Question

cubic & cube root functions unit online practice
complete this assessment to review what you’ve learned. it will not count toward your grade.
which function represents a horizontal compression by a factor of \\(\frac{1}{7}\\) of the function \\(g(x) = -(x - 1)^3 + 2\\)?
(1 point)
\\(\circ\\) \\(j(x) = -\left(\frac{1}{7}(x - 1)\
ight)^3 + 2\\)
\\(\boldsymbol{\odot}\\) \\(m(x) = -(7x - 1)^3 + 2\\) \\(\boldsymbol{\times}\\)
\\(\circ\\) \\(h(x) = -(7(x - 1))^3 + 2\\)
\\(\circ\\) \\(k(x) = -\left(\frac{1}{7}x - 1\
ight)^3 + 2\\)
incorrect

• this would be the answer if \\(g(x)\\) was to be both horizontally compressed and horizontally shifted.

check answer remaining attempts: 2
graphing calculator

Explanation:

Step1: Recall Horizontal Compression Rule

For a function \( y = f(x) \), a horizontal compression by a factor of \( \frac{1}{a} \) (where \( a>1 \)) is given by replacing \( x \) with \( ax \) in the function. So the transformation is \( y = f(ax) \).

Step2: Apply the Rule to \( g(x) \)

Given \( g(x)=-(x - 1)^{3}+2 \), a horizontal compression by a factor of \( \frac{1}{7} \) means we replace \( x \) with \( 7x \) in \( g(x) \) (since \( a = 7 \) for compression factor \( \frac{1}{7} \)).

Substitute \( x \) with \( 7x \) in \( g(x) \):
\( g(7x)=-(7x - 1)^{3}+2 \)? Wait, no. Wait, the original function is \( g(x)=-(x - 1)^{3}+2 \), so when we do horizontal compression, we replace \( x \) in the argument of the function with \( 7x \). Wait, the argument of the cubic function is \( (x - 1) \). So to apply horizontal compression by factor \( \frac{1}{7} \), we replace \( x \) with \( 7x \) in the entire function's input. Wait, no, the correct substitution is: if we have \( y = f(x) \), horizontal compression by \( \frac{1}{a} \) is \( y = f(ax) \). So here, \( f(x)=-(x - 1)^{3}+2 \), so \( f(ax)=-(ax - 1)^{3}+2 \)? No, wait, no. Wait, the inside of the function is \( (x - 1) \). Wait, no, let's re - express. Let \( u=x - 1 \), then \( g(x)=-u^{3}+2 \) where \( u=x - 1 \). A horizontal compression by factor \( \frac{1}{7} \) on \( x \) means that \( x \) is replaced by \( 7x \) (because compression by \( \frac{1}{7} \) in \( x \) - direction is equivalent to scaling \( x \) by 7). So \( u = 7x-1 \)? No, wait, no. Wait, the horizontal compression affects the \( x \) - value in the input of the function. The function is \( g(x)=-(x - 1)^{3}+2 \). So to compress horizontally by \( \frac{1}{7} \), we need to make the input to the function change faster. So the transformation is \( x\to7x \) in the function's argument. Wait, the correct formula: if we have a function \( y = f(x - h)+k \), a horizontal compression by factor \( \frac{1}{a} \) is \( y = f(a(x - h))+k \). Wait, I made a mistake earlier. The horizontal compression of \( y = f(x - h)+k \) by factor \( \frac{1}{a} \) is \( y = f(a(x - h))+k \). So in our case, \( f(x)=-(x)^{3}+2 \) with a horizontal shift of 1 unit to the right. Wait, no, the original function is \( g(x)=-(x - 1)^{3}+2 \), which is \( f(x - 1) \) where \( f(x)=-x^{3}+2 \). So a horizontal compression of \( f(x - 1) \) by factor \( \frac{1}{7} \) is \( f(7(x - 1)) \). Because the rule for horizontal compression on \( y = f(x - h) \) is \( y = f(a(x - h)) \) for compression factor \( \frac{1}{a} \).

So let's re - do this. Let \( f(x)=-x^{3}+2 \), then \( g(x)=f(x - 1) \). A horizontal compression of \( g(x) \) by factor \( \frac{1}{7} \) is \( f(7(x - 1)) \).

Compute \( f(7(x - 1)) \):
\( f(7(x - 1))=- (7(x - 1))^{3}+2=-(7(x - 1))^{3}+2 \)

So the function \( h(x)=-(7(x - 1))^{3}+2 \) is the result of horizontal compression by factor \( \frac{1}{7} \) of \( g(x) \).

Answer:

\( h(x)=-(7(x - 1))^{3}+2 \) (corresponding to the option \( h(x)=-(7(x - 1))^{3}+2 \))