer to do business with companies and brands that communicate socially responsible messages, utilize sustainable manufacturing processes, and practice ethical business standards. 10
Responsibility to Society
A business must also be responsible to society. A business provides a community with jobs, goods, and services. It also pays taxes that go to support schools, hospitals, and better roads. Some companies have taken an additional step to demonstrate their commitment to stakeholders and society as a whole by becoming Certified Benefit Corporations, or B Corps for short. Verified by B Lab, a global nonprofit organization, B Corps meet the highest standards of social and environmental performance, public transparency, and legal accountability and strive to use the power of business to solve social and environmental problems via an impact assessment that rates each company on a possible score of 200 points. To become certified as a Benefit Corporation, companies need to reach a score of at least 80 and must be recertified every two years. There are more than 2,000 companies worldwide that have been certified as B Corps, including Method, W.S. Badger Company, Fishpeople Seafood, LEAP Organics, New Belgium Brewing Company, Ben & Jerry’s, Cabot Creamery Co-op, Comet Skateboards, Etsy, Patagonia, Plum Organics, and Warby Parker. 11
Environmental Protection
Business is also responsible for protecting and improving the world’s fragile environment. The world’s forests are being destroyed fast. Every second, an area the size of a football field is laid bare. Plant and animal species are becoming extinct at the rate of 17 per hour. A continent-size hole is opening up in the earth’s protective ozone shield. Each year we throw out 80 percent more refuse than we did in 1960; as a result, more than half of the nation’s landfills are filled to capacity.
To slow the erosion of the world’s natural resources, many companies have become more environmentally responsible. For example, Toyota now uses renewable energy sources such as solar, wind, geothermal, and water power for electricity to run its facilities. When its new $1 billion North American headquarters opened in Plano, Texas, in May 2017, Toyota said the 2.1 million square-foot campus would eventually be powered by 100% clean energy, helping the auto giant move closer to its goal of eliminating carbon emissions in all of its operations. 12
ETHICS IN PRACTICE
This Fish Story Has a Tasting Ending
Duncan Berry has always been an environmentalist at heart. Brought up on the Oregon coast, he was a sea captain at an early age, spending nearly two decades on the ocean before going on to become a successful entrepreneur in the organic cotton industry. After selling the textile business at the age of 50, he retired back to the Oregon coast to work on a state initiative to preserve marine habitats.
He quickly discovered that the state’s commercial fishing industry had gone into major disrepair since his seafaring adventure years earlier. Berry learned the majority of seafood consumed in the United States was being imported from other countries and more than 90 percent of U.S. seafood was being exported. In addition, great harm was being done to the ocean because it was being overfished.
Although several groups were already working to improve the commercial fishing industry, he observed that one key group was not part of the discussion: consumers. Berry decided a key component of change had to be involving consumers in the process. He spent more than a year meeting with everyone involved in the Oregon fishing industry—from fishermen to processors, distributors, truck drivers, chefs, and consumers—to gain perspective on why the industry was failing. His “aha” moment occurred when he realized the majority of fish is consumed in restaurants because consumers think preparing fish at home is too difficult and time-consuming. That’s when he co-founded Fishpeople Seafood.
Started in 2012, Fishpeople has a mission of changing the way people think about seafood by being transparent about where the seafood comes from, how it is processed, and how it is handled. Berry believes the company’s transparency helps consumers understand how the process translates into sustainable food that tastes good and is good for you. The company makes shelf-stable, ready-to-eat restaurant-quality seafood in the form of soups, meal kits, and fresh and frozen filets, complete with farm-to-table ingredients. On every package there is a code consumers can enter at the company’s website that will tell them everything about the seafood’s origin, down to the fisherman who caught it. Fishpeople also operates a processing plant in Toledo, Oregon, where workers are paid a livable wage and receive health insurance—benefits typically unheard of in the fishing industry.
Fishpeople’s products are available in more than 5,000 stores nationwide, including Walmart, Whole Foods, Costco, Kroger, and other grocery stores and markets. Recently the company announced a merger with Ilwaco Landing Fishermen, which will help further the two groups’ shared vision of supporting local fishermen and providing sustainable seafood to consumers.
Questions for Discussion
- How does Fishpeople’s transparency contribute to the company’s success?
- What responsibility, if any, does Fishpeople have to the local fishing industry?
Sources: Company website, https://fishpeopleseafood.com , accessed June 27, 2017; J. David Santen, Jr., “Adding Value to Oregon Seafood,” Built Oregon, http://builtoregon.com , accessed June 27, 2017; Elizabeth Crawford, “Fishpeople Wants to Fix the ‘Fundamentally Broken’ Seafood Industry with Increased Transparency,” Food Navigator, http://www.foodnavigator-usa.com , May 25, 2017; Fishpeople Seafood Announces Merger with Ilwaco Landing Fishermen,” Tillamook County Pioneer , https://www.tillamookcountypioneer.net , May 22, 2017; Leigh Buchanan, “Why This Entrepreneur Ditched Fashion for the ‘Hunting and Gathering’ Business,” Inc., https://www.inc.com , April 2017 issue; Kate Harrison, “This Former Green Textile Maven Is Making Microwaved Seafood Sustainable,” Forbes, http://www.forbes.com , August 25, 2015.
Corporate Philanthropy
Companies also display their social responsibility through corporate philanthropy. Corporate philanthropy includes cash contributions, donations of equipment and products, and support for the volunteer efforts of company employees. Recent statistics suggest U.S. corporate philanthropy exceeds more than $19 billion annually. 13 American Express is a major supporter of the American Red Cross. The organization relies almost entirely on charitable gifts to carry out its programs and services, which include disaster relief, armed-forces emergency relief, blood and tissue services, and health and safety services. The funds provided by American Express have enabled the Red Cross to deliver humanitarian relief to victims of numerous disasters around the world. 14 When Hurricane Katrina hit the Gulf Coast, Bayer sent 45,000 diabetes blood glucose monitors to the relief effort. Within weeks of the disaster, Abbott, Alcoa, Dell, Disney, Intel, UPS, Walgreens, Walmart, and others contributed more than $550 million for disaster relief. 15
Responsibilities to Investors
Companies’ relationships with investors also entail social responsibility. Although a company’s economic responsibility to make a profit might seem to be its main obligation to its shareholders, some investors increasingly are putting more emphasis on other aspects of social responsibility.
Some investors are limiting their investments to securities (e.g., stocks and bonds) that coincide with their beliefs about ethical and social responsibility. This is called social investing . For example, a social investment fund might eliminate from consideration the securities of all companies that make tobacco products or liquor, manufacture weapons, or have a history of being environmentally irresponsible. Not all social investment strategies are alike. Some ethical mutual funds will not invest in government securities because they help to fund the military; others freely buy government securities, with managers noting that federal funds also support the arts and pay for AIDS research. Today, assets invested using socially responsible strategies total more than $7 trillion. 16
Perhaps partly as the result of the global recession of 2007–2009, over the last several years companies have tried to meet responsibilities to their investors as well as to their other stakeholders. Recent research suggests that now more than ever, CEOs are being held to higher standards by boards of directors, investors, governments, media, and even employees when it comes to corporate accountability and ethical behavior. A recent global study by PwC reveals that over the last several years, there has been a large increase in the number of CEOs being forced out due to some sort of ethical lapse in their organizations. Strategies to prevent such missteps should include establishing a culture of integrity to prevent anyone from breaking the rules, making sure company goals and metrics do not create undue pressure on employees to cut corners, and implementing effective processes and controls to minimize the opportunity for unethical behavior. 17
CONCEPT CHECK
- How do businesses carry out their social responsibilities to consumers?
- What is corporate philanthropy?
- Is a company’s only responsibility to its investors to make a profit? Why or why not?


3.2: Young's Double-Slit Interference
-
- Last updated
- Save as PDF
Learning Objectives
By the end of this section, you will be able to:
- Explain the phenomenon of interference
- Define constructive and destructive interference for a double slit
The Dutch physicist Christiaan Huygens (1629–1695) thought that light was a wave, but Isaac Newton did not. Newton thought that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous reputation, his view generally prevailed; the fact that Huygens’s principle worked was not considered direct evidence proving that light is a wave. The acceptance of the wave character of light came many years later in 1801, when the English physicist and physician Thomas Young (1773–1829) demonstrated optical interference with his now-classic double-slit experiment.
If there were not one but two sources of waves, the waves could be made to interfere, as in the case of waves on water (Figure \(\PageIndex{1}\)). If light is an electromagnetic wave, it must therefore exhibit interference effects under appropriate circumstances. In Young’s experiment, sunlight was passed through a pinhole on a board. The emerging beam fell on two pinholes on a second board. The light emanating from the two pinholes then fell on a screen where a pattern of bright and dark spots was observed. This pattern, called fringes, can only be explained through interference, a wave phenomenon.
