The function h(x) = x^3 + bx^2 + d has a critical point at (2, -4). Determine the constants b and d and find the equation of h(x).
IM SO CONFUSED :(
dh/dx = 0 at critical points
dh/dx = 3 x^2 + 2 b x = 0
x (3x+2b) = 0
so
at (2,4), 3x+2b = 0 = 6 + 2b
so b = -3
so
h = x^3 -3 x^2 + d
at x = 2, h = -4
-4 = 8 - 12 + d
d = 0
so
h = x^3 -3 x^2
To find the constants b and d and the equation of h(x), we can use the information that the function has a critical point at (2, -4).
The critical point of a function occurs when its derivative is equal to zero. So, to find the constants and equation, we need to find the derivative of h(x) and use the critical point to solve for b and d.
First, let's find the derivative of h(x).
h(x) = x^3 + bx^2 + d
To differentiate h(x), we apply the power rule of differentiation:
h'(x) = 3x^2 + 2bx
Now, let's substitute x = 2 into h'(x) to find the slope at x = 2:
h'(2) = 3(2)^2 + 2b(2)
= 12 + 4b
Since the function has a critical point at (2, -4), the slope at x = 2 should be zero.
Therefore, we can set h'(2) equal to zero:
12 + 4b = 0
Solving this equation will give us the value of b:
4b = -12
b = -3
Now that we have obtained the value of b, let's substitute it back into the original equation for h(x):
h(x) = x^3 + bx^2 + d
= x^3 - 3x^2 + d
We still need to find the value of d, so let's substitute x = 2 and y = -4 into h(x):
-4 = (2)^3 - 3(2)^2 + d
-4 = 8 - 12 + d
-4 = -4 + d
By rearranging this equation, we can solve for d:
d = -4 + 4
d = 0
Now we have found the values of both b and d. Plugging these values back into the original equation, we get:
h(x) = x^3 - 3x^2 + 0
= x^3 - 3x^2
So, the equation of h(x) is h(x) = x^3 - 3x^2.