Find all points of the graph f(x) = x^2 +4 that have a tangent that passes through (0,0)

NVM, thought about it upside down

This post was edited by allthelivelongday on Feb 27 2014 03:12pm

Here's how id do it:
-the equation of the tangent for x=a is y=f'(a)(x-a)+f(a)
-the tangent passes through (0,0) iff x=0 and y=0, thus it passes through (0,0) iff 0=-af'(a)+f(a)
-thus, the answers are the solutions to the equation -a^2+4=0 , so (2,8) and (-2,8) are the solutions.

This post was edited by m0hawk on Feb 27 2014 03:55pm

I'm lost even at this lol,

he's skipping some steps but that is 6/7th grade algebra 1, y=mx+b


so, just take your derivative to find slope then apply 6/7th grade algebra 1 equations, y=mx+b