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

1 Answer
Mar 6, 2015

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#).