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.

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

To find the values of a and b, we need to first find the gradient of the tangent line to the curve at the point (2, 8).

The gradient of a curve at any point can be found by taking the derivative of the equation of the curve with respect to x. In this case, the equation of the curve is y = ax^2 + bx.

Taking the derivative of y with respect to x gives us:
dy/dx = 2ax + b

Since the gradient of the tangent line is parallel to the x-axis, it means that the slope of the tangent line is zero. Therefore, dy/dx = 0.

Setting dy/dx = 0, we have:
0 = 2ax + b

Now, we can substitute the coordinates of the given point (2, 8) into the equation to find the values of a and b.

Plugging in x = 2 and y = 8 into the equation, we get:
0 = 2a(2) + b
0 = 4a + b

We now have a system of equations:
0 = 2ax + b
0 = 4a + b

To solve this system, we can use either substitution or elimination.

Let's use elimination to solve it:
Multiply the second equation by -1 to get:
0 = -4a - b

Now, add the two equations together to eliminate b:
0 = 2ax + b
0 = -4a - b
----------------
0 = -2a

Since -2a = 0, we can divide both sides by -2 to solve for a:
a = 0

Substitute a = 0 into one of the equations to find b:
0 = 4(0) + b
0 = 0 + b
b = 0

Therefore, the values of a and b are both zero.

So, a = 0 and b = 0.