Question

graph both of the following functions.
$f(x) = -dfrac{1}{5}x + 5$
$g(x) = dfrac{6}{5}x - 2$

Explanation:

Step1: Analyze \( f(x) = -\frac{1}{5}x + 5 \)

The function is in slope - intercept form \( y=mx + b \), where \( m = -\frac{1}{5} \) (slope) and \( b = 5 \) (y - intercept). To graph it, start by plotting the y - intercept at \( (0,5) \). Then, use the slope: from \( (0,5) \), move down 1 unit and right 5 units (or up 1 unit and left 5 units) to get another point. For example, when \( x = 5 \), \( f(5)=-\frac{1}{5}(5)+5=- 1 + 5 = 4 \), so the point \( (5,4) \) is on the line. Also, when \( x = 0 \), \( y = 5 \); when \( x=-5 \), \( f(-5)=-\frac{1}{5}(-5)+5 = 1 + 5=6 \), so \( (-5,6) \) is on the line.

Step2: Analyze \( g(x)=\frac{6}{5}x - 2 \)

This function is also in slope - intercept form with \( m=\frac{6}{5} \) (slope) and \( b=-2 \) (y - intercept). Plot the y - intercept at \( (0,-2) \). Then, use the slope: from \( (0,-2) \), move up 6 units and right 5 units (or down 6 units and left 5 units). For example, when \( x = 5 \), \( g(5)=\frac{6}{5}(5)-2=6 - 2 = 4 \), so the point \( (5,4) \) is on the line. When \( x = 0 \), \( y=-2 \); when \( x = 5 \), \( y = 4 \); when \( x=-5 \), \( g(-5)=\frac{6}{5}(-5)-2=-6 - 2=-8 \), so \( (-5,-8) \) is on the line.

Step3: Match with the given graph

Looking at the graph, the line with the y - intercept at \( (0,5) \) (or near \( y = 5 \)) and a shallow negative slope should be \( f(x) \), and the line with y - intercept at \( (0,-2) \) and a steeper positive slope should be \( g(x) \). The two lines intersect at some point (we can also find the intersection by setting \( f(x)=g(x) \): \( -\frac{1}{5}x + 5=\frac{6}{5}x-2 \), \( 5 + 2=\frac{6}{5}x+\frac{1}{5}x \), \( 7=\frac{7}{5}x \), \( x = 5 \), then \( y = 4 \), so the intersection point is \( (5,4) \), which matches the points on the graph).

To graph \( f(x) \):

• Plot the y - intercept \( (0,5) \).
• Use the slope \( -\frac{1}{5} \) to find additional points. For example, from \( (0,5) \), moving right 5 units and down 1 unit gives \( (5,4) \), moving left 5 units and up 1 unit gives \( (-5,6) \).

To graph \( g(x) \):

• Plot the y - intercept \( (0,-2) \).
• Use the slope \( \frac{6}{5} \) to find additional points. For example, from \( (0,-2) \), moving right 5 units and up 6 units gives \( (5,4) \), moving left 5 units and down 6 units gives \( (-5,-8) \).

The graph provided has two lines: one with a y - intercept near \( y = 5 \) (shallow negative slope) for \( f(x) \) and one with a y - intercept at \( y=-2 \) (steeper positive slope) for \( g(x) \), and they intersect at \( (5,4) \), which is consistent with our analysis.

Answer:

To graph \( f(x)=-\frac{1}{5}x + 5 \):

1. Plot the y - intercept at \( (0,5) \).
2. Use the slope \( -\frac{1}{5} \) to draw the line (e.g., from \( (0,5) \), move right 5, down 1 to \( (5,4) \)).

To graph \( g(x)=\frac{6}{5}x - 2 \):

1. Plot the y - intercept at \( (0,-2) \).
2. Use the slope \( \frac{6}{5} \) to draw the line (e.g., from \( (0,-2) \), move right 5, up 6 to \( (5,4) \)).

The given graph correctly represents these two lines with \( f(x) \) being the less - steep line (with a y - intercept above the x - axis) and \( g(x) \) being the steeper line (with a y - intercept below the x - axis) intersecting at \( (5,4) \).