Compressing and Stretching Graphs Practice

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

   
   

   What is the equation of ‍$g$?

   Choose 1 answer:
   

   1. $g(x)=\dfrac{2}5x^3+x^2$
   2. $g(x)=\dfrac{2}5x^3+\dfrac{2}5x^2$
   3. $g(x)=\dfrac{5}2x^3+x^2$

   4. $g(x)=\dfrac{5}2x^3+\dfrac{5}2x^2$

2. This is the graph of function $f(x)=|x+3|-3$:

   
   

   What is the graph of $g(x)=3|x+3|-9$?

   Choose 1 answer:

   1. 
      
   2. 
      
   3. 
      

   4. 

3. Function $g$ is a vertically 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. $g(x)=\dfrac{1}2f(x)$
   2. $g(x)=\dfrac{1}4f(x)$
   3. $g(x)=2f(x)$

   4. $g(x)=4f(x)$

4. This is the graph of function $f(x)=\log_2(x)$:

   
   

   What is the graph of $g(x)=\dfrac{1}2\log_2(x)$?

   Choose 1 answer:

   1. 
      
   2. 
      
   3. 
      
   4. 
      

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:transformations/x2ec2f6f830c9fb89:scale/e/scale-functions-vertically
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. The expression $k\cdot f(x)$ when $| k| >1$ is a vertical stretch: The $y$-value of every point on the graph of $y=f(x)$ is multiplied by $k$, so the points get farther away from the $x$-axis.

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

   The graph of $f$ passes through the point $(1, 2)$ and the graph of $g$ passes through the point $(1, 5)$, so $2\cdot k= 5$.

   
   

   We found that ${k=\dfrac{5}2}$. Now let's find the equation of
   $g$.

   $\begin{aligned}g(x)&= {\dfrac{5}2}f(x)\\\\&= {\dfrac{5}2}\left( x^3+x^2 \right)\\\\&=\dfrac{5}2x^3+\dfrac{5}2x^2\end{aligned}$
   

   In conclusion, $g(x)=\dfrac{5}2x^3+\dfrac{5}2x^2$.
   

   ---

2. First, we note that $g(x)= {3}f(x)$.
   
   The expression $k\cdot f(x)$ when $| k| >1$ is a vertical stretch: The $y$-value of every point on the graph of $y=f(x)$ is multiplied by $k$, so the points get farther away from the $x$-axis.

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

   The graph of $f$ passes through the points $(-6,0)$,$(-3,-3)$, and $(0,0)$.

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

   • $(-6,0\cdot {3})=(-6,0)$
   • $(-3,-3\cdot {3})=(-3,-9)$
   • $(0,0\cdot {3})=(0,0)$

   So the correct answer is B.

   
   

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

   ---

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

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

   The graph of $f$ passes through the point $(4, {16})$ and the graph of $g$ passes through the point $(4, 4)$, so ${16}\cdot k= 4$.

   
   

   We found that ${k=\dfrac{1}4}$.

   In conclusion, $g(x)=\dfrac{1}4f(x)$.

   ---

4. First, we note that $g(x)= {\dfrac{1}2}f(x)$. 

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

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

   The graph of $f$ passes through the points $(1,0)$, $(2,1)$, and $(4,2)$.

   
   

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

   • $\left(1,0\cdot {\dfrac{1}2}\right)=(1,0)$
   • $\left(2,1\cdot {\dfrac{1}2}\right)=\left(2,\dfrac{1}2\right)$
   • $\left(4,2\cdot {\dfrac{1}2}\right)=(4,1)$

   So the correct answer is $A$.

   
   

   Notice how the graph of $g$ looks as if we took the graph of $f$ and squashed it towards both sides of the $x$-axis.