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)
Check:
f'(4) = 8 = 32/4
f'(-4) = 24 = 96/4
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.
1 answer
2 answers