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
a + 3b + 3c = -33 (#1)
at (2, -16) ---> -16 = (8/3)a + 2b + c
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
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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