We can analyze double-slit interference with the help of Figure \(\PageIndex{2}\), which depicts an apparatus analogous to Young’s. Light from a monochromatic source falls on a slit \(S_0\). The light emanating from \(S_0\) is incident on two other slits \(S_1\) and \(S_2\) that are equidistant from \(S_0\). A pattern of interference fringes on the screen is then produced by the light emanating from \(S_1\) and \(S_2\). All slits are assumed to be so narrow that they can be considered secondary point sources for Huygens’ wavelets ( The Nature of Light ). Slits \(S_1\) and \(S_2\) are a distance d apart (\(d≤1\,mm\)), and the distance between the screen and the slits is D(≈1m), which is much greater than d.
Since \(S_0\) is assumed to be a point source of monochromatic light, the secondary Huygens wavelets leaving \(S_1\) and \(S_2\) always maintain a constant phase difference (zero in this case because \(S_1\) and \(S_2\) are equidistant from \(S_0\)) and have the same frequency. The sources \(S_1\) and \(S_2\) are then said to be coherent. By coherent waves, we mean the waves are in phase or have a definite phase relationship. The term incoherent means the waves have random phase relationships, which would be the case if \(S_1\) and \(S_2\) were illuminated by two independent light sources, rather than a single source \(S_0\). Two independent light sources (which may be two separate areas within the same lamp or the Sun) would generally not emit their light in unison, that is, not coherently. Also, because \(S_1\) and \(S_2\) are the same distance from \(S_0\), the amplitudes of the two Huygens wavelets are equal.
Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. In the following discussion, we illustrate the double-slit experiment with monochromatic light (single λ) to clarify the effect. Figure \(\PageIndex{3}\) shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.
When light passes through narrow slits, the slits act as sources of coherent waves and light spreads out as semicircular waves, as shown in Figure \(\PageIndex{1a}\). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure \(\PageIndex{1}\). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.
To understand the double-slit interference pattern, consider how two waves travel from the slits to the screen (Figure \(\PageIndex{5}\)). Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the path length difference \(\Delta l\) between the two waves is any half-integral number of wavelengths [(1 / 2)λ, (3 / 2)λ, (5 / 2)λ, etc.], then destructive interference occurs. Similarly, if the path length difference is any integral number of wavelengths (λ, 2λ, 3λ, etc.), then constructive interference occurs. These conditions can be expressed as equations:
\[\underbrace{\Delta l = m \lambda}_{\text{constructive interference}} \nonumber \]
for \(m = 0, \, ±1, \, ±2, \, ±3…\)
\[\underbrace{\Delta l = \left(m + \frac{1}{2}\right)\lambda }_{\text{destructuve interference}} \nonumber \]
for \(m = 0, \, ±1, \, ±2, \, ±3…\)


7.4: Exercises
1. How many distinct partitions of the set \(S=\{a,b,c,d\}\) are there? You do not need to list them. (Yes, you can find this answer online. But I recommend doing the work yourself for practice working with partitions!)
2.
- Let \(n\in \mathbb{Z}^+\text{.}\) Prove that \(\equiv_n\) is an equivalence relation on \(\mathbb{Z}\text{.}\)
- The cells of the induced partition of \(\mathbb{Z}\) are called the residue classes (or congruence classes ) of \(\mathbb{Z}\) modulo \(n\). Using set notation of the form \(\{\ldots,\#, \#,\#,\ldots\}\) for each class, write down the residue classes of \(\mathbb{Z}\) modulo \(4\text{.}\)
3. Let \(G\) be a group with subgroup \(H\text{.}\) Prove that \(\sim_R\) is an equivalence relation on \(G\text{.}\)
4. Find the indices of:
- \(H=\langle (15)(24)\rangle\) in \(S_5\)
- \(K=\langle (2354)(34)\rangle\) in \(S_6\)
- \(A_n\) in \(S_n\)
5. For each subgroup \(H\) of group \(G\text{,}\) (i) find the left and the right cosets of \(H\) in \(G\text{,}\) (ii) decide whether or not \(H\) is normal in \(G\text{,}\) and (iii) find \((G:H)\text{.}\)
Write all permutations using disjoint cycle notation , and write all dihedral group elements using standard form .
- \(H=6\mathbb{Z}\) in \(G=2\mathbb{Z}\)
- \(H=\langle 4\rangle\) in \(\mathbb{Z}_{20}\)
- \(H=\langle (23)\rangle\) in \(G=S_3\)
- \(H=\langle r\rangle\) in \(G=D_4\)
- \(H=\langle f\rangle\) in \(G=D_4\)
6. For each of the following, give an example of a group \(G\) with a subgroup \(H\) that matches the given conditions. If no such example exists, prove that.
- A group \(G\) with subgroup \(H\) such that \(|G/H|=1\text{.}\)
- A finite group \(G\) with subgroup \(H\) such that \(|G/H|=|G|\text{.}\)
- An abelian group \(G\) of order \(8\) containing a non-normal subgroup \(H\) of order 2.
- A group \(G\) of order 8 containing a normal subgroup of order \(2\text{.}\)
- A nonabelian group \(G\) of order 8 containing a normal subgroup of index \(2\text{.}\)
- A group \(G\) of order 8 containing a subgroup of order \(3\text{.}\)
- An infinite group \(G\) containing a subgroup \(H\) of finite index.
- An infinite group \(G\) containing a finite nontrivial subgroup \(H\text{.}\)
7. True/False. For each of the following, write T if the statement is true; otherwise, write F. You do NOT need to provide explanations or show work for this problem. Throughout, let \(G\) be a group with subgroup \(H\) and elements \(a,b\in G\text{.}\)
- If \(a\in bH\) then \(aH\) must equal \(bH\text{.}\)
- \(aH\) must equal \(Ha\text{.}\)
- If \(aH=bH\) then \(Ha\) must equal \(Hb\text{.}\)
- If \(a\in H\) then \(aH\) must equal \(Ha\text{.}\)
- \(H\) must be normal in \(G\) if there exists \(a\in G\) such that \(aH=Ha\text{.}\)
- If \(aH=bH\) then \(ah=bh\) for every \(h\in H\text{.}\)
- If \(G\) is finite, then \(|G/H|\) must be less than \(|G|\text{.}\)
- If \(G\) is finite, then \((G:H)\) must be less than or equal to \(|G|\text{.}\)
8. Let \(G\) be a group of order \(pq\text{,}\) where \(p\) and \(q\) are prime, and let \(H\) be a proper subgroup of \(G\text{.}\) Prove that \(H\) is cyclic.
9. Prove Corollary \(7.3.2\): that is, let \(G\) be a group of prime order, and prove that \(G\) is cyclic.


12.3: The Parabola
- Graph parabolas with vertices at the origin.
- Write equations of parabolas in standard form.
- Graph parabolas with vertices not at the origin.
- Solve applied problems involving parabolas.
Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (Figure \(\PageIndex{1}\)), which focuses light rays from the sun to ignite the flame.
Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs.
Graphing Parabolas with Vertices at the Origin
Previously , we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola (Figure \(\PageIndex{2}\)).
Like the ellipse and hyperbola , the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points \((x,y)\) in a plane that are the same distance from a fixed line, called the directrix , and a fixed point (the focus ) not on the directrix.
Previously, we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola (Figure \(\PageIndex{3}\)). Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum . The endpoints of the latus rectum lie on the curve. By definition, the distanced d from the focus to any point \(P\) on the parabola is equal to the distance from \(P\) to the directrix.
To work with parabolas in the coordinate plane , we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former.
Let \((x,y)\) be a point on the parabola with vertex \((0,0)\), focus \((0,p)\),and directrix \(y=−p\) as shown in Figure \(\PageIndex{4}\). The distanced d from point \((x,y)\) to point \((x,−p)\) on the directrix is the difference of the y -values: \(d=y+p\). The distance from the focus \((0,p)\) to the point \((x,y)\) is also equal to \(d\) and can be expressed using the distance formula .
\[ \begin{align*} d &=\sqrt{{(x−0)}^2+{(y−p)}^2} \\[4pt] &=\sqrt{x^2+{(y−p)}^2} \end{align*} \]
Set the two expressions for \(d\) equal to each other and solve for \(y\) to derive the equation of the parabola. We do this because the distance from \((x,y)\) to \((0,p)\) equals the distance from \((x,y)\) to \((x,−p)\).
\[\sqrt{x^2+{(y−p)}^2}=y+p \]
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
\[ \begin{align*} x^2+{(y−p)}^2 &={(y+p)}^2 \\[4pt] x^2+y^2−2py+p^2 &=y^2+2py+p^2 \\[4pt] x^2−2py &=2py \\[4pt] x^2 &=4py \end{align*} \]
The equations of parabolas with vertex \((0,0)\) are \(y^2=4px\) when the x -axis is the axis of symmetry and \(x^2=4py\) when the y -axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
Table \(\PageIndex{1}\) and Figure \(\PageIndex{5}\) summarize the standard features of parabolas with a vertex at the origin.
