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§ 4.2. Graphing Linear Equations Using Intercepts. Linear Functions. All equations of the form Ax + By = C are straight lines when graphed, as long as A and B are not both zero. Such an equation is called the standard form of the equation of the line .

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4 2

§4.2

Graphing Linear Equations Using Intercepts


4 2

Linear Functions

All equations of the form Ax + By = C are straight lines when graphed, as long as A and B are not both zero.

Such an equation is called the standard form of the equation of the line.

To graph equations of this form, two very important points are used – the intercepts.

Blitzer, Introductory Algebra, 5e – Slide #2 Section 4.2


4 2

x - intercept

An x-intercept of a graph is the x-coordinate of a point where the graph intersects the x-axis. The y-coordinate of the x-intercept is always zero.

The graph of y = 4x – 8 crosses the x-axis at (2, 0) and that point is the x-intercept.

(2, 0)

Blitzer, Introductory Algebra, 5e – Slide #3 Section 4.2


4 2

y- intercept

The y-intercept of a graph is the y-coordinate of a point where the graph intersects the y-axis. The x-coordinate of the y-intercept is always zero.

The graph of y = 3x + 4 crosses the y-axis at (0, 4) and that point is the y-intercept.

(0, 4)

Blitzer, Introductory Algebra, 5e – Slide #4 Section 4.2


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x- and y-Intercepts

Blitzer, Introductory Algebra, 5e – Slide #5 Section 4.2


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Graphing Using Intercepts

EXAMPLE

Graph the equation using intercepts.

SOLUTION

1) Find the x-intercept. Let y = 0 and then solve for x.

Replace y with 0

Multiply and simplify

Divide by -2

The x-intercept is -6, so the line passes through (-6,0).

Blitzer, Introductory Algebra, 5e – Slide #6 Section 4.2


4 2

Graphing Using Intercepts

CONTINUED

2) Find the y-intercept. Let x = 0 and then solve for y.

Replace x with 0

Multiply and simplify

Divide by 4

The y-intercept is 3, so the line passes through (0,3).

Blitzer, Introductory Algebra, 5e – Slide #7 Section 4.2


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Graphing Using Intercepts

CONTINUED

3) Find a checkpoint, a third ordered-pair solution. For our checkpoint, we will let x = 1 (because x = 1 is not the x-intercept) and find the corresponding value for y.

Replace x with 1

Multiply

Add 2 to both sides

Divide by 4 and simplify

The checkpoint is the ordered pair (1,3.5).

Blitzer, Introductory Algebra, 5e – Slide #8 Section 4.2


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Graphing Using Intercepts

CONTINUED

4) Graph the equation by drawing a line through the three points. The three points in the figure below lie along the same line. Drawing a line through the three points results in the graph of .

(0,3)

(1,3.5)

(-6,0)

Blitzer, Introductory Algebra, 5e – Slide #9 Section 4.2


Example use intercepts for graphing 4x 3y 12

Graphing Using Intercepts

ExampleUse intercepts for graphing 4x + 3y = 12

  • Find the x-intercept. Let y = 0 and solve for x.

    4x + 3(0) = 12

    4x = 12

    x = 3

    The x-intercept is 3. The line passes through (3, 0).

  • Find the y-intercept. Let x = 0 and solve for y.

    4(0) + 3y = 12

    3y = 12

    y = 4

    The y-intercept is 4. The line passes through (0, 4).

Blitzer, Introductory Algebra, 5e – Slide #10 Section 4.2


Example continued graph 4x 3y 12

Graphing Using Intercepts

Example , continued: Graph 4x + 3y = 12.

  • Find a checkpoint, a third ordered-pair solution. Let x = 2

    4(2) + 3y = 12

    8 + 3y = 12

    3y = 4

    y = 4/3 ~ 1.33

    A checkpoint is the ordered pair (2, 1.33)

Blitzer, Introductory Algebra, 5e – Slide #11 Section 4.2


Example continued graph 4x 3y 121

Graphing Using Intercepts

Example , continued: Graph 4x + 3y = 12.

  • Graph the equation by drawing a line through the intercepts and checkpoint.

(0, 3)

(2, 1.33)

(4, 0)

Blitzer, Introductory Algebra, 5e – Slide #12 Section 4.2


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Horizontal and Vertical Lines

Blitzer, Introductory Algebra, 5e – Slide #13 Section 4.2


4 2

Horizontal and Vertical Lines

EXAMPLE

Graph the linear equation: y = -4.

SOLUTION

All ordered pairs that are solutions of y = -4 have a value of y that is always -4. Any value can be used for x. Let’s select three of the possible values for x: -3, 1, 6:

(1,-4)

(-3,-4)

(6,-4)

Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution.

Blitzer, Introductory Algebra, 5e – Slide #14 Section 4.2


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Horizontal and Vertical Lines

EXAMPLE

Graph the linear equation: x = 5.

SOLUTION

All ordered pairs that are solutions of x = 5 have a value of x that is always 5. Any value can be used for y. Let’s select three of the possible values for y: -3, 1, 5:

(5,5)

(5,1)

(5,-3)

Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution.

Blitzer, Introductory Algebra, 5e – Slide #15 Section 4.2


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Horizontal Lines

Remember that the graph of y = b is a horizontal line. The y-intercept is b.

y = b

y

(0, b)

x

Blitzer, Introductory Algebra, 5e – Slide #16 Section 4.2


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Horizontal Lines

Example: Consider the graph of y = 2. This is a horizontal line and the y-intercept is 2.

y = 2

1

-3 -2 -11 2 3

y

2

(0, 2)

x

Blitzer, Introductory Algebra, 5e – Slide #17 Section 4.2


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Vertical Lines

Remember that the graph of x = a is a vertical line. The x-intercept is a.

x = a

y

x

(a, 0)

Blitzer, Introductory Algebra, 5e – Slide #18 Section 4.2


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Vertical Lines

Consider the graph of x = -3. The graph is a vertical line. The x-intercept is -3.

x = -3

y

x

-3

Blitzer, Introductory Algebra, 5e – Slide #19 Section 4.2


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Graphing Using Intercepts – In Summary

  • Find the x-intercept. Let y = 0 and solve for x.

  • Find the y-intercept. Let x = 0 and solve for y.

  • Find a checkpoint, a third ordered-pair solution.

  • Graph the equation by drawing a line through the three points.

Blitzer, Introductory Algebra, 5e – Slide #20 Section 4.2


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