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

The points are (-1, 1) and (7, 49).
Method: Find the general form of the equation of a line tangent to the graph of
${\displaystyle y=x^2}$. 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 ${\displaystyle (a,a^2)}$ (The y-coordinate must satisfy ${\displaystyle y=x^2}$ 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, ${\displaystyle a^2}$) the slope of the tangent is m=2a.
Use your favorite tehnique to find the equation of the line through (a, ${\displaystyle a^2}$) with slope 2a.
The tangent line has equation: ${\displaystyle y=2ax-a^2}$.
(One way to find the line: start with ${\displaystyle y-a^2=2a(x-a)}$, so ${\displaystyle y-a^2=2ax-2a^2}$.
Add ${\displaystyle a^2}$ to both sides to get ${\displaystyle y=2ax-a^2}$.)
We have been asked to make the point (3, -7) lie on the line. So we need,
${\displaystyle (-7)=2a\left(3\right)-a^2}$. Now, solve for a
${\displaystyle -7=6a-a^2}$ if and only if ${\displaystyle a^2-6x-7=0}$.
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).