Question

(see image for problem)a. the a -term:b. the y -intercept:c. the common ratio:d. the x -intercept:e. the end behavior:

Answer 1

The general form of an exponential function is expressed as

$\begin{gathered} y=ab^x\text{ ----- equation 1} \\ \text{where} \\ a\text{ is the y-intercept of the function} \\ b\text{ is the common ration of the function} \end{gathered}$

In the function

$h(x)\text{ = }-2(\frac{1}{3})^x\text{ ------ equation 2}$

A) a-term:

In, the h(x) function, the a-term is -2.

B) the y-intercept:

The y-intercept of the function is obtained as the value of h(x), when x equals zero.

thus,

$\begin{gathered} \text{when x = 0,} \\ h(x)\text{ = -2(}\frac{1}{3})^0\text{ = -2}\times1 \\ \Rightarrow h(x)\text{ = }-2 \end{gathered}$

thus, the y-intercept is -2.

C) the common ratio

In equation 1, b is the common ratio of the exponential function. In comparison with equation 2, we have

$b\text{ = }\frac{1}{3}$

Thus, the common ratio of the function is

$\frac{1}{3}$

D) the x-intercept:

The x-intercept of the function is obtained as the value of x when h(x) equals zero.

thus,

$\begin{gathered} \text{when h(x) = 0} \\ 0\text{ = -2(}\frac{1}{3})^x \\ \Rightarrow0=-2(3^{-1})^x \\ 0=3^{-x} \\ take\text{ the log of both sides,} \\ \log \text{ 0 = log}3^{-x} \\ \infty\text{ = -xlog3} \\ \Rightarrow x=\text{ }\frac{\infty}{-\log 3} \\ x\text{ = }\infty \\ \end{gathered}$

thus, the x-intercept is at ∞ (infinity).

E) the end behaviour:

The end behavoiur of the function is the behaviour of the h(x) function as x approaches plus infinity or negative infinity.

Thus,

$\begin{gathered} \lim _{x\to-\infty}y\text{ = }\lim _{x\to-\infty}(-2(\frac{1}{3})^x)\text{ } \\ =-2\cdot\lim _{x\to\: -\infty\: }\mleft(\mleft(\frac{1}{3}\mright)^x\mright) \\ =-2\cdot\infty \\ \Rightarrow-\infty \end{gathered}$$\begin{gathered} \lim _{x\to\infty}y\text{ = }\lim _{x\to\infty}(-2(\frac{1}{3})^x)\text{ } \\ =-2\cdot\lim _{x\to\infty\: }\mleft(\mleft(\frac{1}{3}\mright)^x\mright) \\ =-2\cdot\: 0 \\ \Rightarrow0 \\ \end{gathered}$

thus, as x tends to negative infinity, h(x) tends to negative infinity. when x tends to positive infinity, h(x) tends to zero.

Sketch of the h(x) graph on a coordinate plane:

The sketch of the h(x) function is as shown below: