Question

Which function would horizontally stretch g(x)=−x3+1
by a factor of 2?(1 point)
Responses

j(x)=−(12x)3+1
j times x is equal to negative open paren 1 half x close paren cubed plus 1

h(x)=−(2x)3+1
h of x is equal to negative 2 x cubed plus 1

k(x)=−x3+2
k times x is equal to negative x cubed plus 2

b(x)=−(0.2x)3+1

Answers

GPT-4o mini
To horizontally stretch a function \( g(x) \) by a factor of \( k \), you replace \( x \) in the function with \( \frac{x}{k} \). In this case, to stretch the function \( g(x) = -x^3 + 1 \) horizontally by a factor of 2, you would replace \( x \) with \( \frac{x}{2} \).

Let's apply this transformation:

1. Start with the original function: \( g(x) = -x^3 + 1 \).
2. Replace \( x \) with \( \frac{x}{2} \):
\[
g\left(\frac{x}{2}\right) = -\left(\frac{x}{2}\right)^3 + 1
\]
3. Simplifying this:
\[
g\left(\frac{x}{2}\right) = -\frac{x^3}{8} + 1
\]

Now, let's examine the provided options:

1. **j(x) = −(12x)³ + 1**: This represents a vertical compression, not a horizontal stretch.
2. **h(x) = −(2x)³ + 1**: This represents a horizontal compression by a factor of 0.5, not a stretch.
3. **k(x) = −x³ + 2**: This is a vertical shift, not a stretch.
4. **b(x) = −(0.2x)³ + 1**: This corresponds to replacing \( x \) with \( 5x \) (since \( 0.2 = \frac{1}{5} \)), making it a horizontal stretch by a factor of 5.

None of the given functions are correct. The function that would be correct for a horizontal stretch by a factor of 2 is typically denoted as:

\[
j(x) = -\left(\frac{x}{2}\right)^3 + 1
\]

However, if one of the options were \( j(x) = -\left(\frac{1}{2}x\right)^3 + 1 \), this would be equivalent to the correct function for a horizontal stretch by a factor of 2.

Considering the closest one for a horizontal stretch, option **j(x)** looks potentially like a misrepresentation. If you would like, I can clarify further.

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