the slope of a line joining that point to some other point on the curve, call it $x_{2}$, is just the rise over the run so that

$$\begin{aligned}
                \mbox{ slope} &= &\frac
{y(x_{2})-y(x_{1})}{x_{2}-x_{1}} \label{eq:q1}
 = \frac{cx_{2}^{2}-cx_{1}^{2}}{x_2-x_1} \nonumber \\ &=
&c(x_{2}+x_{1})\end{aligned}$$

We wanted to find the slope of the quadratic at a point whose $x-$ coordinate is equal to $x_{1}$. But the process of taking the slope involves two points because the slope is the rise of $y$ between two points divided by the run along $x$ between the same two points. So we have a result that depends, not only on the point with $x=x_{1}$, where we want the slope, but also on some other point with $x=x_{2}$. We have, in other words, the slope of a line that cuts the curve at two points.

But we only want the tangent line to touch the curve at the one point, namely where $x=x_1$. What to do? Choose $x_{2}$=$x_{1}$! Then Eq. \[eq:q1\] becomes: $$\hbox{\rm slope} =2cx_{1}           \label{eq:q2}$$

Eq.\[eq:q2\] shows that it is easy to find the slope of a quadratic function $y(x) = x^{2}$ at a point $x = x_{1}$. The slope is just the ratio of the change in $y$ 