EXPLAIN 2 Graphing Linear Functions Given in Standard Fo Read the section below and complete the following activity (adapted from Lesson 5.1). Graphing in Standard Form e graph of a linear equation is a line To granh a line, t is necessary to plot only two points. However, it is a o nltea to plot a third point as a check For linear equations written in standard form, two very useful points piot are where the line crosses each axis The r-coordinate of the point where the line crosses the x-axis is caned the x-intercept and is found by substituting 0 for y in the equation of the line. The y-coordinate of the pomt where the line crosses the y-axis is called the v-intercept and is found by substituting 0 for x in the equation of the line. Example Graph 3x+ 5y = 15 Step 1- Make a table of values. Each row in the table of values makes an ordered pair (x, y). Step 2 - Substitute 0 for x in the equation and solve for y. 3(0) + 5y = 15 5y = 15 y y = 3 (0, 3) satisfies the equation Step 4 - Choose another value for x and solve for y. Choose convenient values if possible such as Step 3 - Next substitute 0 for y in the equation and solve for x. -5 in this case. 3x + 5(0) = 15 3(-5) +5y = 15 -15 +5y = 15 5y = 15 + 15 5y = 30 y = 6 3x = 15 x = 5 (5,0) satisfies the equation (-5, 6) satisfies the equation ep 5 - Plot the first two points on the coordinate grid. Then plot the third point. All three points should lie on a straight line. Draw a line through the points. (0, 3), (5,0), (-5, 6) 9. 5' 4- 3 2. 3 4 -6 -5 -4 -3-2 -10 -1 -2 -3 -4' -5- -6-

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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O Read the section below and complete the following activity (adapted from Lesson 5.1).
EXPLAIN 2
Graphing Linear Functions Given in Standard Fo
Graphing in Standard Form
goodpiror a linear equation is a line. To graph a line, it is necessary to plot only two points. However, it is a
to nlca to plot a third point as a check. For linear equations written in standard form, two very useful points
to plot are where the line crosses each avis The racoordinate of the point where the line crosses the x-axis is
ted the X-intercept and is found by substituting 0 for y in the equation of the line. The y-coordinate of the
Pomt where the line crosses the y-axis is called the v-intercept and is found by substituting 0 for x in the
equation of the line.
Example
Graph 3x + 5y = 15
Step 1 - Make a table of values. Each row in the table
of values makes an ordered pair (x, y).
Step 2 - Substitute 0 for x in the equation and solve
for y.
3(0) +5y = 15
5y = 15
y
y = 3
(0,3) satisfies the equation
Step 3 - Next substitute 0 for y in the equation and
Step 4 - Choose another value for x and solve for y.
Choose convenient values if possible such as
-5 in this case.
solve for x.
3x + 5(0) = 15
3(-5) +5y = 15
-15 + 5y = 15
5y = 15 + 15
5y = 30
y = 6
3x = 15
x = 5
(5,0) satisfies the equation
(-5, 6) satisfies the equation
Step 5 - Plot the first two points on the coordinate
grid. Then plot the third point. All three
points should lie on a straight line. Draw a
line through the points. (0,3), (5, 0), (-5, 6)
61
4
31
2.
-6 -5 -4
-3 -2
1
2
34
5 6
-2
-3"
-4"
-5-
-6-
Transcribed Image Text:O Read the section below and complete the following activity (adapted from Lesson 5.1). EXPLAIN 2 Graphing Linear Functions Given in Standard Fo Graphing in Standard Form goodpiror a linear equation is a line. To graph a line, it is necessary to plot only two points. However, it is a to nlca to plot a third point as a check. For linear equations written in standard form, two very useful points to plot are where the line crosses each avis The racoordinate of the point where the line crosses the x-axis is ted the X-intercept and is found by substituting 0 for y in the equation of the line. The y-coordinate of the Pomt where the line crosses the y-axis is called the v-intercept and is found by substituting 0 for x in the equation of the line. Example Graph 3x + 5y = 15 Step 1 - Make a table of values. Each row in the table of values makes an ordered pair (x, y). Step 2 - Substitute 0 for x in the equation and solve for y. 3(0) +5y = 15 5y = 15 y y = 3 (0,3) satisfies the equation Step 3 - Next substitute 0 for y in the equation and Step 4 - Choose another value for x and solve for y. Choose convenient values if possible such as -5 in this case. solve for x. 3x + 5(0) = 15 3(-5) +5y = 15 -15 + 5y = 15 5y = 15 + 15 5y = 30 y = 6 3x = 15 x = 5 (5,0) satisfies the equation (-5, 6) satisfies the equation Step 5 - Plot the first two points on the coordinate grid. Then plot the third point. All three points should lie on a straight line. Draw a line through the points. (0,3), (5, 0), (-5, 6) 61 4 31 2. -6 -5 -4 -3 -2 1 2 34 5 6 -2 -3" -4" -5- -6-
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