Chapter 2 Section 3. Graph linear functions EXAMPLE 1 Graph the equation. Compare the graph with the graph of y = x. a.a. y = 2x b.b. y = x + 3 SOLUTION.

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Presentation transcript:

Chapter 2 Section 3

Graph linear functions EXAMPLE 1 Graph the equation. Compare the graph with the graph of y = x. a.a. y = 2x b.b. y = x + 3 SOLUTION a.a. The graphs of y = 2x and y = x both have a y- intercept of 0, but the graph of y = 2x has a slope of 2 instead of 1.

Graph linear functions EXAMPLE 1 b.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 form EXAMPLE 2 Graph y = – x – 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 form EXAMPLE 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). –

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 = x both 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 GUIDED PRACTICE for Examples 1 and 2 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 GUIDED PRACTICE for Examples 1 and 2 Graph the equation. Compare the graph with the graph of y = x. 3. y = 4x The graphs of y = 4x and y = x both have a y- intercept of 0, but the graph of y = 4x has a slope of 4 instead of 1.

GUIDED PRACTICE for Examples 1 and 2 Graph the equation 4. y = –x y = x

GUIDED PRACTICE for Examples 1 and 2 Graph the equation 6. y = x – y = 5 + x

GUIDED PRACTICE for Examples 1 and 2 Graph the equation 8. f (x) = 1 – 3x9. f (x) = 10 – x

Biology Solve a multi-step problem EXAMPLE 3 Graph the equation. Describe what the slope and y-intercept represent in this situation. Use the graph to estimate the body length of a calf that is 10 months old. The body length y (in inches) of a walrus calf can be modeled by y = 5x + 42 where x is the calf’s age (in months).

SOLUTION Solve a multi-step problem EXAMPLE 3 STEP 1 Graph the equation. STEP 2 Interpret the slope and y- intercept. The slope, 5,represents the calf’s rate of growth in inches per month. The y-intercept, 42, represents a newborn calf’s body length in inches.

Solve a multi-step problem EXAMPLE 3 Estimate the body length of the calf at age 10 months by starting at 10 on the x-axis and moving up until you reach the graph. Then move left to the y-axis.At age 10 months, the body length of the calf is about 92 inches. STEP 3

GUIDED PRACTICE for Example 3 WHAT IF? In Example 3, suppose that the body length of a fast-growing calf is modeled by y = 6x Repeat the steps of the example for the new model.

STEP 2 SOLUTION STEP 1 Graph the equation. GUIDED PRACTICE for Example 3 The y-intercept, 48, represents the length of the newborn calf’s body. The slope, 6, represents the calf’s growth rate in inches per month. At age 10 months, the body length of the calf is about 108 inches. STEP 3

Graph an equation in standard form EXAMPLE 4 Graph 5x + 2y = 10. STEP 1 The equation is already in standard form. Identify the x-intercept. STEP 2 5x + 2(0) = 10 x = 2 Let y = 0. Solve for x. SOLUTION The x-intercept is 2. So, plot the point (2, 0).

Graph an equation in standard form EXAMPLE 4 Identify the y-intercept. STEP 3 5(0) + 2y = 10 y = 5 Let y = 0. Solve for y. The y-intercept is 5. So, plot the point (0, 5). Draw a line through the two points. STEP 4

Graph horizontal and vertical lines EXAMPLE 5 Graph (a) y = 2 and (b) x = –3. a. The graph of y = 2 is the horizontal line that passes through the point (0, 2). Notice that every point on the line has a y-coordinate of 2. SOLUTION b. The graph of x = –3 is the vertical line that passes through the point (–3, 0). Notice that every point on the line has an x-coordinate of –3.

GUIDED PRACTICE for Examples 4 and 5 Graph the equation. 11.2x + 5y = x – 2y = 12

GUIDED PRACTICE for Examples 4 and 5 Graph the equation. 13. x = 114. y = –4