Q. 129 - Precalculus Enhanced with Graphing Utilities

Question

Which of the following equations might have the graph shown? (More than one answer is possible.)

$$\begin{gathered} \left(a\right) 2x + 3y = 6 \\ \left(b\right) -2x + 3y = 6 \\ \left(c\right) 3x - 4y = -12 \\ \left(d\right) x - y = 1 \\ \left(e\right) x - y = -1 \\ \left(f\right) y = 3x - 5 \\ \left(g\right) y = 2x + 3 \\ \left(h\right) y = -3x + 3 \end{gathered}$$

Step-by-Step Solution

Answer

The equation which describes the given line is

$$\begin{gathered} \left(b\right) -2x + 3y = 6 \\ \left(c\right) 3x - 4y = -12 \\ \left(e\right) x - y = -1 \\ \left(g\right) y = 2x + 3 \end{gathered}$$

Step 1: Given information

We are given graph and several equations

Step 2: Explanation for correct options

Consider the option

$$\begin{gathered} b)-2x+3y=6 \\ y=\frac{2}{3}x+2 \end{gathered}$$

and 

$$\begin{gathered} c)3x-4y=-12 \\ 4y=3x+12 \\ y=\frac{3}{4}x+3 \end{gathered}$$

also

$$\begin{gathered} e)x-y=-1 \\ y=x+1 \end{gathered}$$

Similarly

$g)y=2x+3$

Each and every equation has positive y-intercept similarly from the graph we can say that the line has positive y-intercept

Step 3: Consider the equations

$$\begin{gathered} \left(a\right) 2x + 3y = 6 \\ y=-\frac{2}{3}+2 \\ \left(d\right) x - y = 1 \\ y=x-1 \\ \left(f\right) y = 3x - 5 \\ \left(h\right) y = -3x + 3 \end{gathered}$$

In the equation a) and h) the slopes are negative where as in option d) and f) the y-intercept is negative hence this does not match the given graph of lines

Step 4: Conclusion

The equations that can have the given graph as 

$$\begin{gathered} \left(b\right) -2x + 3y = 6 \\ \left(c\right) 3x - 4y = -12 \\ \left(e\right) x - y = -1 \\ \left(g\right) y = 2x + 3 \end{gathered}$$