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
calculus - Determine the equation of a curve in the xy-plane that passes through...
calculus - the tangent line to the curve y=x^2 at the point (a,a^2) passes ...
calculus - find an equation of the curve that passes through the point (1,1) and...
calculus - Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=...
calculus - Find the equation of a straight line that passes through a point (1,5...
calculus - is the answer to find an equation of the curve that passes through ...
Algebra II - What is the equation of the line, in point -slope form, that ...
Calculus - Sketch a graph of the parabola y=x^2+3. On the same graph, plot the ...
calculus - Find a function f such that the curve y = f(x) satisfies y'' = 12x, ...
Calculus - Hi, I'm absolutely stuck on this problem. This is for a newton's ...