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
|---|---|---|---|---|
| x -axis | \(y^2=4px\) | \((p, 0)\) | \(x=−p\) | \((p, \pm 2p)\) |
| y -axis | \(x^2=4py\) | \((0, p)\) | \(y=−p\) | \((\pm 2p, p)\) |
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum (Figure \(\PageIndex{5}\)). When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum , these lines intersect on the axis of symmetry, as shown in Figure \(\PageIndex{6}\).
- Determine which of the standard forms applies to the given equation: \(y^2=4px\) or \(x^2=4py\).
-
Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
-
If the equation is in the form \(y^2=4px\), then
- the axis of symmetry is the \(x\)-axis, \(y=0\)
- set \(4p\) equal to the coefficient of \(x\) in the given equation to solve for \(p\). If \(p>0\), the parabola opens right. If \(p<0\), the parabola opens left.
- use \(p\) to find the coordinates of the focus, \((p,0)\)
- use \(p\) to find the equation of the directrix, \(x=−p\)
- use \(p\) to find the endpoints of the latus rectum, \((p,\pm 2p)\). Alternately, substitute \(x=p\) into the original equation.
-
If the equation is in the form \(x^2=4py\),then
- the axis of symmetry is the \(y\)-axis, \(x=0\)
- set \(4p\) equal to the coefficient of \(y\) in the given equation to solve for \(p\). If \(p>0\), the parabola opens up. If \(p<0\), the parabola opens down.
- use \(p\) to find the coordinates of the focus, \((0,p)\)
- use \(p\) to find equation of the directrix, \(y=−p\)
- use \(p\) to find the endpoints of the latus rectum, \((\pm 2p,p)\)
-
If the equation is in the form \(y^2=4px\), then
- Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Graph \(y^2=24x\). Identify and label the focus, directrix, and endpoints of the latus rectum .
Solution
The standard form that applies to the given equation is \(y^2=4px\). Thus, the axis of symmetry is the x -axis. It follows that:
- \(24=4p\), so \(p=6\). Since \(p>0\), the parabola opens right
- the coordinates of the focus are \((p,0)=(6,0)\)
- the equation of the directrix is \(x=−p=−6\)
- the endpoints of the latus rectum have the same x -coordinate at the focus. To find the endpoints, substitute \(x=6\) into the original equation: \((6,\pm 12)\)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure \(\PageIndex{7}\)).
Graph \(y^2=−16x\). Identify and label the focus, directrix, and endpoints of the latus rectum .
- Answer
-
- Focus: \((−4,0)\)
- Directrix: \(x=4\)
- Endpoints of the latus rectum: \((−4,\pm 8)\)
Graph \(x^2=−6y\). Identify and label the focus, directrix, and endpoints of the latus rectum .
Solution
The standard form that applies to the given equation is \(x^2=4py\). Thus, the axis of symmetry is the \(y\)-axis. It follows that:
- \(−6=4p\),so \(p=−\dfrac{3}{2}\). Since \(p<0\), the parabola opens down.
- the coordinates of the focus are \((0,p)=(0,−\dfrac{3}{2})\)
- the equation of the directrix is \(y=−p=\dfrac{3}{2}\)
- the endpoints of the latus rectum can be found by substituting \(y=\dfrac{3}{2}\) into the original equation, \((\pm 3,−\dfrac{3}{2})\)
Next we plot the focus, directrix, and latus rectum , and draw a smooth curve to form the parabola.
Graph \(x^2=8y\). Identify and label the focus, directrix, and endpoints of the latus rectum .
- Answer
-
- Focus: \((0,2)\)
- Directrix: \(y=−2\)
- Endpoints of the latus rectum: \((\pm 4,2)\).
Writing Equations of Parabolas in Standard Form
In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.
-
Determine whether the axis of symmetry is the \(x\)- or \(y\)-axis.
- If the given coordinates of the focus have the form \((p,0)\), then the axis of symmetry is the \(x\)-axis. Use the standard form \(y^2=4px\).
- If the given coordinates of the focus have the form \((0,p)\), then the axis of symmetry is the \(y\)-axis. Use the standard form \(x^2=4py\).
- Multiply \(4p\).
- Substitute the value from Step 2 into the equation determined in Step 1.
What is the equation for the parabola with focus \((−\dfrac{1}{2},0)\) and directrix \(x=\dfrac{1}{2}\)?
Solution
The focus has the form \((p,0)\), so the equation will have the form \(y^2=4px\).
- Multiplying \(4p\), we have \(4p=4(−\dfrac{1}{2})=−2\).
- Substituting for \(4p\), we have \(y^2=4px=−2x\).=
Therefore, the equation for the parabola is \(y^2=−2x\).
What is the equation for the parabola with focus \(\left(0,\dfrac{7}{2}\right)\) and directrix \(y=−\dfrac{7}{2}\)?
- Answer
-
\(x^2=14y\).
Graphing Parabolas with Vertices Not at the Origin
Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated \(h\) units horizontally and \(k\) units vertically, the vertex will be \((h,k)\). This translation results in the standard form of the equation we saw previously with \(x\) replaced by \((x−h)\) and \(y\) replaced by \((y−k)\).
To graph parabolas with a vertex \((h,k)\) other than the origin, we use the standard form \({(y−k)}^2=4p(x−h)\) for parabolas that have an axis of symmetry parallel to the \(x\)-axis, and \({(x−h)}^2=4p(y−k)\) for parabolas that have an axis of symmetry parallel to the \(y\)-axis. These standard forms are given below, along with their general graphs and key features.
Table \(\PageIndex{2}\) and Figure \(\PageIndex{11}\) summarize the standard features of parabolas with a vertex at a point \((h,k)\).
| Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
|---|---|---|---|---|
| \(y=k\) | \({(y−k)}^2=4p(x−h)\) | \((h+p, k)\) | \(x=h−p\) | \((h+p, k\pm 2p)\) |
| \(x=h\) | \({(x−h)}^2=4p(y−k)\) | \((h, k+p)\) | \(y=k−p\) | \((h\pm 2p, k+p)\) |
- Determine which of the standard forms applies to the given equation: \({(y−k)}^2=4p(x−h)\) or \({(x−h)}^2=4p(y−k)\).
-
Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
-
If the equation is in the form \({(y−k)}^2=4p(x−h)\),then:
- use the given equation to identify \(h\) and \(k\) for the vertex, \((h,k)\)
- use the value of \(k\) to determine the axis of symmetry, \(y=k\)
- set \(4p\) equal to the coefficient of \((x−h)\) in the given equation to solve for \(p\). If \(p>0\),the parabola opens right. If \(p<0\), the parabola opens left.
- use \(h\), \(k\), and \(p\) to find the coordinates of the focus, \((h+p, k)\)
- use \(h\) andp p to find the equation of the directrix, \(x=h−p\)
- use \(h\), \(k\), and \(p\) to find the endpoints of the latus rectum, \((h+p,k\pm 2p)\)
-
If the equation is in the form \({(x−h)}^2=4p(y−k)\),then:
- use the given equation to identify \(h\) and \(k\) for the vertex, \((h,k)\)
- use the value of \(h\) to determine the axis of symmetry, \(x=h\)
- set \(4p\) equal to the coefficient of \((y−k)\) in the given equation to solve for \(p\). If \(p>0\), the parabola opens up. If \(p<0\), the parabola opens down.
- use \(h\), \(k\), and \(p\) to find the coordinates of the focus, \((h, k+p)\)
- use \(k\) and \(p\) to find the equation of the directrix, \(y=k−p\)
- use \(h\), \(k\), and \(p\) to find the endpoints of the latus rectum, \((h\pm 2p, k+p)\)
-
If the equation is in the form \({(y−k)}^2=4p(x−h)\),then:
- Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Graph \({(y−1)}^2=−16(x+3)\). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.
Solution
The standard form that applies to the given equation is \({(y−k)}^2=4p(x−h)\). Thus, the axis of symmetry is parallel to the \(x\)-axis. It follows that:
- the vertex is \((h,k)=(−3,1)\)
- the axis of symmetry is \(y=k=1\)
- \(−16=4p\),so \(p=−4\). Since \(p<0\), the parabola opens left.
- the coordinates of the focus are \((h+p,k)=(−3+(−4),1)=(−7,1)\)
- the equation of the directrix is \(x=h−p=−3−(−4)=1\)
- the endpoints of the latus rectum are \((h+p,k\pm 2p)=(−3+(−4),1\pm 2(−4))\), or \((−7,−7)\) and \((−7,9)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure \(\PageIndex{10}\)).
Graph \({(y+1)}^2=4(x−8)\). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum .
- Answer
-
- Vertex: \((8,−1)\)
- Axis of symmetry: \(y=−1\)
- Focus: \((9,−1)\)
- Directrix: \(x=7\)
- Endpoints of the latus rectum : \((9,−3)\) and \((9,1)\).
Graph \(x^2−8x−28y−208=0\). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum .
