# At what points on the graph of y = x^2 does the tangent line pass through (3, −7)?

Mar 6, 2015

The points are $\left(- 1 , 1\right)$ and $\left(7 , 49\right)$.

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 $\left(3 , - 7\right)$.

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: $y ' = 2 x$, so at the point $\left(a , {a}^{2}\right)$ the slope of the tangent is $m = 2 a$.

Use your favorite tehnique to find the equation of the line through $\left(a , {a}^{2}\right)$ with slope $2 a$.

The tangent line has equation: $y = 2 a x - {a}^{2}$.

(One way to find the line: start with $y - {a}^{2} = 2 a \left(x - a\right)$, so $y - {a}^{2} = 2 a x - 2 {a}^{2}$. Add ${a}^{2}$ to both sides to get $y = 2 a x - {a}^{2}$.)

We have been asked to make the point $\left(3 , - 7\right)$ lie on the line. So we need, $\left(- 7\right) = 2 a \left(3\right) - {a}^{2}$. Now, solve for $a$.

$- 7 = 6 a - {a}^{2}$ if and only if ${a}^{2} - 6 x - 7 = 0$.
Solve by factoring: $\left(a + 1\right) \left(a - 7\right) = 0$, which requires $a = - 1$ or $a = 7$.

The points we are looking for, then, are $\left(- 1 , 1\right)$ and $\left(7 , 49\right)$.

You can now check the answers by verifying that the point $\left(3 , - 7\right)$ lies on the lines:
$y = - 2 x - 1$ (the tangent when $a = - 1$),
and also on $y = 14 x - 49$ (the tangent when $a = 7$).