the gradient of a curve y=f(x) is given by dy/dx =ax(1-x) where a is a constant. the gradient of the curve at the point (2,-3) is -12 find the value of a and the equation of the curve.

i integrated it and subbed in the point(2,-3) to get 4a/2 -8a/3 +c =-3 i also know ax-ax^2=-12

2. Re: gradient of a curve

as u know the gradient of the curve at the point (2,-3) is -12, you can substitute x=2 and dy/dx=-12 into the equation of dy/dx to solve for a. good luck continuing from there

3. Re: gradient of a curve

$\dfrac{dy}{dx}=\left . a x(1-x)\right|_{x=2} = -12$

$a(2)(1-2) = -2a = -12$

$a=6$

$\dfrac{dy}{dx}=6x(1-x) = 6x - 6x^2$

$y(x) = \int~\dfrac{dy}{dx}~dx = 3x^2 -2x^3 + y_0$

now we know that $(2,-3)$ is on the curve so

$-3 = 3(2^2)-2(2^3) + y_0$

$-3 = 12 - 16 + y_0$

$y_0 = 1$

$y(x) = 3x^2 -2x^3 +1$