Chapter 6.3, Problem 14E

To determine

To find: the nature of the graph formed by the points, $\left(x,f(x)\right)=\left\{\left(0,6\right),\left(1,6\right),(2,1.5),(3,0.75),(-1,12)\right\}$ . And verify whether it represent linear or exponential function.

Answer to Problem 14E

Given graph not alinearnora exponential function.

  

Explanation of Solution

Given information :

Given points $\left(x,f(x)\right)=\left\{\left(0,6\right),\left(1,6\right),(2,1.5),(3,0.75),(-1,12)\right\}$ .

Calculation:

Given points of the graph are, $\left(x,f(x)\right)=\left\{\left(0,6\right),\left(1,6\right),(2,1.5),(3,0.75),(-1,12)\right\}$ .

Therefore the difference is,

  $3-6=-3,1.5-3=-1.5,0.75-1.5=0.75,12-0.75=11.25$

So there is no constant difference between terms. It is not representing a linear graph.

Further the common ratio is,

  $\frac{3}{6}=\frac{1}{2},\frac{1.5}{3}=\frac{1}{2},\frac{0.75}{1.5}=\frac{1}{2},\frac{12}{0.75}=16$

That is common ratio of DE is different. So if the slope of DE is equal as any one slope of AB, BC, CD then it will be a exponential function.

  $\begin{array}{l}\text{Slope of B}C=\frac{1.5-6}{2-1}=\frac{-4.5}{1}=-4.5\\ \text{Slope of }DE=\frac{12-0.75}{-1-3}=2.8125\end{array}$

Hence given graph is not exponential.

Graph: