Question

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

Answer 1

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 `(a,a^2)` (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(x-a)`, so `y-a^2=2ax-2a^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, `(-7)=2a(3)-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`).