Compressing and Stretching Graphs Horizontally Practice

1. This is the graph of function $f(x)=\log_3(x)$:

   
   

   What is the graph of $g(x)=\log_3\left( \dfrac{x}{3} \right)$?

   Choose 1 answer:
   

   1. 
   2. 
   3. 

   4. 

2. This is the graph of function $f$:

   
   

   Function $g$ is defined as $g(x)=f\left( \dfrac{1}3x \right)$.
   

   What is the graph of $g$?
   

   Choose 1 answer:
   

   1. 
   2. 
   3. 

   4. 

3. $f(x)=|x+2|-2$ and $g$ is a horizontally scaled version of $f$. The functions are graphed where $f$ is solid and $g$ is dashed.

   
   

   
   

   
   

   Choose 1 answer:
   

   1. $g(x)=|4x+2|-2$
      
   2. $g(x)=\left|\dfrac{1}4 x+2\right|-2$
      
   3. $g(x)=|2x+2|-2$
      

   4. $g(x)=\left|\dfrac{1}2 x+2\right|-2$

4. Function $g$ is a horizontally scaled version of function $f$. The functions are graphed where $f$ is solid and $g$ is dashed.

   
   

   What is the equation of $g$ in terms of $f$?

   Choose 1 answer:
   

   1. $g(x)=f(2x)$
      
   2. $g(x)=f\left( \dfrac{1}2 x \right)$
      
   3. $g(x)=f(4x)$
      
   4. $g(x)=f\left( \dfrac{1}4 x \right)$

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:transformations/x2ec2f6f830c9fb89:scale/e/scale-functions-horizontally
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1. 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.

   ---

2. 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.

   ---

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

   ---

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