Calc I derivatives, I know the answers, but how did I get there?

1. Sep 7, 2010

i2c

1. The problem statement, all variables and given/known data
Question #2
http://img830.imageshack.us/img830/8185/0000825.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

I can of course find the derivatives for both functions. When I set them equal to each other I get x=1/2 that's where the functions have parallel tangent lines for the same x value, however (I think that) the functions have parallel tangents everywhere, just at different x values. So I figured out that at any point x on x^2, there will be a parallel tangent from -x^2+2x-5 at 1-x

I guess you have to find the equation of a tangent line that intersects with both graphs only once, no more no less, but I don't know how to write that out and solve for it. The answers are the tangent lines when x = -1 and 2 on x^2

Last edited by a moderator: May 4, 2017
2. Sep 7, 2010

vela

Staff Emeritus
Label the point where the line intersects the upper parabola (a, a2) and the point where it intersects the lower parabola (b, -b2+2b-5). So those are two points on the line, and you also found that b=1-a. That should be enough to figure out the equation of the line, I think.