graphing linear equations

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Graphing Linear Equations What we should learn

Graphing linear equations in one variable Graphing linear equations in two variables Graphing using intercepts Slope Graphing using slope Solutions of linear equations Graphs of absolute value Solving absolute value equations.

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The graph of a point in two variables x and y is the ordered pair (x, y), where x and y are the coordinates of the point.

x, the first number, tells how far to the left or right the point is from the vertical axis. y, the second number, tells how far up or down the point is from the vertical axis.

For instance, the point (–1, 3) is the point that is 1 to left of vertical axis and is three up from horizontal axis.

4.1 Graphing Linear Equations

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y

x1

2

–3 2 3

x-axis

y-axis

origin

Quadrant IQuadrant II

Quadrant IIIQuadrant IV

(x, y) coordinate system terms

(3,2)(-3,2)

(-3,-3)(1,-2)


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The graph of an equation in two variables x and y is the set of all points (x, y) whose coordinates satisfy the equation.

For instance, the point (–1, 3) is on the graph of 2y – x = 7 because the equation is satisfied when –1 is substituted for x and 3 is substituted for y. That is,

2y – x = 7 Original Equation

2(3) – (–1) = 7 Substitute for x and y.

7 = 7 Equation is satisfied.


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To sketch the graph of an equation,

1. Find several solution points of the equation by substituting various values for x and solving the equation for y.

2. Plot the points in the coordinate plane.

3. Connect the points using straight lines or smooth curves.


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Example: Sketch the graph of y = –2x + 3.

1. Find several solution points of the equation.

x y = –2x + 3 (x, y)

–2 y = –2(–2) + 3 = 7 (–2, 7)

–1 y = –2(–1) + 3 = 5 (–1, 5)

0 y = –2(0) + 3 = 3 (0, 3)

1 y = –2(1) + 3 = 1 (1, 1)

2 y = –2(2) + 3 = –1 (2, –1)


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Example: Sketch the graph of y = x - 1.

1. Find several solution points of the equation.

x y = –2x + 3 (x, y)

–2 y = (–2) - 1 = -3 (–2, -3)

–1 y = (–1) - 1 = -2 (–1, -2)

0 y = (0) -1 = -1 (0, -1)

1 y = (1) - 1 = 0 (1, 0)

2 y = (2) - 1 = 1 (2, 1)


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Equations of the form ax + by = c are called linear equations in two variables.

The point (0,4) is the y-intercept.

The point (6,0) is the x-intercept.

x

y

2-2

This is the graph of the equation 2x + 3y = 12.

(0,4)

(6,0)

4.3 Quick Graphs Using Intercepts

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Equations of the form ax + by = c are called linear equations in two variables.

The point (0,-2) is the y-intercept.

The point (3,0) is the x-intercept.

x

y

This is the graph of the equation 2x - 3y = 6.

(0,-2)

(3,0)

4.3 Quick Graphs Using Intercepts

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y

x2-2

The slope of a line is

a number, m,

which measures its

steepness.m = 0

m = 2m is undefined

m =1

2

m = -1

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4.4 Slope

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x

y

x2 – x1

y2 – y1

change in y

change in x

The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula

The slope is the change in y divided by the change in x as we move along the line from (x1, y1) to (x2, y2).

y2 – y1

x2 – x1

m = , (x1 ≠ x2 ).

(x1, y1)

(x2, y2)

4.4 Slope

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Example: Find the slope of the line passing through the points (2, 3) and (4, 5).

Use the slope formula with x1= 2, y1 = 3, x2 = 4, and y2 = 5.

y2 – y1

x2 – x1

m =

5 – 3

4 – 2= =

22

= 1 2

2(2, 3)

(4, 5)

x

y

4.4 Slope

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Example: Find the slope of the line passing through the points (-1, 2) and (3, 1). Use the slope formula with x1= -1, y1 = 2, x2 = 3, and y2 = 1.

y2 – y1

x2 – x1

m =(-1, 2)

(3, 1)

x

y

4.4 Slope

m 1 2

3 ( 1)

1

4

1

4-1

4

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A linear equation written in the form y = mx + b is in slope-intercept form.

To graph an equation in slope-intercept form:

1. Write the equation in the form y = mx + b. Identify m and b.

The slope is m and the y-intercept is (0, b).

2. Plot the y-intercept (0, b).

3. Starting at the y-intercept, find another point on the line using the slope.

4. Draw the line through (0, b) and the point located using the slope.

4.5 Slope-Intercept Form

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1

Example: Graph the line y = 2x – 4.

2. Plot the y-intercept, (0, - 4).

1. The equation y = 2x – 4 is in the slope-intercept form. So, m = 2 and b = - 4.

3. The slope is 2.

The point (1, -2) is also on the line.

