Question

Figuring out End Behavior, Decreasing and Increasing Interval, and asymptote

Answer 1

- The function given is:

`f(x)=2^(x+1)-3`

- The best way to understand the solution is to plot the function.

- The plot of the function is given below:

- From the above, we can already make the following deductions depicted below:

- Thus, we can answer the questions asked as follows:

Growth or Decay:

- This is clearly a GROWTH function as it increases from left to right.

- The reason for this increase is the term

`2^(x+1)`

- 2 is the rate of change and since it is greater than 1, the function is increasing. If it were less than 1, it would be decay.

Domain:

- The domain is the set of all x-values for which the function is defined. The function has no breaks from left to right, thus, we can conclude that the domain is

`Domain:(-\infty,\infty)`

Range:

- The range is the set of all y-values for which the function is defined. The function has a minimum value of -3, which is also the Horizontal asymptote of the function. This is because,

`\begin{gathered} \text{ For any exponential function} \\ f(x)=ab^x+c \\ c\text{ is the horizontal asymptote of the function and the minimum value of the function} \\ \\ \text{ Also, if we substitute }x=-\infty\text{ into the function give,} \\ f(x)=2^(-\infty+1)-3 \\ f(x)=(1)/(2^(\infty))-3=0-3=-3 \\ \\ \text{ Thus, when we move further and further towards the left of the graph, the value of} \\ \text{ f\lparen x\rparen becomes -3, just as we established earlier} \end{gathered}`

Y-intercept:

- The y-intercept is simply where the graph crosses the y-axis. This also happens to be where x = 0 in the graph as well.

- From the figure above, we have that the y-intercept is (0, -1)

- But we can also get it from the equation by substituting the value of x = 0 into the function

`\begin{gathered} f(x)=2^(x+1)-3 \\ \text{ put }x=0 \\ f(x)=2^(0+1)-3 \\ f(x)=2-3 \\ f(x)=-1 \\ \text{ Thus, when }x=0,f(x)=-1 \\ \\ \text{ Thus, the y-intercept is }(0,-1) \end{gathered}`

Increasing Interval and Decreasing Interval:

- The interval for which the function is increasing is done by testing values of x and checking out the trend of the function. Usually, we test a negative value, x = 0, a positive value, then, negative and positive infinity.

- But since we have the graph plotted above, we can easily see that the graph is ALWAYS INCREASING over the interval

`x\in(-\infty,\infty)`

End behavior:

- The end behavior of the graph is simply the values of y as x tends to negative and positive infinity.

- Thus, we have:

`\begin{gathered} f(x)=2^(x+1)-3 \\ \text{ Put }x=-\infty, \\ f(-\infty)=2^(-\infty+1)-3 \\ f(-\infty)=-3 \\ \\ \text{ Put }x=\infty \\ f(\infty)=2^(\infty+1)-3 \\ f(\infty)=\infty-3 \\ f(\infty)=\infty \\ \\ \text{ Thus, the End behavior of the function is:} \\ As\text{ }x\to\infty,f(x)\to\infty \\ As\text{ }x\to-\infty,f(x)\to-3 \end{gathered}`