Question

Which function would horizontally stretch g(x)=−x3+1 by a factor of 2? (1 point) Responses j(x)=−(12x)3+1 j left parenthesis x right parenthesis equals negative left parenthesis start fraction 1 over 2 x right parenthesis superscript 3 baseline plus 1 m(x)=−(0.2x)3+1 m left parenthesis x right parenthesis equals negative left parenthesis 0.2 x right parenthesis superscript 3 baseline plus 1 h(x)=−(2x)3+1 h left parenthesis x right parenthesis equals negative left parenthesis 2 x right parenthesis superscript 3 baseline plus 1 k(x)=−x3+2

Answer 1

To horizontally stretch the function \( g(x) = -x^3 + 1 \) by a factor of 2, we need to replace \( x \) with \( \frac{x}{2} \) in the function. This is because a horizontal stretch by a factor of \( a \) is achieved by replacing \( x \) with \( \frac{x}{a} \).

So, for a factor of 2, we replace \( x \) with \( \frac{x}{2} \):

\[
g\left(\frac{x}{2}\right) = -\left(\frac{x}{2}\right)^3 + 1 = -\frac{x^3}{8} + 1
\]

Looking at the options provided:

1. \( j(x) = -\left(\frac{1}{2} x\right)^3 + 1 \)
2. \( m(x) = -\left(0.2x\right)^3 + 1 \)
3. \( h(x) = -\left(2x\right)^3 + 1 \)
4. \( k(x) = -x^3 + 2 \)

The correct function that represents the horizontal stretch by a factor of 2 is:

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

So the answer is:

**j(x) = −(12x)³ + 1**