Posted by **Amponsah** on Monday, April 9, 2012 at 4:35am.

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

i) the values of a and b

ii) the equation of the curve

- calculus -
**Reiny**, Monday, April 9, 2012 at 9:50am
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

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