Solution
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \({(x−h)}^2=4p(y−k)\). Thus, the axis of symmetry is parallel to the \(y\)-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable \(x\) in order to complete the square.
\[ \begin{align*} x^2−8x−28y−208&=0 \\[4pt] x^2−8x &=28y+208 \\[4pt] x^2−8x+16 &=28y+208+16 \\[4pt] (x−4)^2 &=28y+224 \\[4pt] (x−4)^2 &=28(y+8) \\[4pt] (x−4)^2&= 4⋅7⋅(y+8) \end{align*}\]
It follows that:
- the vertex is \((h,k)=(4,−8)\)
- the axis of symmetry is \(x=h=4\)
- since \(p=7\), \(p>0\) and so the parabola opens up
- the coordinates of the focus are \((h,k+p)=(4,−8+7)=(4,−1)\)
- the equation of the directrix is \(y=k−p=−8−7=−15\)
- the endpoints of the latus rectum are \((h\pm 2p,k+p)=(4\pm 2(7),−8+7)\), or \((−10,−1)\) and \((18,−1)\)
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola (Figure \(\PageIndex{14}\)).
Graph \({(x+2)}^2=−20(y−3)\). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum .
- Answer
-
- Vertex: \((−2,3)\)
- Axis of symmetry: \(x=−2\)
- Focus: \((−2,−2)\)
- Directrix: \(y=8\)
- Endpoints of the latus rectum : \((−12,−2)\) and \((8,−2)\).
Solving Applied Problems Involving Parabolas
As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors , such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus (Figure \(\PageIndex{16}\)). This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror.
Parabolic mirrors have the ability to focus the sun’s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds.
A cross-section of a design for a travel-sized solar fire starter is shown in Figure \(\PageIndex{17}\). The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds.
- Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane.
- Use the equation found in part (a) to find the depth of the fire starter.
Solution
- The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form \(x^2=4py\),where \(p>0\). The igniter, which is the focus, is \(1.7\) inches above the vertex of the dish. Thus we have \(p=1.7\).
\[\begin{align*} x^2&=4py\qquad \text{Standard form of upward-facing parabola with vertex } (0,0)\\ x^2&=4(1.7)y\qquad \text{Substitute } 1.7 \text{ for } p\\ x^2&=6.8y\qquad \text{Multiply.} \end{align*}\]
- The dish extends \(\dfrac{4.5}{2}=2.25\) inches on either side of the origin. We can substitute \(2.25\) for \(x\) in the equation from part (a) to find the depth of the dish.
\[\begin{align*} x^2&=6.8y\qquad \text{ Equation found in part } (a)\\ {(2.25)}^2&=6.8y\qquad \text{Substitute } 2.25 \text{ for } x\\ y&\approx 0.74\qquad \text{Solve for } y \end{align*}\]
The dish is about \(0.74\) inches deep.
Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of \(1600\) mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed \(320\) mm from the base.
- Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x -axis as its axis of symmetry).
- Use the equation found in part (a) to find the depth of the cooker.
- Answer a
-
\(y^2=1280x\)
- Answer b
-
The depth of the cooker is \(500\) mm
Key Equations
| Parabola, vertex at origin, axis of symmetry on x -axis | \(y^2=4px\) |
| Parabola, vertex at origin, axis of symmetry on y -axis | \(x^2=4py\) |
| Parabola, vertex at \((h,k)\),axis of symmetry on x -axis | \({(y−k)}^2=4p(x−h)\) |
| Parabola, vertex at \((h,k)\),axis of symmetry on y -axis | \({(x−h)}^2=4p(y−k)\) |
Key Concepts
- A parabola is the set of all points \((x,y)\) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
- The standard form of a parabola with vertex \((0,0)\) and the x -axis as its axis of symmetry can be used to graph the parabola. If \(p>0\), the parabola opens right. If \(p<0\), the parabola opens left. See Example \(\PageIndex{1}\).
- The standard form of a parabola with vertex \((0,0)\) and the y -axis as its axis of symmetry can be used to graph the parabola. If \(p>0\), the parabola opens up. If \(p<0\), the parabola opens down. See Example \(\PageIndex{2}\).
- When given the focus and directrix of a parabola, we can write its equation in standard form. See Example \(\PageIndex{3}\).
- The standard form of a parabola with vertex \((h,k)\) and axis of symmetry parallel to the \(x\)-axis can be used to graph the parabola. If \(p>0\), the parabola opens right. If \(p<0\), the parabola opens left. See Example \(\PageIndex{4}\).
- The standard form of a parabola with vertex \((h,k)\) and axis of symmetry parallel to the \(y\)-axis can be used to graph the parabola. If \(p>0\), the parabola opens up. If \(p<0\), the parabola opens down. See Example \(\PageIndex{5}\).
- Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See Example \(\PageIndex{6}\).


3.4: Find the Equation of a Line
By the end of this section, you will be able to:
- Find an equation of the line given the slope and y-intercept
- Find an equation of the line given the slope and a point
- Find an equation of the line given two points
- Find an equation of a line parallel to a given line
- Find an equation of a line perpendicular to a given line
How do online companies know that “you may also like” a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.
The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section, we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.
Find an Equation of the Line Given the Slope and y -Intercept
We can easily determine the slope and intercept of a line if the equation is written in slope-intercept form, \(y=mx+b\). Now we will do the reverse—we will start with the slope and y -intercept and use them to find the equation of the line.
Find the equation of a line with slope \(−9\) and y -intercept \((0,−4)\).
SolutionSince we are given the slope and y -intercept of the line, we can substitute the needed values into the slope-intercept form, \(y=mx+b\).
| Name the slope. | \( m = {\color{red}{-9}} \) |
| Name the y -intercept. | y=intercept is \( (0, {\color{Cerulean}{-4}} ) \) |
| Substitute the values into \(y=mx+b\). | \( y = {\color{red}{m}}x + \color{Cerulean}{b} \) |
| \( y = {\color{red}{-9}}x + \color{Cerulean}{(-4)} \) | |
| \( y=-9x-4\) |
Find the equation of a line with slope \(\dfrac{2}{5}\) and y -intercept \((0,4)\).
- Answer
-
\(y=\frac{2}{5}x+4\)
Find the equation of a line with slope \(−1\) and y -intercept \((0,−3)\).
- Answer
-
\(y=−x−3\)
Sometimes, the slope and intercept need to be determined from the graph.
Find the equation of the line shown in the graph.
Solution
We need to find the slope and y -intercept of the line from the graph so we can substitute the needed values into the slope-intercept form, \(y=mx+b\).
To find the slope, we choose two points on the graph.
The y -intercept is \((0,−4)\) and the graph passes through \((3,−2)\).
| Find the slope, by counting the rise and run. | \( m=\dfrac{\text{rise}}{\text{run}} \) |
| \( m = {\color{red}{\dfrac{2}{3}}} \) | |
| Find the y -intercept. | y=intercept is \( (0, {\color{Cerulean}{-4}} ) \) |
| Substitute the values into \(y=mx+b\). | \( y = {\color{red}{m}}x + \color{Cerulean}{b} \) |
| \( y = {\color{red}{\dfrac{2}{3}}}x \color{Cerulean}{-4} \) |
Find the equation of the line shown in the graph.
- Answer
-
\(y=\frac{3}{5}x+1\)
Find the equation of the line shown in the graph.
- Answer
-
\(y=\frac{4}{3}x−5\)
Find an Equation of the Line Given the Slope and a Point
Finding an equation of a line using the slope-intercept form of the equation works well when you are given the slope and y -intercept or when you read them off a graph. But what happens when you have another point instead of the y -intercept?
We are going to use the slope formula to derive another form of an equation of the line.
Suppose we have a line that has slope m and that contains some specific point \((x_1,y_1)\) and some other point, which we will just call \((x,y)\). We can write the slope of this line and then change it to a different form.
\( \begin{array} {llll} {} &{m} &= &{\dfrac{y-y_1}{x-x_1}} \\ {\text{Multiply both sides of the equation by }x−x_1.} &{m(x-x_1)} &= &{\left( \dfrac{y−y_1}{x−x_1} \right)(x−x_1)} \\ {\text{Simplify.}} &{m(x-x_1)} &= &{y-y_1} \\ {\text{Rewrite the equation with theyterms on the left.}} &{y-y_1} &= &{m(x-x_1)} \\ \end{array} \)
This format is called the point-slope form of an equation of a line.
The point-slope form of an equation of a line with slope m and containing the point \((x_1,y_1)\) is:
\[y−y_1=m(x−x_1) \nonumber\]
We can use the point-slope form of an equation to find an equation of a line when we know the slope and at least one point. Then, we will rewrite the equation in slope-intercept form. Most applications of linear equations use the the slope-intercept form.
How to Find an Equation of a Line Given a Point and the Slope
Find an equation of a line with slope \(m=−\frac{1}{3}\) that contains the point \((6,−4)\). Write the equation in slope-intercept form.
