calculus
posted by Amponsah .
A curve passes through the point (1,11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,16) is parallel to the xaxis. Find
i) the values of a and b
ii) the equation of the curve

since dy/dx = ax^2 + b
the equation must have been
y = (1/3)a x^3 + bx + c
at (1,11) > 11 = (1/3) a + b + c
or
a + 3b + 3c = 33 (#1)
at (2, 16) > 16 = (8/3)a + 2b + c
or
8a + 6b + 3c = 48 (#2)
#2  #1 :
7a + 3b = 15 (#3)
We also know that at (2,16), the slope is zero
ax^2 + b = 0
4a + b = 0
b = 4a
sub into #3
7a  12a = 15
5a=15
a=3
then b= 12
and in #1
3 36+ 3c = 33
3c = 0
c = 0
function is y = (1/3)(3)x^3  12x
y = x^3  12x