Chapter 5.3, Problem 48AYU

To determine

The graph of function $f(x)=-3^x+1$ using transformation, domain, range, horizontal asymptote, $y$ - intercept of function.

Answer to Problem 48AYU

Solution:

The graph of $f(x)=-3^x+1$ is

  

Domain of the function $f(x)=-3^x+1$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=-3^x+1$ is $\left(-\infty ,1\right)$

Horizontal asymptote of $f(x)=-3^x+1$ is $y=1$

  $y$ - intercept of $f(x)=-3^x+1$ is $0$

Explanation of Solution

Given information:

The function $f(x)=-3^x+1$

To graph a function, begin with a graph of $f(x)=3^x$

Find some points satisfying $f(x)=3^x$

  $\begin{array}{cccccc}x& -1& 0& 1& 2& -2\\ f\left(x\right)& \begin{array}{l}f\left(-1\right)=3^{-1}\\ =\frac{1}{3}\end{array}& \begin{array}{l}f\left(0\right)=3^0\\ =1\end{array}& \begin{array}{l}f\left(1\right)=3^1\\ =3\end{array}& \begin{array}{l}f\left(2\right)=3^2\\ =9\end{array}& \begin{array}{l}f\left(-2\right)=3^{-2}\\ =\frac{1}{9}\end{array}\\ \left(x,f\left(x\right)\right)& \left(-1,\frac{1}{3}\right)& \left(0,1\right)& \left(1,3\right)& \left(2,9\right)& \left(-2,\frac{1}{9}\right)\end{array}$

Plot these points on the graph.

Therefore, the graph is

  

To graph a function $f(x)=-3^x$ , reflect the graph of $f(x)=3^x$ about $x$ axis.

The graph of $f(x)=-3^x$ is

  

To graph the function $f(x)=-3^x+1$ , shift the graph of $f(x)=-3^x$ by $1$ unit upward.

The graph of $f(x)=-3^x+1$ is

  

From the graph, domain of the function $f(x)=-3^x+1$ is $\left(-\infty ,\infty \right)$

Range of the function is the set of values of the dependent variable for which a function is defined.

Therefore, range of $f(x)=-3^x+1$ is $\left(-\infty ,1\right)$

As $x\to -\infty$ , $f(x)$ approaches to $1$

Therefore, $f(x)=-3^x+1$ has horizontal asymptote $y=1$

To find $y$ - intercept, substitute $x=0$ in $f(x)=-3^x+1$

  $\Rightarrow f(0)=-3^0+1=-1+1=0$

Therefore, $y$ - intercept is $0$