Solution| Step 1. Identify the slope. | The slope is given. | \( \color{Cerulean}{m=-\dfrac{1}{3}} \) |
| Step 2. Identify the point. | The point is given. | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{6} & \color{red}{-4} \end{pmatrix} \) |
| Step 3. Substitute the values into the point-slope form, \( y−y_1=m(x−x_1) \) | Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \)
\( y−{\color{red}{(-4)}}={\color{Cerulean}{-\dfrac{1}{3}}}(x−{\color{red}{6}}) \) \(y +4 = -\dfrac{1}{3}x+2 \) |
| Step 4. Write the equation in slope intercept form. | \(y = -\dfrac{1}{3}x-2 \) |
Find the equation of a line with slope \(m=−\frac{2}{5}\) and containing the point \((10,−5)\).
- Answer
-
\(y=−\frac{2}{5}x−1\)
Find the equation of a line with slope \(m=−\frac{3}{4}\), and containing the point \((4,−7)\).
- Answer
-
\(y=−\frac{3}{4}x−4\)
We list the steps for easy reference.
- Identify the slope.
- Identify the point.
- Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form.
Find an equation of a horizontal line that contains the point \((−2,−6)\). Write the equation in slope-intercept form.
SolutionEvery horizontal line has slope 0. We can substitute the slope and points into the point-slope form, \(y−y_1=m(x−x_1)\).
| Identify the slope. | \( \color{Cerulean}{m=0} \) |
| Identify the point. | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{-2} & \color{red}{-6} \end{pmatrix} \) |
| Substitute the values into \(y−y_1=m(x−x_1) \). Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \)
\( y−{\color{red}{(-6)}}={\color{Cerulean}{0}}(x−{\color{red}{-2}}) \) \(y + 6 = 0 \) |
| Write in slope-intercept form. | It is in y -form, but could be written \(y=−6\). |
Did we end up with the form of a horizontal line, \(y=a\)?
Find the equation of a horizontal line containing the point \((−3,8)\).
- Answer
-
\(y=8\)
Find the equation of a horizontal line containing the point \((−1,4)\).
- Answer
-
\(y=4\)
Find an Equation of the Line Given Two Points
When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.
So far, we have two options for finding an equation of a line: slope-intercept or point-slope. When we start with two points, it makes more sense to use the point-slope form.
But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.
How to Find the Equation of a Line Given Two Points
Find an equation of a line that contains the points \((−3,−1)\) and \((2,−2)\) Write the equation in slope-intercept form.
Solution| Step 1. Find the slope using the given points. | Find the slope of the line through \( (-3. -1) \) and \( (2, -2) \). |
\( m = \dfrac{y_2-y_1}{x_2-x_1} \\ \)
\( m = \dfrac{-2-(-1)}{2-(-3)} \\ \) \( m = \dfrac{-1}{5} \\ \) \( \color{Cerulean}{m = -\dfrac{1}{5}} \) |
| Step 2. Choose one point. | Choose either point. | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{2} & \color{red}{-2} \end{pmatrix} \) |
| Step 3. Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\). | Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \\ \)
\( y−{\color{red}{(-2)}}={\color{Cerulean}{-\dfrac{1}{5}}}(x−{\color{red}{2}}) \\ \) \(y +2 = -\dfrac{1}{5}x+\dfrac{2}{5} \) |
| Step 4. Write the equation in slope-intercept form. | \(y = -\dfrac{1}{5}x-\dfrac{8}{5} \) |
Find the equation of a line containing the points \((−2,−4)\) and \((1,−3)\).
- Answer
-
\(y=\frac{1}{3}x−\frac{10}{3}\)
Find the equation of a line containing the points \((−4,−3)\) and \((1,−5)\).
- Answer
-
\(y=−\frac{2}{5}x−\frac{23}{5}\)
The steps are summarized here.
- Find the slope using the given points. \(m=\frac{y_2−y_1}{x_2−x_1}\)
- Choose one point.
- Substitute the values into the point-slope form: \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form.
Find an equation of a line that contains the points \((−3,5)\) and \((−3,4)\). Write the equation in slope-intercept form.
SolutionAgain, the first step will be to find the slope.
Find the slope of the line through\((−3,5)\)and\((−3,4)\).
\[m=\frac{y_2−y_1}{x_2−x_1} \nonumber\]
\[m=\frac{4−5}{−3−(−3)} \nonumber\]
\[m=\frac{−1}{0} \nonumber\]
The slope is undefined.
This tells us it is a vertical line. Both of our points have an x -coordinate of \(−3\). So our equation of the line is \(x=−3\). Since there is no y , we cannot write it in slope-intercept form.
You may want to sketch a graph using the two given points. Does your graph agree with our conclusion that this is a vertical line?
Find the equation of a line containing the points \((5,1)\) and \((5,−4)\).
- Answer
-
\(x=5\)
Find the equation of a line containing the points \((−4,4)\) and \((−4,3)\).
- Answer
-
\(x=−4\)
We have seen that we can use either the slope-intercept form or the point-slope form to find an equation of a line. Which form we use will depend on the information we are given.
| To Write an Equation of a Line | ||
|---|---|---|
| If given: | Use: | Form: |
| Slope and y -intercept | slope-intercept | \(y=mx+b\) |
| Slope and a point | point-slope | \(y−y_1=m(x−x_1)\) |
| Two points | point-slope | \(y−y_1=m(x−x_1)\) |
Find an Equation of a Line Parallel to a Given Line
Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point-slope equation.
First, let’s look at this graphically.
This graph shows \(y=2x−3.\) We want to graph a line parallel to this line and passing through the point \((−2,1)\).
We know that parallel lines have the same slope. So the second line will have the same slope as \(y=2x−3\). That slope is \(m_∥=2\). We’ll use the notation \(m_∥\) to represent the slope of a line parallel to a line with slope m . (Notice that the subscript || looks like two parallel lines.)
The second line will pass through \((−2,1)\) and have \(m=2\).
To graph the line, we start at \((−2,1)\) and count out the rise and run.
With \(m=2\) (or \(m=\frac{2}{1}\)), we count out the rise 2 and the run 1. We draw the line, as shown in the graph.
Do the lines appear parallel? Does the second line pass through\((−2,1)\)?
We were asked to graph the line, now let’s see how to do this algebraically.
We can use either the slope-intercept form or the point-slope form to find an equation of a line. Here we know one point and can find the slope. So we will use the point-slope form.
How to Find the Equation of a Line Parallel to a Given Line and a Point
Find an equation of a line parallel to \(y=2x−3\) that contains the point \((−2,1)\). Write the equation in slope-intercept form.
Solution| Step 1. Find the slope of the given line. | The given line is in slope-intercept form, \( y = 2x - 3 \) | \( m = 2 \) |
| Step 2. Find the slope of the parallel line. | Parallel lines have the same slope. | \(m_∥=\color{Cerulean}{2}\) |
| Step 3. Identify the point. | The given point is \( (-2, 1) \). | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{-2} & \color{red}{1} \end{pmatrix} \) |
| Step 4. Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\). | Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \\ \)
\( y−{\color{red}{(1)}}={\color{Cerulean}{2}}(x−{\color{red}{(-2)}}) \\ \) \(y - 1 = 2(x + 2) \\ \) \(y - 1 = 2x + 4 \) |
| Step 5. Write the equation in slope-intercept form. | \(y = 2x + 5 \) |
Look at graph with the parallel lines shown previously. Does this equation make sense? What is the y -intercept of the line? What is the slope?
Find an equation of a line parallel to the line \(y=3x+1\) that contains the point \((4,2)\). Write the equation in slope-intercept form.
- Answer
-
\(y=3x−10\)
Find an equation of a line parallel to the line \(y=12x−3\) that contains the point \((6,4)\).
Write the equation in slope-intercept form.
- Answer
-
\(y=\frac{1}{2}x+1\)
- Find the slope of the given line.
- Find the slope of the parallel line.
- Identify the point.
- Substitute the values into the point-slope form: \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form.
Find an Equation of a Line Perpendicular to a Given Line
Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point-slope equation, like we did with parallel lines.
This graph shows \(y=2x−3\). Now, we want to graph a line perpendicular to this line and passing through \((−2,1)\).
We know that perpendicular lines have slopes that are negative reciprocals.
We’ll use the notation \(m_⊥\) to represent the slope of a line perpendicular to a line with slope m . (Notice that the subscript \(⊥\) looks like the right angles made by two perpendicular lines.)
\[y=2x−3 \quad \text{perpendicular line} \nonumber\]
\[m=2 \quad m_⊥=−\dfrac{1}{2} \nonumber\]
We now know the perpendicular line will pass through \((−2,1)\) with \(m_⊥=−\dfrac{1}{2}\).
To graph the line, we will start at \((−2,1)\) and count out the rise \(−1\) and the run \(2\). Then we draw the line.
Do the lines appear perpendicular? Does the second line pass through\((−2,1)\)?
We were asked to graph the line, now, let’s see how to do this algebraically.
We can use either the slope-intercept form or the point-slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point-slope form.
