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
Answer this Question
calculus - Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=...
Calculus - Maths - I got a few questions. Hope ya'll can help out. 1) for F(X...
Maths - a curve ahs parametric equations x=t^2 and y= 1-1/2t for t>0. i)find ...
Calculus - Damon - Find the line which passes through the point (0, 1/4) and is ...
Math - How can I determine a value of gradient from a graph if the line is ...
calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
Maths: Intergration - The curve with equation y = ax^2 + bx + c passes through ...
calculus - the tangent line to the curve y=x^2 at the point (a,a^2) passes ...
Calculus - A curve passes through the point (7,6) and has the property that the ...
Calculus Help Please!!! Check - A curve passes through the point (0, 2) and has...