Question

Which of the following cubic functions has a point of inflection of (−4,2)?(1 point)
Responses

h(x)=(x−4)3+2
h of x is equal to open paren x minus 4 close paren cubed plus 2

h(x)=3(x−4)3−2
h of x is equal to 3 times open paren x minus 4 close paren cubed minus 2

h(x)=2(x+4)3−2
h of x is equal to 2 times open paren x plus 4 close paren cubed minus 2

h(x)=−2(x+4)3+2

Answer 1

To determine which cubic function has a point of inflection at \((-4, 2)\), we need to check the candidates for having their second derivative equal to zero at \(x = -4\) and also ensure the function value at \(x = -4\) is equal to 2.

### Step 1: Finding the second derivative

We need to analyze each function to find the second derivative.

1. **Function**: \( h(x) = (x - 4)^3 + 2 \)

- First derivative:
\[
h'(x) = 3(x - 4)^2
\]

- Second derivative:
\[
h''(x) = 6(x - 4)
\]
- Setting \(h''(x) = 0\):
\[
6(x - 4) = 0 \implies x = 4 \quad \text{(not at } x = -4\text{)}
\]

2. **Function**: \( h(x) = 3(x - 4)^3 - 2 \)

- First derivative:
\[
h'(x) = 9(x - 4)^2
\]

- Second derivative:
\[
h''(x) = 18(x - 4)
\]
- Setting \(h''(x) = 0\):
\[
18(x - 4) = 0 \implies x = 4 \quad \text{(not at } x = -4\text{)}
\]

3. **Function**: \( h(x) = 2(x + 4)^3 - 2 \)

- First derivative:
\[
h'(x) = 6(x + 4)^2
\]

- Second derivative:
\[
h''(x) = 12(x + 4)
\]
- Setting \(h''(x) = 0\):
\[
12(x + 4) = 0 \implies x = -4 \quad \text{(is at } x = -4\text{)}
\]

4. **Function**: \( h(x) = -2(x + 4)^3 + 2 \)

- First derivative:
\[
h'(x) = -6(x + 4)^2
\]

- Second derivative:
\[
h''(x) = -12(x + 4)
\]
- Setting \(h''(x) = 0\):
\[
-12(x + 4) = 0 \implies x = -4 \quad \text{(is at } x = -4\text{)}
\]

### Step 2: Evaluating function values at \(x = -4\)

Now, we need to check that the function value at \(x = -4\) is equal to 2 for both of the functions that have points of inflection there.

1. For \( h(x) = 2(x + 4)^3 - 2 \):
\[
h(-4) = 2(-4 + 4)^3 - 2 = 2(0) - 2 = -2 \quad \text{(not equal to 2)}
\]

2. For \( h(x) = -2(x + 4)^3 + 2 \):
\[
h(-4) = -2(-4 + 4)^3 + 2 = -2(0) + 2 = 2 \quad \text{(equal to 2)}
\]

### Conclusion

Thus, the cubic function that has a point of inflection at \((-4, 2)\) is:

\[
\boxed{h(x) = -2(x + 4)^3 + 2}
\]