How to Find the Equation of a Line Perpendicular to a Given Line and a Point
Find an equation of a line perpendicular to \(y=2x−3\) that contains the point \((−2,1)\). Write the equation in slope-intercept form.
Solution| Step 1. Find the slope of the given line. | The given line is in slope-intercept form, \( y = 2x - 3 \) | \( m = 2 \) |
| Step 2. Find the slope of the perpendicular line. | The slopes of perpendicular lines are negative reciprocals. | \(m_⊥=\color{Cerulean}{−\dfrac{1}{2}}\) |
| Step 3. Identify the point. | The given point is \( (-2, 1) \). | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{-2} & \color{red}{1} \end{pmatrix} \) |
| Step 4. Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\). | Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \\ \)
\( y−{\color{red}{(1)}}={\color{Cerulean}{−\dfrac{1}{2}}}(x−{\color{red}{(-2)}}) \\ \) \(y - 1 = −\dfrac{1}{2}(x + 2) \\ \) \(y - 1 = −\dfrac{1}{2}x - 1 \) |
| Step 5. Write the equation in slope-intercept form. | \(y = −\dfrac{1}{2}x \) |
Find an equation of a line perpendicular to the line \(y=3x+1\) that contains the point \((4,2)\). Write the equation in slope-intercept form.
- Answer
-
\(y=−\frac{1}{3}x+\frac{10}{3}\)
Find an equation of a line perpendicular to the line \(y=12x−3\) that contains the point \((6,4)\). Write the equation in slope-intercept form.
- Answer
-
\(y=−2x+16\)
- Find the slope of the given line.
- Find the slope of the perpendicular line.
- Identify the point.
- Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form.
Find an equation of a line perpendicular to \(x=5\) that contains the point \((3,−2)\). Write the equation in slope-intercept form.
SolutionAgain, since we know one point, the point-slope option seems more promising than the slope-intercept option. We need the slope to use this form, and we know the new line will be perpendicular to \(x=5\). This line is vertical, so its perpendicular will be horizontal. This tells us the \(m_⊥=0\).
| Step 1. Identify the point. | The given point is \( (3,−2) \). | \( \begin{pmatrix} \color{red}{x_1}, & \color{red}{y_1} \\ \color{red}{3} & \color{red}{−2} \end{pmatrix} \) |
| Step 2. Identify the slope of the perpendicular line. | \(m_⊥=\color{Cerulean}{0}\) | |
| Step 3. Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\). | Simplify. |
\( y−{\color{red}{y_1}}={\color{Cerulean}{m}}(x−{\color{red}{x_1}}) \\ \)
\( y−{\color{red}{(-2)}}={\color{Cerulean}{0}}(x−{\color{red}{(3)}}) \) |
| Step 4. Simplify |
\(y + 2 = 0 \\ \)
\(y = −2 \) |
Sketch the graph of both lines. On your graph, do the lines appear to be perpendicular?
Find an equation of a line that is perpendicular to the line \(x=4\) that contains the point \((4,−5)\). Write the equation in slope-intercept form.
- Answer
-
\(y=−5\)
Find an equation of a line that is perpendicular to the line \(x=2\) that contains the point \((2,−1)\). Write the equation in slope-intercept form.
- Answer
-
\(y=−1\)
In the Example above, we used the point-slope form to find the equation. We could have looked at this in a different way.
We want to find a line that is perpendicular to \(x=5\) that contains the point \((3,−2)\). This graph shows us the line \(x=5\) and the point \((3,−2)\).
We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through \((3,−2)\).
Do the lines appear perpendicular?
If we look at a few points on this horizontal line, we notice they all have y -coordinates of \(−2\). So, the equation of the line perpendicular to the vertical line \(x=5\) is \(y=−2\).
Find an equation of a line that is perpendicular to \(y=−3\) that contains the point \((−3,5)\). Write the equation in slope-intercept form.
SolutionThe line \(y=−3\) is a horizontal line. Any line perpendicular to it must be vertical, in the form \(x=a\). Since the perpendicular line is vertical and passes through \((−3,5)\), every point on it has an x -coordinate of \(−3\). The equation of the perpendicular line is \(x=−3\).
You may want to sketch the lines. Do they appear perpendicular?
Find an equation of a line that is perpendicular to the line \(y=1\) that contains the point \((−5,1)\). Write the equation in slope-intercept form.
- Answer
-
\(x=−5\)
Find an equation of a line that is perpendicular to the line \(y=−5\) that contains the point \((−4,−5)\). Write the equation in slope-intercept form.
- Answer
-
\(x=−4\)
Access these online resources for additional instruction and practice with finding the equation of a line.
Key Concepts
-
How to find an equation of a line given the slope and a point.
- Identify the slope.
- Identify the point.
- Substitute the values into the point-slope form,\( y−y_1=m(x−x_1)\).
-
Write the equation in slope-intercept form.
-
How to find an equation of a line given two points.
- Find the slope using the given points. \(m=\frac{y_2−y_1}{x_2−x_1}\)
- Choose one point.
- Substitute the values into the point-slope form: \(y−y_1=m(x−x_1)\).
-
Write the equation in slope-intercept form.
To Write an Equation of a Line If given: Use: Form: Slope and y -intercept slope-intercept \(y=mx+b\) Slope and a point point-slope \(y−y_1=m(x−x_1)\) Two points point-slope \(y−y_1=m(x−x_1)\)
-
How to find an equation of a line parallel to a given line.
- Find the slope of the given line.
- Find the slope of the parallel line.
- Identify the point.
- Substitute the values into the point-slope form: \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form
-
How to find an equation of a line perpendicular to a given line.
- Find the slope of the given line.
- Find the slope of the perpendicular line.
- Identify the point.
- Substitute the values into the point-slope form, \(y−y_1=m(x−x_1)\).
- Write the equation in slope-intercept form.
Glossary
point-slope form
The point-slope form of an equation of a line with slope m and containing the point \((x_1,y_1)\) is \(y−y_1=m(x−x_1)\).


6.15: Unveiling the Truth - Heteronormativity and Cisnormativity
Sexual orientation and gender identity often fall under the same umbrella when it comes to understanding human sexuality, but as we covered in both this chapter and the previous one, they are two different aspects of the self. Sexual orientation describes a person’s patterns of emotional, romantic, and/or sexual attraction to another person, while gender identity refers to an individual’s own gender. Gender identity can be the same as someone’s assigned sex, but it can also be different. Both gender identity and sexual orientation are larger than a binary, and may shift throughout a person’s lifetime.
Heteronormativity is the assumption that all people are heterosexual, and that there are no other types of sexuality. (Cis)normativity has more to do with gender, in that it is the assumption that all individuals identify with a role that corresponds to the sex assigned to them at birth (cisgendered). Understanding human sexuality through a Cisgendered, heteronormative lens automatically, labels LGBTQIA+ people as abnormal. These types of microaggressions are destructive whether intentional or unintentional, and discount the reality of entire groups of people. Microaggressions maintain heteronormativity and/or cisnormativity by invalidating non-hetero or trans folx lived experiences. An example of this could be is when a cisgendered, heterosexual female asks a cis-gendered lesbian female if they have a boyfriend. Another could be if a cisgendered male expresses hostility over a transgender male who uses a men’s bathroom. The previous example is an act of a microaggression, but tragically, in some cases trans males have been assaulted while just trying to use the restroom.
Heteronormativity and cisnormativity are built into the fabric of American culture, which can cause many queer identifying people to hide their true selves from their friends and family.  The hope is that one day in the near future, it’ll be okay to be who you are without worry of backlash and discrimination, but until then, the need for solidarity is critical. There are solutions for overcoming the harmful effects of heteronormativity and cisnormativity. Speaking truth to power is one way to move the conversation forward. Allies are critical in the struggle. Here are a few organizations to check out: PFLAG | , About GLAAD , LGBTQ Youth Resources | Lesbian, Gay, Bisexual, and Transgender Health | CDC , LGBTQIA+ Resources - Positive Images
If you are a person who is questioning your sexual identity or gender identity, surrounding yourself with open minded people who are a part of the LGBTQIA+ community, or those who affirm your own identity will help you feel supported. Remember, one of the goals of this textbook is radical self-love so be sure to practice self-love now!  Sometimes, your chosen family can be who you need to turn to if your blood family is unable to accept you just exactly as you are. Just know that you are valued, and there is a community out there ready to invite you in.
 Was it always this bad? Is it worse now? What is the history of Sex Ed in the U.S.? Watch and see: Sex Education In America: A Brief History


9.1: Overview and Purpose
Overview and Purpose
For this lab students will create a 3D model of a fetal pig in order to study mammalian anatomy and compare it to human anatomy.


