posted by Ifi on .
Let f(x) be the parabola -x^2+16x-16. find the point (x,y) on f such that tangent line to f at (x,y) passes through the origin.
My ans is (8,48). First i do f'x then i find x and lastly i find y. I do this because the max/min point is the point where tangent line pass through. Pls correct me if im wrong.
No, (8,48) is the max, so the tangent there is horizontal. It will not pass through the origin.
f'(x) = -2x+16
you want to find (h,k) where the line joining (0,0) and (h,k) has slope -2h+16
k/h = -2h+16
But, k = -h^2+16h-16, so
-h^2+16h-16 = h(-2h+16)
-h^2 + 16h - 16 = -2h^2 + 16h
h^2 - 16 = 0
h = 4,-4, so k = 32,-96
The points are thus (-4,-96) and (4,32)
f'(4) = 8 = 32/4
f'(-4) = 24 = 96/4