Chapter 1.2, Problem 16E

To determine

Complete the table given below for the equation $y+1=\frac{3}{4}x$ and use the resulting solution points to sketch the graph of the equation.

$x$	$-2$	$0$	$1$	$\frac{4}{3}$	$2$
$y$
$\left(x,y\right)$

Answer to Problem 16E

The table is

$x$	$-2$	$0$	$1$	$\frac{4}{3}$	$2$
$y$	$-\frac{5}{2}$	$-1$	$-\frac{1}{4}$	$0$	$\frac{1}{2}$
$\left(x,y\right)$	$\left(-2,-\frac{5}{2}\right)$	$\left(0,-1\right)$	$\left(1,-\frac{1}{4}\right)$	$\left(\frac{4}{3},0\right)$	$\left(2,\frac{1}{2}\right)$

The graph of the equation is:

  

Explanation of Solution

Given:

The equation $y+1=\frac{3}{4}x$ and table to fill

$x$	$-1$	$0$	$1$	$2$	$\frac{5}{2}$
$y$
$\left(x,y\right)$

Concept Used:

The graph of an equation is the smooth curve of all the points satisfying the equation.

Calculation:

For the givenequation $y+1=\frac{3}{4}x$

At $x=-2$

  $\begin{array}{c}y+1=\frac{3}{4}\left(-2\right)\\ y+1=-\frac{3}{2}\\ y=-\frac{3}{2}-1\\ y=\frac{-5}{2}\end{array}$

At $x=0$

  $\begin{array}{c}y+1=\frac{3}{4}\left(0\right)\\ y+1=0\\ y=-1\end{array}$

At $x=1$

  $\begin{array}{c}y+1=\frac{3}{4}\left(1\right)\\ y+1=\frac{3}{4}\\ y=\frac{3}{4}-1\\ y=\frac{-1}{4}\end{array}$

At $x=\frac{4}{3}$

  $\begin{array}{c}y+1=\frac{3}{4}\left(\frac{4}{3}\right)\\ y+1=1\\ y=1-1\\ y=0\end{array}$

At $x=2$

  $\begin{array}{c}y+1=\frac{3}{4}\left(2\right)\\ y+1=\frac{3}{2}\\ y=\frac{3}{2}-1\\ y=\frac{1}{2}\end{array}$

The table is

$x$	$-2$	$0$	$1$	$\frac{4}{3}$	$2$
$y$	$-\frac{5}{2}$	$-1$	$-\frac{1}{4}$	$0$	$\frac{1}{2}$
$\left(x,y\right)$	$\left(-2,-\frac{5}{2}\right)$	$\left(0,-1\right)$	$\left(1,-\frac{1}{4}\right)$	$\left(\frac{4}{3},0\right)$	$\left(2,\frac{1}{2}\right)$

Joining the points and plotting the graph, we have

  

Conclusion:

The table is

$x$	$-2$	$0$	$1$	$\frac{4}{3}$	$2$
$y$	$-\frac{5}{2}$	$-1$	$-\frac{1}{4}$	$0$	$\frac{1}{2}$
$\left(x,y\right)$	$\left(-2,-\frac{5}{2}\right)$	$\left(0,-1\right)$	$\left(1,-\frac{1}{4}\right)$	$\left(\frac{4}{3},0\right)$	$\left(2,\frac{1}{2}\right)$

The graph of the equation is: