The points are (-1, 1) and (7, 49).
Method: Find the general form of the equation of a line tangent to the graph of
. 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 (The y-coordinate must satisfy 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, ) the slope of the tangent is m=2a.
Use your favorite tehnique to find the equation of the line through (a, ) with slope 2a.
The tangent line has equation: .
(One way to find the line: start with , so .
Add to both sides to get .)
We have been asked to make the point (3, -7) lie on the line. So we need,
. Now, solve for a
if and only if .
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).
Not exactly what you’re looking for?