# At what points on the graph of y=x^{2} does the

At what points on the graph of $y={x}^{2}$ does the tangent line pass through (3, -7)?
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The points are (-1, 1) and (7, 49).
Method: Find the general form of the equation of a line tangent to the graph of
$y={x}^{2}$. Then find the particular points that satisfy the condition: the tangent line passes through (3, -7).
I will continue to use x and y as variables.
Consider a particular value of x, calls it a.
The point an the graph at x=a has coordinates $\left(a,{a}^{2}\right)$ (The y-coordinate must satisfy $y={x}^{2}$ in order to be on the graph.)
The slope of the tangent to the graph is determined by differentiating: y'=2x, so at the point (a, ${a}^{2}$) the slope of the tangent is m=2a.
Use your favorite tehnique to find the equation of the line through (a, ${a}^{2}$) with slope 2a.
The tangent line has equation: $y=2ax-{a}^{2}$.
(One way to find the line: start with $y-{a}^{2}=2a\left(x-a\right)$, so $y-{a}^{2}=2ax-2{a}^{2}$.
Add ${a}^{2}$ to both sides to get $y=2ax-{a}^{2}$.)
We have been asked to make the point (3, -7) lie on the line. So we need,
$\left(-7\right)=2a\left(3\right)-{a}^{2}$. Now, solve for a
$-7=6a-{a}^{2}$ if and only if ${a}^{2}-6x-7=0$.
Solve by factoring: (a+1)(a-7)=0, which requires a=-1 or a=7.
The points we are looking for, then, are (-1, 1) and (7, 49).
You can now check the answers by verifying that the point (3, -7) lies on the lines:
y=-2x-1 (the tangent when a=-1),
and also on y=14x-49 (the tangent when a=7).