1= change in y

change in xm = 2

4. Start at the point (0, 4). Count 1 unit to the right and 2 units up to locate a second point on the line.

2

x

y

5. Draw the line through (0, 4) and (1, -2).

(0, - 4)

(1, -2)


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A linear equation written in the form y – y1 = m(x – x1) is in point-slope form. The graph of this equation is a line with slope m passing through the point (x1, y1).

Example:

The graph of the equation

y – 3 = - (x – 4) is a line

of slope m = - passing

through the point (4, 3).

1

2 1

2

(4, 3)

m = -1

2

x

y

4

4

8

8


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Example: Write the slope-intercept form for the equation of the line through the point (-2, 5) with a slope of 3.

Use the point-slope form, y – y1 = m(x – x1), with m = 3 and (x1, y1) = (-2, 5).

y – y1 = m(x – x1) Point-slope form

y – y1 = 3(x – x1) Let m = 3.

y – 5 = 3(x – (-2)) Let (x1, y1) = (-2, 5).

y – 5 = 3(x + 2) Simplify.

y = 3x + 11 Slope-intercept form


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Example: Write the slope-intercept form for the equation of the line through the points (4, 3) and (-2, 5).

y – y1 = m(x – x1) Point-slope form

Slope-intercept formy = - x + 13

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3

2 1 5 – 3 -2 – 4

= - 6

= - 3

Calculate the slope.m =

Use m = - and the point (4, 3).y – 3 = - (x – 4)1

3 3

1


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State the slope and y-intercept y = -2x + 3 x + y = 10 2x = 4 y = 5 x + 2y – 5 = 0 y = x2 + 4


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Two lines are parallel if they have the same slope.

If the lines have slopes m1 and m2, then the lines are parallel whenever m1 = m2.

Example: The lines y = 2x – 3 and y = 2x + 4 have slopes m1 = 2 and m2 = 2.

The lines are parallel.

x

y

y = 2x + 4

(0, 4)

y = 2x – 3

(0, -3)


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Two lines are perpendicular if their slopes are negative reciprocals of each other.If two lines have slopes m1 and m2, then the lines are

perpendicular whenever

The lines are perpendicular.

1m1

m2= - or m1m2 = -1.

y = 3x – 1

x

y

(0, 4)

(0, -1)

y = - x + 41

3Example:

The lines y = 3x – 1 and

y = - x + 4 have slopes

m1 = 3 and m2 = - .

1

3 1

3


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4.6 Solutions and x-Intercepts

Use a Graphic Check of a Solution

1. Write the equation in the form ax + b = 0.

2. Sketch the graph of y = ax + b.

3. The solution of ax + b = 0 is the x-intercept of y = ax + b.

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Solve 2x + 4 = -2 add -4 to both sides

2x + 4 – 4 = -2 - 4 combine like terms

2x = -6 divide both sides by -2

x = -3

Graph y = 2x + 6

Get 2x + 4 = -2 equal to 0 2x + 6 = 0

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Graph y = 2x + 6

x = -3 is the x-intercept and the solution to

2x + 6 = 0

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Solve 3x + 1 = -5 subtract 1 from both sides

3x + 1 – 1 = -5 - 1 combine like terms

3x = -6 divide both sides by 3

x = -2

Graph y = 3x + 6

Get 3x + 1 = -5 equal to 0 3x + 6 = 0

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Graph y = 3x + 6

x = -2 is the x-intercept and the solution to

3x + 6 = 0

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4.7 Graph Absolute Value

Graph y = |x| x y

-3 3

-2 2

-1 1

0 0

1 1

2 2

3 3

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Graph y = 2|x| - 1

x y

-3 5

-2 3

-1 1

0 -1

1 1

2 3

3 5

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Graph y = -|x| + 2

x y

-3 -1

-2 0

-1 1

0 2

1 1

2 0

3 -1

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4.8 Absolute Value Equations

Let x be a variable of an algebraic expression and let a be real number such that a > 0. The solutions of the equation |x| = a are given by x = a and x = -a

example |x| = 10 this is equivalent to the two equations x = 10 and x = -10

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Example solve |x + 3| = 5 x + 3 = 5 or x + 3 = -5 x = 2 or x = -8 Solution set is {2, -8}

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Example solve |2x - 3| - 5 = 8 |2x – 3| = 13 add 5 to both sides 2x – 3 = 13 or 2x – 3 = -13 2x = 16 2x = -10 x = 8 or x = -5 The solution is { 8, -5}

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Example solve |3x - 1| - 4 = -2 |3x – 1| = 2 add 4 to both sides 3x – 1 = 2 or 3x – 1 = -2 3x = 3 3x = -1 x = 1 or x = -1/3 The solution is { -1/3, 1}

graphing linear equations

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