Question

1. $g(x)=\frac{-x - 5}{3}$

Explanation:

Step1: Rewrite the function

Rewrite \( g(x)=\frac{-x - 5}{3} \) as \( g(x)=-\frac{1}{3}x-\frac{5}{3} \). This is in the slope - intercept form \( y = mx + b \), where the slope \( m=-\frac{1}{3} \) and the y - intercept \( b =-\frac{5}{3}\approx - 1.67 \).

Step2: Find two points on the line

• Using the y - intercept: When \( x = 0 \), \( y=-\frac{5}{3}\approx - 1.67 \). So one point is \( (0,-\frac{5}{3}) \).
• Using the slope: The slope \( m =-\frac{1}{3}=\frac{\text{change in }y}{\text{change in }x} \). Starting from the point \( (0,-\frac{5}{3}) \), if we increase \( x \) by 3 (the denominator of the slope), the change in \( y \) is - 1 (the numerator of the slope). So when \( x=0 + 3=3 \), \( y=-\frac{5}{3}-1=-\frac{5 + 3}{3}=-\frac{8}{3}\approx - 2.67 \). So another point is \( (3,-\frac{8}{3}) \).

Step3: Plot the points and draw the line

Plot the points \( (0,-\frac{5}{3}) \) and \( (3,-\frac{8}{3}) \) on the coordinate plane. Then draw a straight line passing through these two points. The line should have a negative slope (going down from left to right) with a relatively gentle slope (since the absolute value of the slope \( \frac{1}{3}<1 \)) and cross the y - axis below the origin.

(If we want to check with another point, for example, when \( y = 0 \), we solve \( 0=-\frac{1}{3}x-\frac{5}{3} \). Multiply both sides by 3: \( 0=-x - 5 \), then \( x=-5 \). So the x - intercept is at \( (-5,0) \). Plotting this point \( (-5,0) \) also lies on the line we draw.)

Answer:

To graph \( g(x)=\frac{-x - 5}{3} \) (or \( y =-\frac{1}{3}x-\frac{5}{3} \)):

1. Identify the slope \( m =-\frac{1}{3} \) and y - intercept \( b=-\frac{5}{3} \).
2. Plot the y - intercept \( (0,-\frac{5}{3}) \).
3. Use the slope to find another point (e.g., from \( (0,-\frac{5}{3}) \), move 3 units right and 1 unit down to get \( (3,-\frac{8}{3}) \) or use the x - intercept \( (-5,0) \)).
4. Draw a straight line through the plotted points.