6.8: Factoring Trinomials with Leading Coefficient Other Than 1
The Method of Factorization
In the last section we saw that we could easily factor trinomials of the form \(x^2+bx+c\) by finding the factors of the constant \(c\) that add to the coefficient of the linear term \(b\), as shown in the following example:
Factor \(x^2 - 4x - 21\)
The third term of the trinomial is \(-21\). We seek two numbers whose:
(a) product is \(-21\)
(b) sum is \(-4\)
The required numbers are \(-7\) and \(+3\)
\(x^2 - 4x - 21 = (x-7)(x+3)\)
The problem of factoring the polynomial \(ax^2 + bx + c, a\not = 1\), becomes more involved. We will study two methods of factoring such polynomials. Each method produces the same result, and you should select the method you are most comfortable with. The first method is called the trial and error method and requires some educated guesses. We will examine two examples (Sample Sets A and B). Then, we will study a second method of factoring. The second method is called the collect and discard method , and it requires less guessing than the trial and error method. Sample Set C illustrates the use of the collect and discard method.
The Trial and Error Method of Factoring \(ax^2 + bx + c\)
Trial and Error Method
Consider the product
Examining the trinomial \(20x^2 + 23x + 6\), we can immediately see some factors of the first and last terms.
| \(20x^2\) | \(6\) |
| \(20x, x\) | \(6, 1\) |
| \(10x, 2x\) | \(3, 2\) |
| \(5x, 4x\) |
Our goal is to choose the proper combination of factors of the first and last terms that yield the middle term \(23x\).
Notice that the middle term comes from the
sum
of the
outer
and
inner
products in the multiplication of the two binomials.
This fact provides us a way to find the proper combination.
Look for the combination that when multiplied and then added yields the middle term.
The proper combination we're looking for is
Sample Set A
Factor \(6x^2 + x - 12\)
Factor the first and last terms:
Thus, \(3x\) and \(3\) are to be multiplied, \(2x\) and \(-4\) are to be multiplied.
\(\begin{array}{flushleft}
6x^2 + x - 12&=&( ) ( )&\text{Put the factors of the leading term in immediately.}\\
&=&(3x)(2x)&\text{Since } 3x \text{ and } 3 \text{ are to be multiplied, they must be located in different binomials.}\\
&=&(3x)(2x + 3)&\text{Place the } -4 \text{ in the remaining set of parentheses.}\\
&=&(3x-4)(2x+3)\\
6x^2 + x - 12&=&(3x-4)(2x+3)
\end{array}\)
Check:
\(\begin{array}{flushleft}
(3x-4)(2x+3)&=&6x^2 + 9x - 8x - 12\\
&=&6x^2 + x - 12
\end{array}\)
Factor \(8x^2 - 20x - 27\)
Find the factors of the first and last terms.
Thus, the \(4x\) and \(-9\) are to be multiplied, and \(2x\) and \(3\) are to be multiplied.
\(\begin{array}{flushleft}
8x^2 - 30x - 27&=& (4x)(2x)\\
&=&(4x)(2x-9)\\
&=&(4x + 3)(2x-9)
\end{array}\)
Check:
\(\begin{array}{flushleft}
(4x+3)(2x-9)&=&8x^2 - 36x + 6x - 27\\
&=&8x^2-30x-27
\end{array}\)
Factor \(15x^2 + 44x + 32\)
Before we start finding the factors of the first and last terms, notice that the constant term is \(+32\). Since the product is
positive
, the two factors we are looking for
must
have the same sign. They must both be positive or both be negative. Now the middle term, \(+44x\), is preceded by a positive sign. We know that the middle term comes from the
sum
of the outer and inner products. If these two numbers are to sum to a positive number, they must both be positive themselves. If they were negative, their sum would be negative. Thus, we can conclude that the two factors of \(+32\) that we are looking for are both positive numbers. This eliminates several factors of \(32\) and lessens our amount of work.
Factor the first and last terms.
After a few trials we see that \(5x\) and \(4\) are to be multiplied, and \(3x\) and \(8\) are to be multiplied.
\(15x^2 + 44x + 32 = (5x + 8)(3x + 4)\)
Factor \(18x^2 - 56x + 6\)
We see that each term is even, so we can factor out \(2\).
\(2(9x^2 - 28x + 3)\)
Notice that the constant term is positive. Thus, we know that the factors of 3 that we are looking for must have the same sign. Since the sign of the middle term is negative, both factors must be negative.
Factor the first and last terms.
There are not many combinations to try, and we find that \(9x\) and \(-3\) are to be multiplied and \(x\) and \(-1\) are to be multiplied.
\(\begin{array}{flushleft}
18x^2 - 56x + 6 &=& 2(9x^2 - 28x + 3)\\
&=&2(9x-1)(x-3)
\end{array}\)
If we had not factored out the \(2\) first, we would have gotten the factorization
The factorization is not complete since one of the factors may be factored further.
\(\begin{array}{flushleft}
18x^2 - 56x + 6&=&(9x-1)(2x-6)\\
&=&(9x-1)\cdot 2(x-3)\\
&=&2(9x-1)(x-3)&\text{(By the commutative property of multiplication)\)
\end{array}\)
The results are the same, but it is much easier to factor a polynomial after all common factors have been factored out first.
Factor \(3x^2 + x - 14\)
There are no common factors. We see that the constant term is negative. Thus, the factors of \(-14\) must have different signs. Factor the first and last terms.
After a few trials, we see that \(3x\) and \(-2\) are to be multiplied.
\(3x^2 + x - 14 = (3x + 7)(x - 2)\)
Factor \(8x^2 - 26xy + 15y^2\).
We see that the constant term is positive and that the middle term is preceded by a minus sign.
Hence, the factors of \(15y^2\) that we are looking for must both be negative.
Factor the first and last terms.
After a few trials, we see that \(4x\) and \(-5y\) are to be multiplied and \(2x\) and \(-3y\) are to be multiplied.
\(8x^2 - 26xy + 15y^2 = (4x - 3y)(2x - 5y)\)
Practice Set A
Factor the following, if possible.
\(2x^2+13x−7\)
- Answer
-
\((2x−1)(x+7)\)
\(3x^2+x−4\)
- Answer
-
\((3x+4)(x−1)\)
\(4a^2−25a−21\)
- Answer
-
\((4a+3)(a−7)\)
\(16b^2−22b−3\)
- Answer
-
\((8b+1)(2b−3)\)
\(10y^2−19y−15\)
- Answer
-
\((5y+3)(2y−5)\)
\(6m^3+40m^2−14m\)
- Answer
-
\(2m(3m−1)(m+7)\)
\(14p^2+31pq−10q^2\)
- Answer
-
\((7p−2q)(2p+5q)\)
\(−24w^2z^2+14wz^3−2z^4\)
- Answer
-
\(−2z^2(4w−z)(3w−z)\)
\(3x^2+6xy+2y^2\)
- Answer
-
not factorable
As you get more practice factoring these types of polynomials you become faster at picking the proper combinations. It takes a lot of practice!
There is a shortcut that may help in picking the proper combinations. This process does not always work, but it seems to hold true in many cases. After you have factored the first and last terms and are beginning to look for the proper combinations, start with the
intermediate
factors and not the extreme ones.
Sample Set B
Factor \(24x^2 - 41x + 12\)
Factor the first and last terms.
| \(24x^2\) | \(12\) |
| \(24x, x\) | \(-12, -1\) |
| \(12x, 2x\) | \(-12, -1\) |
| \(8x, 3x\) | \(-6, -2\) |
| \(6x, 4x\) |
Rather than starting with the \(24x, x\) and \(-12, -1\), pick some intermediate values, \(8x\) and \(3x\), the \(6x\) and \(4x\), or the \(-6\) and \(-2\), or the \(-4\) and \(-3\).
\(24x^2 - 41x + 12 = (8x-3)(3x-4)\)
Practice Set B
Factor \(48x^2+22x−15\).
- Answer
-
\((6x+5)(8x−3)\)
Factor \(54y^2+39yw−28w^2\).
- Answer
-
\((9y−4w)(6y+7w)\)
The Collect and Discard Method of Factoring \(ax^2+bx+c\)
Collect and Discard Method
Consider the polynomial \(6x^2+x−12\). We begin by identifying \(a\) and \(c\). In this case, \(a=6\) and \(c=−12\). We start out as we would with \(a=1\).
\(6x^2+x−12: (6x)(6x)\)
Now, compute \(a \cdot c\).
\(a \cdot c=(6)(−12)=−72\)
Find the factors of \(−72\) that add to \(1\), the coefficient of \(x\), the linear term. The factors are \(9\) and \(−8\). Include these factors in the parentheses.
\(6x^2+x−12: (6x+9)(6x−8)\)
But we have included too much. We must eliminate the surplus. Factor each parenthesis.
\(6x^2+x−12: 3(2x+3) \cdot 2(3x−4)\)
Discard the factors that multiply to \(a=6\). In this case, \(3\) and \(2\). We are left with the proper factorization.
\(6x^2+x−12=(2x+3)(3x−4)\)
Sample Set C
Factor \(10x^2+23x−5\).
Identify \(a=10\) and \(b=−5\).
