f'(x) = 2x
So, the tangent line at (h,k) = y-k = 2h(x-h)
So, we need lines such that the slope of the line through (h,k) and (3,7) is 2h
Do your lines fit?
(15-7)/(4-3) = 8 = 2*4
(3-7)/(2-3) = 4 = 2*2
Looks good to me.
"find the point(s) on the parabola f(x) = x^2 - 1 whose tangent lines pass through the point (3,7)."
I got (4,15) and (2,3). Would that be correct? I think my x is right but I'm not sure if the y is right because I just plugged it back into the equation x^2 - 1. Thanks! :D
1 answer
1 answer