Compressing and Stretching Graphs Horizontally Practice

First, we note that $g(x)=f\left( {\dfrac{1}3}x\right)$. 

The expression $f( k\cdot x)$ when $| k|$ is a horizontal stretch: The $x$-value of every point on the graph of $y=f(x)$ is divided by $k$, so the points get further away from the $y$-axis.

Let's use this information to determine how the graph of $g$ should look.

The graph of $f$ passes through the points $(1,0)$ and
$(3,1)$.

So the graph of $g$ should pass through the following points:

• $\left(1\div {\dfrac{1}3},0\right)=(3,0)$

• $\left(3\div {\dfrac{1}3},1\right)=(9,1)$

So the correct answer is A.

Notice how the graph of $g$ looks as if we took the graph of $f$ and stretched it away from the $y$-axis.

We are given that $g(x)=f\left( {\dfrac{1}3}x \right)$.

The expression $f( k\cdot x)$ when $| k|$ is a horizontal stretch: The $x$-value of every point on the graph of $y=f(x)$ is divided by $k$, so the points get further away from the $y$-axis.

Let's use this information to determine how the graph of $g$ should look.

The graph of $f$ passes through the points $(-3,3)$, $(0,0)$, and $(3,3)$.

So the graph of $g$ should pass through the following points:

• $\left(-3\div {\dfrac{1}3},3\right)=(-9,3)$
• $\left(0\div {\dfrac{1}3},0\right)=(0,0)$
• $\left(3\div {\dfrac{1}3},3\right)=(9,3)$

So the correct answer is B.

Notice how the graph of $g$ looks as if we took the graph of $f$ and stretched it away from both sides of the $y$-axis.

The expression $f( k\cdot x)$ when $| k|>1$ is a horizontal squash (or compression): The $x$-value of every point on the graph of $y=f(x)$ is divided by $k$, so the points get closer to the $y$-axis.

The graph of $g$ is a squashed version of the graph of $f$, so $g(x)=f( k\cdot x)$ for some value of $k$. Let's find that value and then the expression for $g$.

The graph of $f$ passes through the point $( {-4},0)$ and the graph of $g$ passes through the point $( {-2},0)$, so ${-4}\div k= {-2}$.

We found that ${k=2}$. Now let's find the equation of $g$.

$\begin{aligned}g(x)&=f\left( {2} x \right)\\\\&=\left|( {2} x)+2\right|-2\\\\&=|2x+2|-2\end{aligned}$

In conclusion, $g(x)=|2x+2|-2$.

The expression $f( k\cdot x)$ when $| k|$ is a horizontal stretch: The $x$-value of every point on the graph of $y=f(x)$ is divided by $k$, so the points get further away from the $y$-axis.

The graph of $g$ is a stretched version of the graph of $f$, so $g(x)=f( k\cdot x)$ for some value of $k$. Let's find that value.

The graph of $f$ passes through the point $( 4,5)$ and the graph of $g$ passes through the point $( 8,5)$, so $4\div k= 8$.

We found that ${k=\dfrac{1}2}$.

In conclusion, $g(x)=f\left(\dfrac 12 x\right)$.