The gradient of the tangent to a curve y=ax^2+bx at the point (2,8)is the parallel to x-axis. Find the values of a and b.

Answer 1

you know that f'(2) = 0

2ax+b = 0 at (2,8)

4a+b = 0
4a+2b = 8
So, b=8
a = -2

y = -2x^2 + 8x

see

http://www.wolframalpha.com/input/?i=-2x%5E2+%2B+8x

or, without calculus, you know that

-b/2a = 2 since the vertex is there.
so b = -4a

y(2) = 8, so
4a + (-4a)(2) = 8
a = -2
so b = 8