The gradient of a curve at (1,-11) is ax^2 + b at all points. The tangent at (2,-16) is parallel to the x-axis. Find a and b and the equation of the curve
so y = (1/3)ax^3 + bx + c
points (1,-11) and (2, -16) lie on it, so
-11 = (1/3)a + b + c
or
a + 3b + 3c = -33 **
-16 = (1/3)(a)(8) + 2b + c = 0
8a + 6b + 3c = -48 ***
subtract *** from **
7a + 3b = -15 ****
we also know that at (2,-16) , y' = 0
ax^2 + b = 0 at (2,-16)
4a + b = 0
b = -4a
sub into ****
7a + 3(-4a) = -15
-5a=-15
a = 3
b = -12
in **
3 - 36 + 3c = -33
3c = 0
c = 0
y = x^3 - 12x
check my arithmetic