\(10x^2+23x−5; (10x)(10x)\)
Compute
\(a \cdot c=(10)(−5)=−50\)
Find the factors of \(−50\) that add to \(+23\), the coefficient of \(x\), the linear term. The factors are \(25\) and \(−2\). Place these numbers into the parentheses.
\(10x^2+23x−5: (10x+25)(10x−2)\)
We have collected too much. Factor each set of parentheses and eliminate the surplus.
\(10x^2+23x−5: (5)(2x+5)⋅(2)(5x−1)\)
Discard the factors that multiply to \(a=10\). In this case, \(5\) and \(2\).
\(10x^2+23x−5=(2x+5)(5x−1)\)
Factor \(8x^2−30x−27\).
Identify \(a=8\) and \(c=−27\).
\(8x^2−30x−27: (8x)(8x)\)
Compute
\(a \cdot c=(8)(−27)=−216\)
Find the factors of \(−216\) that add to \(−30\), the coefficient of \(x\), the linear term. This requires some thought. The factors are \(−36\) and \(6\). Place these numbers into the parentheses.
\(8x^2−30x−27: (8x−36)(8x+6)\)
We have collected too much. Factor each set of parentheses and eliminate the surplus.
\(8x^2−30x−27: (4)(2x−9)⋅(2)(4x+3)\)
Discard the factors that multiply to \(a=8\). In this case, \(4\) and \(2\).
\(8x^2−30x−27=(2x−9)(4x+3)\)
Factor \(18x^2−5xy−2y^2\).
Identify \(a=18\) and \(c=−2\).
\(18x^2−5xy−2y^2: (18x)(18x)\)
Compute
\(a \cdot c=(18)(−2)=−36\)
Find the factors of \(−36\) that add to \(−5\), the coefficient of \(xy\). In this case, \(−9\) and \(4\). Place these numbers into the parentheses, affixing \(y\) to each.
\(18x^2−5xy−2y^2: (18x−9y)(18x+4y)\)
We have collected too much. Factor each set of parentheses and eliminate the surplus.
\(18x^2−5xy−2y^2: (9)(2x−y) \cdot (2)(9x+2y)\)
Discard the factors that multiply to \(a=18\). In this case, \(9\) and \(4\).
\(18x^2−5xy−2y^2=(2x−y)(9x+2y)\)
Practice Set C
Factor \(6x^2+7x−3\).
- Answer
-
\((3x−1)(2x+3)\)
Factor \(14x^2−31x−10\).
- Answer
-
\((7x+2)(2x−5)\)
Factor \(48x^2+22x−15\).
- Answer
-
\((6x+5)(8x−3)\)
Factor \(10x^2−23xw+12w^2\).
- Answer
-
\((5x−4w)(2x−3w)\)
Exercises
Factor the following problems, if possible.
\(x^2+3x+2\)
- Answer
-
\((x+2)(x+1)\)
\(x^2+7x+12\)
\(2x^2+7x+5\)
- Answer
-
\((2x+5)(x+1)\)
\(3x^2+4x+1\)
\(2x^2+11x+12\)
- Answer
-
\((2x+3)(x+4)\)
\(10x^2+33x+20\)
\(3x^2−x−4\)
- Answer
-
\((3x−4)(x+1)\)
\(3x^2+x−4\)
\(4x^2+8x−21\)
- Answer
-
\((2x−3)(2x+7)\)
\(2a^2−a−3\)
\(9a^2−7a+2\)
- Answer
-
not factorable
\(16a^2+16a+3\)
\(16y^2−26y+3\)
- Answer
-
\((8y−1)(2y−3)\)
\(3y^2+14y−5\)
\(10x^2+29x+10\)
- Answer
-
\((5x+2)(2x+5)\)
\(14y^2+29y−15\)
\(81a^2+19a+2\)
- Answer
-
not factorable
\(24x^2+34x+5\)
\(24x^2−34x+5\)
- Answer
-
\((6x−1)(4x−5)\)
\(24x^2−26x−5\)
\(24x^2+26x−5\)
- Answer
-
\((6x−1)(4x+5)\)
\(6a^2+13a+6\)
\(6x^2+5xy+y2\)
- Answer
-
\((3x+y)(2x+y)\)
\(6a^2−ay−y^2\)
For the following problems, the given trinomial occurs when solving the corresponding applied problem. Factor each trinomial. You do not need to solve the problem.
\(5r^2−24r−5\)
It takes 5 hours to paddle a boat 12 miles downstream and then back. The current flows at the rate of 1 mile per hour. At what rate was the boat paddled?
- Answer
-
\((5r+1)(r−5)\)
\(x^2+5x−84\)
The length of a rectangle is 5 inches more than the width of the rectangle. If the area of the rectangle is 84 square inches, what are the length and width of the rectangle?
\(x^2+24x−145\)
A square measures 12 inches on each side. Another square is to be drawn around this square in such a way that the total area is 289 square inches. What is the distance from the edge of the smaller square to the edge of the larger square? (The two squares have the same center.)
- Answer
-
\((x+29)(x−5)\)
\(x^2+8x−20\)
A woman wishes to construct a rectangular box that is open at the top. She wishes it to be 4 inches high and have a rectangular base whose length is three times the width. The material used for the base costs $2 per square inch, and the material used for the sides costs $1.50 per square inch. The woman will spend exactly $120 for materials. Find the dimension of the box (length of the base, width of the base, and height).
For the following problems, factor the trinomials if possible.
\(16x^2−8xy−3y^2\)
- Answer
-
\((4x+y)(4x−3y)\)
\(6a^2+7ab+2b^2\)
\(12a^2+7ab+12b^2\)
- Answer
-
not factorable
\(9x^2+18xy+8y^2\)
\(8a^2+10ab−6b^2\)
- Answer
-
\(2(4a^2+5ab−3b^2)\)
\(12a^2+54a−90\)
\(12b^4+30b^2a+12a^2\)
- Answer
-
\(6(2b^2+a)(b^2+2a)\)
\(30a^4b^4−3a^2b^2−6c^2\)
\(3a^6−3a^3b^2−18b^4\)
- Answer
-
\(3(a^3+2b^2)(a^3−3b^2)\)
\(20a^2b^2+2abc^2−6a^2c^4\)
\(14a^2z^2−40a^3z^2−46a^4z^2r\)
- Answer
-
\(2a^2z^2(7−20a−23a^2) \text{ or} −2a^2z^2(23a^2+20a−7)\)
Exercises for Review
Simplify \((a^3b^6)^4\)
Find the product \(x^2(x-3)(x+4)\)
- Answer
-
\(x^4+x^3−12x^2\)
Find the product \((5m-3n)^2\)
Solve the equation \(5(2x−1)−4(x+7)=0\).
- Answer
-
\(x = \dfrac{11}{2}\)
Factor \(x^5 - 8x^4 + 7x^3\)


5.6: Addressing Vaccine Hesitancy
Although parents use the internet and social media for medical information, they still look to healthcare providers for trusted advice. It is important that health professionals have the knowledge, attitude, and skills to counsel clients who have questions about vaccines. Respectful dialogue between parents and their child’s healthcare provider is crucial to address vaccine hesitancy. Most often, parents express concerns related to vaccine safety and efficacy, and therefore healthcare providers are an important informational resource. It is important to clearly identify the risks and benefits of vaccines.
Growing Your Communication Skills
Effective communication skills can be developed with practice. When learning about the power of one-to-one conversations , you can make use of a number of techniques to strengthen your skills. First, it is important to look within and understand your personal bias. You should have a good sense of population diversity and societal beliefs. Always aim for simplicity and tailor information to match your client’s needs. Use story-telling to make an impact and understand that an emotional perception may be stronger than logic or fact. People may not remember what you said, but they remember what they thought you said.
When Communicating with Parents…
- Be presumptive that their child will be vaccinated.
- Maintain trust. Be non-judgmental.
- Take time to listen. Keep lines of communication open.
- Validate parental concerns and correct misinformation.
- Acknowledge benefits and risks of vaccines in contrast to risk of disease.
- Frame data clearly and positively.
- Use clear language that is relevant to your client’s context.
- Show parents how the stress of vaccination for their baby/child can be lessened.
- Encourage clients to seek credible sources for more information.
- Use every visit as an opportunity to discuss vaccines.
When Communicating with Adults…
- Use every visit as an opportunity to discuss vaccines.
- Include vaccination in adult health education and assessment.
- Be presumptive that the client will be vaccinated.
- Be non-judgmental.
- Listen to your client’s concerns.
- Validate concerns and correct misinformation.
- Frame data clearly and positively.
- Acknowledge benefits and risks of vaccines in contrast to risk of disease.
- Use clear language that is relevant to your client’s context.
- Encourage clients to seek credible sources for more information.
- Address waning immunity in diseases more common to adults.
- Assess client risks based on occupation, travel habits, underlying illness, lifestyle, and/or age.
- Acknowledge that vaccines for adults may not prevent the onset of disease but will at least reduce disease severity.