3.2 Graphing Linear Equations in Two Variables

Report
3.2 Graphing Linear
Equations in Two
Variables
Objective 1
Graph linear equations by plotting
ordered pairs.
Slide 3.2-3
Graph linear equations by plotting ordered pairs.
Infinitely many ordered pairs satisfy a linear equation in two variables. We
find these ordered-pair solutions by choosing as many values of x (or y) as
we wish and then completing each ordered pair.
Some solutions of the equation x + 2y = 7 are graphed below.
Slide 3.2-4
Graph linear equations by plotting ordered pairs. (cont’d)
Notice that the points plotted in the previous graph all appear to lie on a
straight line, as shown below.
Every point on the line represents a solution of the equation
x + 2y = 7, and every solution of the equation corresponds to a point on the
line.
The line gives a “picture” of all the solutions of the equation x + 2y = 7. Only
a portion of the line is shown, but it extends indefinitely in both directions,
suggested by the arrowheads.
The line is called the
graph of the equation,
and the process of
plotting the ordered
pairs and drawing the
line through the
corresponding points is
called graphing.
Slide 3.2-5
Graph linear equations by plotting ordered pairs. (cont’d)
Graph of a Linear Equation
The graph of any linear equation in two variables is a straight line.
Notice the word line appears in the name “linear equation.”
Since two distinct points determine a line, we can graph a straight line by finding any
two different points on the line. However, it is a good idea to plot a third point as a
check.
Slide 3.2-6
CLASSROOM
EXAMPLE 1
Graph
Graphing a Linear Equation
5x  2 y  10.
Solution:
5  0  2 y  10
2 y 1 0

2
2
y  5
5x  2  0  10
 0, 5
5 x 1 0

5
5
x  2
 2, 0
5  4  2 y  10
20  2 y  20  10  20
2 y 10

2
2
y5
 4, 5
When graphing a linear equation, all three points should lie on the same straight
line. If they don’t, double-check the ordered pairs you found.
Slide 3.2-7
CLASSROOM
EXAMPLE 2
Graph
Graphing a Linear Equation
2
y  x  2.
3
Solution:
2
y   0  2
3
y  2
 0, 2
2
02  x22
3
2 3
3
 2  x 
3 2
2
x3
2
4  2  x  2  2
3
2
3
 2  x
3
2
 3, 0
x  3
 3, 4
Slide 3.2-8
Objective 2
Find intercepts.
Slide 3.2-9
Find intercepts.
In the previous example, the graph intersects (crosses) the y-axis at (0,−2)
and the x-axis at (3,0). For this reason (0,−2) is called the y-intercept and
(3,0) is called the x-intercept of the graph.
The intercepts are particularly useful for graphing linear equations. They are
found by replacing, in turn, each variable with 0 in the equation and solving
for the value of the other variable.
Finding Intercepts
To find the x-intercept, let y = 0 and
solve for x. Then (x,0) is the x-intercept.
To find the y-intercept, let x = 0 and
solve for y. Then (0, y) is the yintercept.
Slide 3.2-10
CLASSROOM
EXAMPLE 3
Finding Intercepts
Find the intercepts for the graph of 5x + 2y = 10. Then draw the graph.
Solution:
5x  2  0  10
5 x 10

5
5
x2
x-intercept:
 2, 0 
5  0  2 y  10
2 y 10

2
2
y5
y-intercept:
 0,5
When choosing x- or y-values to find ordered pairs to plot, be careful to
choose so that the resulting points are not too close together. This may result
in an inaccurate line.
Slide 3.2-11
Objective 3
Graph linear equations
of the form Ax + By = 0.
Slide 3.2-12
Graph linear equations of the form Ax + By = 0.
Line through the Origin
If A and B are nonzero real numbers, the graph of a linear equation of the
form
Ax  By  0
passes through the origin (0,0).
A second point for a linear equation that passes through the origin can be
found as follows:
1.Find a multiple of the coefficients of x and y.
2.Substitute this multiple for x.
3.Solve for y.
4.Use these results as a second ordered pair.
Slide 3.2-13
CLASSROOM
EXAMPLE 4
Graphing an Equation with x- and y-Intercepts (0, 0)
Graph 4x − 2y = 0.
Solution:
12
1
4  6  2 y  0
24  2 y  24  0  24
2 y 24

2
2
y  12
4x  2  2  0
4x  4  4  0  4
4 x 4

4
4
x  1
Slide 3.2-14
Objective 4
Graph linear equations of the form y = k
or x = k.
Slide 3.2-15
Graphing linear equations of the form y = k or x = k.
The equation y = − 4 is the linear equation in which the coefficient of x is 0.
Also, x = 3 is a linear equation in which the coefficient of y is 0. These
equations lead to horizontal straight lines and vertical straight lines,
respectively.
Horizontal Line
The graph of the linear equation y = k, where k is a real number, is the
horizontal line with y-intercept (0, k). There is no y-intercept (unless the
vertical line is the y-axis itself).
Vertical Line
The graph of the linear equation x = k, where k is a real number, is the
vertical line with x-intercept (k, 0). There is no x-intercept (unless the
horizontal line is the x-axis itself).
The equations of horizontal and vertical lines are often confused with each
other. Remember that the graph of y = k is parallel to the x-axis and that of x = k
is parallel to the y-axis (for k ≠ 0).
Slide 3.2-16
CLASSROOM
EXAMPLE 5
Graphing an Equation of the Form y = k (Horizontal Line)
Graph y = − 5.
Solution:
The equation states that every value of y = − 5.
Slide 3.2-17
CLASSROOM
EXAMPLE 6
Graphing an Equation of the Form x = k (Vertical Line)
Graph x − 2 = 0.
Solution:
After 2 is added to each side the equation states that every value
of x = 2.
Slide 3.2-18
Graphing linear equations of the form y = k or x = k. (cont’d)
Slide 3.2-19
Graphing linear equations of the form y = k or x = k. (cont’d)
Slide 3.2-20
Objective 5
Use a linear equation to model data.
Slide 3.2-21
CLASSROOM
EXAMPLE 7
Using a Linear Equation to Model Credit Card Debt
Use (a) the graph and (b) the equation from Example 7 to approximate credit
card debt in 2005.
y  32.0 x  684
Solution:
2005
About 850 billion dollars using the
graph to estimate.
y  32.0 5  684
y  160  684
y  844
Exactly 844 billion dollars using the equation.
Slide 3.2-22

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