Graphing Linear Equations
To graph a line (linear equation), we first want to make sure the equation is in slope intercept form
(y=mx+b). We will then use the slope and the y-intercept to graph the line.
Slope (m): Measures the steepness of a non-vertical line. It is sometimes refereed to as the rise/run
or
. It’s how fast and in what direction y changes compared to x.
y-intercept: The y-intercept is where a line passes through the y axis. It is always stated as an
ordered pair (x,y). The x coordinate is always zero. The y coordinate can be taken from the “b” in
y=mx+b.
Graphing The Linear Equation:
y = 3x - 5
1) Find the slope:
m=3m= 3 . = y .
1
x
2) Find the y-intercept: x = 0 , b = -5  (0, -5)
3) Plot the y-intercept
4) Use slope to find the next point: Start at (0,-5)
m = 3 . = y .  up 3 on the y-axis
1
x
 right 1 on the x-axis
(1,-2) Repeat: (2,1) (3,4) (4,7)
5) To plot to the left side of the y-axis, go to y-int. and
do the opposite. (Down 3 on the y, left 1 on the x)
(-1,-8) Repeat: (-2,-11) (-3,-14)
6) Connect the dots.
Do Now on GP:
1) y = 2x + 1
2) y = -4x + 5
3) y = ½ x – 3
4) y = - ⅔x + 2
Finding the equation of a line in slope intercept form (y=mx + b)
Example: Find the equation in slope intercept form of the line formed by (3,8) and (-2,
-7).
A. Find the slope (m):
B. Use m and one point to find b:
m = y2 – y1
y = mx + b
x2 – x1
m= 3
x= 3
y=8
m = -7 – 8 .
-2 – 3
-7 = 3(-2) + b
-7 = -6 + b
+6 +6
-1 = b
m = -15 .
-5
m= 3
III.
y = 3x – 1
Special Slopes
A. Zero Slope
* No change in Y
* Equation will be Y =
* Horizontal Line
B. No Slope (undefined slope)
* No change in X
* Equation will be X =
* Vertical Line
Find equation in slope intercept form and graph:
1) (3,-2)(-6,-8)
6) m= 4 (-2,-5)
12) 16x -4y =36
2) (-6,10) (9,-10)
7) m= ⅔ (-6,-7)
13) 8x+24y = 96
3) (3,7) (3,-7)
8) m= -3/2 (8,-1)
14) y-7=2(x+1)
4) (7,-6)(-3,4)
9) m = 0 (4,3)
15) y+5=(2/5)(x-10)
5) (5,-9)(-5,-9)
10) m = undefined (-6, 5)
16) y-7= ¾ (x+12)
11) m=-3 (-4,19)
17) y-2=-3(x-2)
IV. Parallel and Perpendicular Lines:
A. Parallel Lines
* Do not intersect
* Have same slopes
B. Perpendicular lines
* Intersect to form right angles (90˚)
* Slopes are negative reciprocals.
(Invert fraction and change sign)
(Products of slopes is –1)
Do in NB:
For the given line, find a line that is parallel and passes through the given point.
Then, find the equation of a line that is perpendicular and passes through the given
point.
Given Line:
Parallel:
Perpendicular:
7) y = ⅓ x + 4
(6,1)
( -2,10)
8) y = 4x – 5
(2,13)
(8.-5)
9) y = -⅔ x + 2
(-9,-11)
(4,-1)
10) –5x + 6
(4,-27)
(-10,6)
Practice Problems: a) Use the two points to find the equation of the line.
b) For the line found in part a, find a line that is parallel and passes through
the given point.
c) Find the equation of a line that is perpendicular and passes through the
given point.
Given Line:
1) (-5, 13) (3, -3)
2) (-6,0) (3,6)
3) (2,6)(-3,-19)
4) (-4,3) (-8,6)
5) (2,-5) (-2, -5)
6) (-9,-11)(6,9)
7) (8,-3) (-4,9)
8) (3,6)(3,-6)
9) (4,-3)(-6,-8)
10) (2,4)(-6,-12)
Parallel:
(4,-10)
(6,3)
(5,14)
(-4, 10)
(8,-2)
(-3,-9)
(-2, 14)
(11,-3)
(6,7)
(-3,-5)
Perpendicular:
(2,7)
(6,-7)
(5,5)
(-6,-8)
(4,-3)
(-4,10)
(6,-4)
(5,2)
(-5,0)
(-8,4)
11) Find the equation of the line parallel to y = 3x – 2, passing through (-2, 1).
12) Find the equation of the line perpendicular to y = -½x – 5, passing through (-2, -10)
13) Find the equation of the line parallel to y = -¼ x + 2, passing through (-8, 7)
14) Find the equation of the line perpendicular to y = (3/2)x + 6, passing through (-6, 1)
15) Find the equation of the line parallel to y = -5, passing through (2,7)
16) Find the equation of the line perpendicular to y = 5, passing through (6, -4).
17) Find the equation of the line parallel to x= 8, passing through (4, -9)
18) Find the equation of the line perpendicular to x = -3, passing through (6, -7).
Solve each system graphically:
19)
y = -4x -5
y = 2x -7
23)
y-2= (3/5)(x-10)
y+11 =2(x+7)
20)
6x + 3y =21
12x + 16y = -48
24)
6x + 9y = 45
9x +15y = 75
21)
12x – 6y = -6
16x -8y = 40
25)
x=5
y-12 = -3(x+2)
22)
y= -4
x=7
26)
9x – 18y = 126
y = -4
Graphing the Parabola
a) Graph y = -3x2 + x + 2
b) Is the vertex a minimum or maximum?
c) Find the axis of symmetry, the vertex, and the roots.
1)
The first thing you should do is put the equation into the calculator. While you are in the y= screen,
you might as well put in y = 0 for the second line (you will need it to find the roots).
Min or Max?
LOOK AT THE PICTURE
Axis of Symmetry:
[a is the coefficient of x2 (-3)and b is the coef. of x (1)]
x = -b .
2a
x = -(1) .
2(-3)
x = .16
Finding the vertex: If the axis of symm. is an integer, then you can find the ordered
pair for the vertex in the table. If not, we will use the calculator.
2nd CALC 1 (for value)
Type in the value for x (.16 ) and hit ENTER
The ordered pair for the vertex will appear on the bottom of the screen. (.16, 2.08)
Finding the roots:
The roots are where the equation or y = 0. (Where it crosses the x-axis).
This is why we put y = 0 into the calculator. The x-axis is y = 0.
There are usually 2 roots. To find the first:
2 CALC 5 (for intersect) ENTER ENTER ENTER
The ordered pair will be at the bottom.
(-.66, 0)
nd
To find the 2nd root:
2 CALC 5 (for intersect) DO NOT HIT ENTER!!!!!
You must move the cursor using the blue left and right arrows (not up and down) close to the 2nd
root. Once close:
ENTER ENTER ENTER The ordered pair will be at the bottom. (1,0)
nd
Graph the parabola:
x
y
-2
-1
0
1
2
-12
-2
2
0
-8
Plot the vertex. Estimate as best you
can.
Plot the points from the
table.
Draw as smooth a curve as possible.
Label with the equation.