graph linear functions example 1 graph the equation. compare the graph with the graph of y = x. a.a....

Graph linear functions

EXAMPLE 1

Graph the equation. Compare the graph with thegraph of y = x.a. y = 2x b. y = x + 3

SOLUTION

a.

The graphs of y = 2x and y = xboth have a y-intercept of 0, but the graph of y = 2x has a slope of 2 instead of 1.

Graph linear functionsEXAMPLE 1

b.

The graphs of y = x + 3 and y = x both have a slope of 1, but the graph of y = x + 3 has a y-intercept of 3 instead of 0.

Graph an equation in slope-intercept formEXAMPLE 2

Graph y = – x – 1.23

SOLUTION

The equation is already in slope-intercept form.

STEP 1

Identify the y-intercept. The y-intercept is – 1, so plot the point (0, – 1) where the line crosses the y-axis.

STEP 2

Graph an equation in slope-intercept formEXAMPLE 2

STEP 3

Identify the slope. The slope is – , or , so plot

a second point on the line by starting at (0, – 1) and then moving down 2 units and right 3 units. The second point is (3, – 3).

– 23

23

Graph an equation in slope-intercept form

EXAMPLE 2

Draw a line through the two points.

STEP 4

SOLUTION

GUIDED PRACTICE for Examples 1 and 2

1. y = –2x

The graphs of y = –2x and y = xboth have a y-intercept of 0, but the graph of y = –2x has a slope of –2 instead of 1.

Graph the equation. Compare the graph with the graph of y = x.

SOLUTION


2. y = x – 2

Graph the equation. Compare the graph with the graph of y = x.

The graphs of y = x – 2 and y = x both have a slope of 1, but the graph of y = x – 2 has a y-intercept of –2 instead of 0.

SOLUTION


Graph the equation. Compare the graph with the graph of y = x.3. y = 4x

The graphs of y = 4x and y = xboth have a y-intercept of 0, but the graph of y = 4x has a slope of 4 instead of 1.


SOLUTION

The equation is already in slope-intercept form.

STEP 1

Identify the y-intercept. The y-intercept is +2, so plot the point (0, +2) where the line crosses the y-axis.

STEP 2

Graph the equation4. y = – x + 2


STEP 3Identify the slope. The slope is –1 so plot a second point on the line by starting at (0, 2) and then moving down 1 unit and right 1 unit. The second point is (1, 1).

Draw a line through the two points.STEP 4


SOLUTION

The equation is already in slope-intercept form.STEP 1


STEP 2

Graph the equation5. y = x + 4 2

5


STEP 3


Identify the slope. The slope is so plot a second point on the line by starting at (0, 4) and then moving up 2 unit and right 5 unit. The second point is (5, 6).

25


SOLUTION


Identify the y-intercept. The y-intercept is –3, so plot the point (0, –3) where the line crosses the y-axis.

STEP 2

Graph the equation6. y = x – 3 1

2


STEP 3


Identify the slope. The slope is so plot a second point on the line by starting at (0, –3) and then moving up 1 unit and right 2 unit. The second point is (–2 , 2).

12


SOLUTION



STEP 2

Graph the equation7. y = 5 + x


STEP 3Identify the slope. The slope is 1 so plot a second point on the line by starting at (0, 5) and then moving up 1 unit and right 1 unit. The second point is (1, 6).



SOLUTION



STEP 2

Graph the equation8. f (x) = 1 – 3x


SOLUTION



STEP 2

Graph the equation9. f (x) = 10 – x

graph linear functions example 1 graph the equation. compare the graph with the graph of y = x. a.a....

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