<s>[INST] Find the points in the graph (my solution) - high school. my math problem is on Swedish so i'll try my best to translate it so you can understand. I'd appreciate it if someone could point out if I did anything wrong and if there is anything that I should add that's essential to these kind of problems.
(I'm a high school student in my second year)


The function $f(x) = x^3 - 6x^2 + 9x + 2$ gives us the points A and B
  on the graph and the second function, $g(x) = x^2 - 7x + 14$ gives C
  and D, where C is a mutual point for these two functions. C has a
  gradient that is -3.


To start off with, I set the x for $f(x)$ to 0 in order to see where on the y-axis the point A crosses.
$f(0) = 0^3 - 6*0^2 + 9*0 + 2 = 2$
$A(0,2)$
The next step is to figure out B. To be able to get the coordinates for this point, I have to derivate $f(x)$, and then set $f'(x)=0$.
$f'(x) = 3x^2 - 12x + 9 = 0$
$3x^2 - 12x + 9 = 0$
Dividing everything with 3 to use the pq-formula.
$x^2 - 4x + 3 = 0$
$x = 2  \pm{\sqrt {2^2-3}}$
$x_1 = 2+1 = 3$
$x_2 = 2-1 = 1$
With these 2 x-values I have to choose one that gives the biggest value for $f(x)$ point B is a Maximum point on the graph that has been provided. I can find this out with the help of Second Derivative.
$f''(x) = 6x - 12$
$f''(3) = 6*3 - 12 = 6$ This is a minimum point because of $f'(3) = 0$ and $f''(3) > 0$.
$f''(1) = 6*1 - 12 = -6$ This is the x-value that gives the maximum point that I am looking for. $f'(1) = 0$ and $f''(1) < 0$
$f(1) = 1^3 - 6*1^2 + 9*1 + 2 = 6$
$B(1,6)$
As stated earlier, C is a mutual point with a gradient -3. Derivating both functions and setting them equal to -3 and solving these will give us which x-value they have in common.
$f'(x) = 3x^2 - 12x + 9 = -3$
$3x^2 - 12x + 9 = -3$
+3 on both sides to get everything on one side then dividing everything with 3 to get x^2 alone so I can use pq-formula.
$x^2 - 4x + 4 = 0$
$x = 2  \pm {\sqrt {2^2-4}}$
$x = 2$
On to the second function: $g(x) = x^2 - 7x + 14$ I set $g'(x) = -3$
By derivating this function and moving -7 to the right side will give us:
$2x = 4$
Dividing both sides with 2
$x = 2$
$f(2) = 2^3 - 6*2^2 + 9*2 + 2 = 4$
$g(2) = 2^2 - 7*2 + 14 = 4$
$f(2)=g(2)=4$
$C(2,4)$
To get the last point (D) I do the exact same way as I did for point B. I derivate the function and set g'(x) = 0 to solve x. To confirm that D is a Min.Point I derivate the first derivative and this gives me $g''(x) = 2$. We can see that $g''(x) > 0$.
$g'(x) = 2x - 7 = 0$
$2x = 7$
$x = 3.5$
$g(3.5) = 3.5^2 - 7*3.5 + 14 = 1.75$
$D(3.5;1.75)$
Answer:
$A(0.2)$
$B(1.6)$
$C(2.4)$
$D(3.5;1.75)$
$${}$$
 [/INST] My solution above is correct.
Answer:
$A(0.2)$
$B(1.6)$
$C(2.4)$
$D(3.5;1.75)$
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