The gradient of a curve at the point (-2, 0) is (3x^{2} – 4x). Find the equation of the curve.
-
A.
(y = 6x – 4) -
B.
(y = 6x^{2} – 4x + 12) -
C.
(y = x^{3} – 2x^{2}) -
D.
(y = x^{3} – 2x^{2} + 16)
Correct Answer: Option D
Explanation
The gradient of a curve is gotten by differentiating the equation of the curve. Therefore, given the gradient, integrate to get the equation of the curve back.
(frac{mathrm d y}{mathrm d x} = 3x^{2} – 4x)
(y = int {(3x^{2} – 4x)} mathrm {d} x = frac{3x^{2+1}}{2+1} – frac{4x^{1+1}}{1+1} + c)
= (x^{3} – 2x^{2} + c )
To find c (the constant of integration), when x = -2, y = 0
(0 = (-2^{3}) – 2(-2^{2}) + c)
(0 = -8 – 8 + c implies c = 16)
(therefore y = x^{3} – 2x^{2} + 16)