Chapter SH, Problem 9.9.13EP

To determine

The graph of $y={\left(\frac{1}{2}\right)}^{x+1}$ is to be drawn.

Explanation of Solution

Given:

  $y={\left(\frac{1}{2}\right)}^{x+1}$

Graph:

  

Interpretation:

The given equation is:

  $y={\left(\frac{1}{2}\right)}^{x+1}$

At $x=2$

  $\begin{array}{l}y={\left(\frac{1}{2}\right)}^{x+1}\\ y={\left(\frac{1}{2}\right)}^{2+1}\\ y=0.125\end{array}$

At $x=1$

  $\begin{array}{l}y={\left(\frac{1}{2}\right)}^{x+1}\\ y={\left(\frac{1}{2}\right)}^{1+1}\\ y=0.25\end{array}$

At $x=0$

  $\begin{array}{l}y={\left(\frac{1}{2}\right)}^{x+1}\\ y={\left(\frac{1}{2}\right)}^{0+1}\\ y=0.5\end{array}$

At $x=-2$

  $\begin{array}{l}y={\left(\frac{1}{2}\right)}^{x+1}\\ y={\left(\frac{1}{2}\right)}^{-2+1}\\ y=2\end{array}$

At $x=-1$

  $\begin{array}{l}y={\left(\frac{1}{2}\right)}^{x+1}\\ y={\left(\frac{2}{3}\right)}^{-1+1}\\ y=1\end{array}$

So, the co-ordinates to draw eq. $y={\left(\frac{1}{2}\right)}^{x+1}$ are

  $\left(2,0.125\right),\left(1,0.25\right),\left(0,0.5\right),\left(-2,2\right),\left(-1,1\right)$