The function h(x) = x^3+bx+d has a critical point at (2,-4). Determine the constants b and d and find the equation of h(x). Show your work.

well, you know that critical points occur when h'(x)=0. So, we want

3x^2+b = 0 at x=2.
So, b = -12

That means that

h(x) = x^3-12x+d
h(2) = 8-24+d = -4
So, d = 12, giving us

h(x) = x^3-12x+12

The graph is at

http://www.wolframalpha.com/input/?i=x^3-12x%2B12+for+x%3D0..3

Looks like there's a minimum at (2,-4) all right.

h ' (x) = 3x^2 + b

= 0 at a critical point
if (2,-4) is a critical point,
3(4) + b = 0
b = -12
but (2,-4) also satisfies h(x)
-4 = 8 - 12(2) + c
c = 12

thus h(x) = x^3 - 12x + 12

To find the constants b and d and determine the equation of h(x) given that it has a critical point at (2, -4), we can use the derivative of the function.

First, let's find the derivative, h'(x), of the function h(x). Since the critical point occurs at (2, -4), the derivative at x = 2 must be equal to 0.

h'(x) = 3x^2 + b

Now, let's substitute x = 2 into the derivative and set it equal to 0:

0 = 3(2)^2 + b
0 = 12 + b

To solve for b, we subtract 12 from both sides of the equation:

b = -12

Now that we have found b = -12, let's substitute this value back into the original function to find d:

h(x) = x^3 + bx + d
h(x) = x^3 - 12x + d

Since we know that the critical point occurs at (2, -4), we can substitute x = 2 and y = -4 into the equation:

-4 = (2)^3 - 12(2) + d
-4 = 8 - 24 + d
-4 = -16 + d

To solve for d, we add 16 to both sides of the equation:

d = 12

Therefore, the constants b and d are b = -12 and d = 12, respectively. The equation of h(x) is:

h(x) = x^3 - 12x + 12