Transcript 7_3 graph using intercepts_TROUT09

Warm Up #10
1.) Graph 5x + 7y =35
2.) Graph y= 2x -3
Graph 5x + 7y =35
Solve for “y” X -values Y= -5/7x +5 Y- values
7y = -5x +35
-5/7(7)+5 0
Y = -5/7 x + 5 7
(X, Y)
(7, 0)
0
-5/7(0)+5 5
(0,5)
-7
-5/7(-7)+5 10
(-7, 10)
Find 3 points using the table, and graph the
line of the equation. y = 2x - 3
1
0
-1
-1
-3
-5
7.3 Linear Equations and Their
Graphs
Linear Equations
(Graphs are straight lines)
1. Equation is linear only if the each 3
1

3x
1
variable has an exponent of “1”. x
2. (exponent in denominator is not linear)
3. Products of variables not linear, ie (x)(y)
y – 3x = -7
y = 2x + 1
y = x2 + 1
5y = 14
xy = 2
x
y
2
2
y
x
y
4
3
x - intercept
2
B
1
-6
-4
-2
2
4
X
-1
y - intercept
-2
A
-3
-4
6
Graphing using Intercepts
1) Let x=0 and determine the y-intercept.
2) Let y=0 and determine the x-intercept
3) Plot both points. Connect them with a line.
Graph 4x + 3y = 12 using intercepts
6
Find x-intercept
5
4x + 3(0) = 12
4x
= 12
x
4
Find y-intercept
4(0) + 3y = 12
3y = 12
3
2
=3
y=4
1
-8
-6
-4
-2
2
-1
-2
-3
-4
4
6
8
Graph 2x + 3y = 12 using intercepts
6
5
4
x
0
y
4
6
0
3
2
1
-8
-6
-4
-2
2
-1
-2
-3
-4
4
6
8
Graph 3x + 5y = 15 using intercepts
6
5
4
x
0
y
3
5
0
3
2
1
-8
-6
-4
-2
2
-1
-2
-3
-4
4
6
8
Graph 5x - 2y = 10 using intercepts
6
5
x y
0 5
2 0
4
3
2
1
-8
-6
-4
-2
2
-1
-2
-3
-4
-5
4
6
8
Graph 2y = 3x - 6 using intercepts
6
x y
0 3
2 0
5
4
3
2
1
-8
-6
-4
-2
2
-1
-2
-3
-4
4
6
8
Horizontal and Vertical Lines
• The graph of y= # is HORIZONTAL
• The graph x =# is VERTICAL
Graph 4y = 16 using 3-points
6
5
4
3
2
-8
x
0
3
6
1
y
-6
-4
-2
2
-1
-2
-3
-4
4
6
8
Graph 3x = 18 using 3-points
6
5
4
3
2
x
-8
-6
y
0
3
-4
1
-4
-2
2
-1
-2
-3
-4
4
6
8
Differences between graphing by
using a table and graphing by
finding the x and y intercepts
• When graphing by a table you need to solve
for y (Slope Intercept Form y=mx+b)
• When graphing by finding the x and y
intercepts you do not have to solve for y
(Standard Form Ax +By =C)
Assignment
Page 316
( 16 – 42 even and 45-49 all)