Graphing Exponential Functions Practice

1. Use the graph below to sketch a graph of
    
   $y=2\cdot \left(\dfrac{1}3\right)^{x+2}-1$.

   
   

2. The graph of $y=3^x$ is shown below.

   
   

   Which of the following is the graph of $y=-2\cdot 3^{5-x}$?

   Choose 1 answer:
   

   1. 
   2. 
   3. 

   4. 

3. Use the graph below to sketch a graph of
    
   $y=5\cdot \left(\dfrac{1}2\right)^{-x-1}-2$.

   
   

4. The graph of $y=3^x$ is shown below.

   
   

   Which of the following is the graph of $y=-1\cdot 2^{x+3}+4$?

   Choose 1 answer:
   

   1. 
   2. 
   3. 
   4. 

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

1. The strategy

   We need to determine the position of the asymptote, and find two points on the graph.

   Determining the horizontal asymptote

   Let's think about the end behavior of $y=2\cdot \left(\dfrac{1}3\right)^{x+2}-1$.

   Notice that for this function, as $x\rightarrow +\infty$, $y\rightarrow -1$, and as $x\rightarrow -\infty$, $y\rightarrow \infty$. The former statement indicates that the horizontal asymptote is $y=-1$.

   In fact, the horizontal asymptote of the graph of any exponential function of the form $y=a\cdot b^{x-c}+d$ is $y=d$. 

   Finding two points on the graph

   Now let's find two points on the graph of $y=2\cdot \left(\dfrac{1}3\right)^{x+2}-1$ by substituting in values for $x$ and solving for $y$.

   The graph passes through $(-2,1)$ and $(-3 ,5)$. 

   In conclusion, the graph has a horizontal asymptote at $y=-1$, and it passes through $(-2,1)$ and $(-3,5)$. Therefore, the correct graph looks as follows:

   
   

   ---

2. The graph of $y=3^x$ can be transformed to get the graph of $y=-2\cdot 3^{5-x}$. 

   Let's write $y=-2\cdot 3^{5-x}$ as $y=-2\cdot 3^{-(x-5)}$ so that it is easier to see the transformations from the graph of $y=3^x$.

   Note it is important to follow order of operations when building the function, as function transformations are not always commutative. 

   Replacing $x$ with $x-5$ shifts the graph of $y=3^x$ to the right $5$ units.

   
   

   Multiplying $x-5$ by $-1$ reflects the graph of $y=3^{x-5}$ over the line $x=5$.

   
   

   Multiplying $3^{5-x}$ by $-2$ reflects the graph of
   $y=3^{5-x}$ across the $x$-axis and stretches it vertically by a factor of $2$.

   
   

   Note that the graph of the function passes through $(4,3)$ before the transformation, and $(4, -6 )$ after the transformation. This is because each $y$-coordinate is multiplied by $-2$.

   The correct graph is graph D:

   

   ---

3. The strategy

   We need to determine the position of the asymptote, and find two points on the graph.

   Determining the horizontal asymptote

   Let's think about the end behavior of $y=5\cdot \left(\dfrac{1}2\right)^{-x-1}-2$.

   Notice that for this function, as $x\rightarrow +\infty$, $y\rightarrow +\infty$, and as $x\rightarrow -\infty$, $y\rightarrow -2$. The latter statement indicates that the horizontal asymptote is $y=-2$. 

   In fact, the horizontal asymptote of the graph of any exponential function of the form $y=a\cdot b^{x-c}+d$ is $y=d$. 

   Finding two points on the graph

   Now let's find two points on the graph of $y=5\cdot \left(\dfrac{1}2\right)^{-x-1}-2$ by substituting in values for $x$ and solving for$y$.

   The graph passes through $(-1,3)$ and $(0 ,8)$. 

   In conclusion, the graph has a horizontal asymptote at $y=-2$, and it passes through $(-1,3)$ and $(0,8)$. Therefore, the correct graph looks as follows:

   
   

   ---

4. The graph of $y=2^x$ can be transformed to get the graph of $y=-1\cdot 2^{x+3}+4$. 

   Note it is important to follow order of operations when building the function, as function transformations are not always commutative. 

   Replacing $x$ with $x+3$ shifts the graph of $y=2^x$ to the left $3$ units.

   
   

   Multiplying $2^{x+3}$ by $-1$ reflects the graph of
   $y=2^{x+3}$ across the $x$-axis.

   
   

   Adding $4$ to the function shifts the graph of $y=-1\cdot 2^{x+3}$ up $4$ units.

   
   

   Since the graph of $y=-1\cdot 2^{x+3}$ had a horizontal asymptote of $y=0$, the graph of $y=-1\cdot 2^{x+3}+4$ has a horizontal asymptote of $y=4$.

   The $x$-intercept of the above graph is $(-1,0)$ and the $y$-intercept is $(0,-4)$.

   We can verify this algebraically to check our work. 

   The correct graph is graph C: