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 x-axis. Find

Question

i) the values of a and b
ii) the equation of the curve

Answer 1

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

Answer 2

I like the way you solve it