Graphing Linear Equations



Intro. To Graphing Linear Equations

The Coordinate Plane

A. The coordinate plane has 4 quadrants.

B. Each point in the coordinate plain has an x coordinate (the abscissa) and a y coordinate (the ordinate). The point is stated as an ordered pair (x,y).

C. Horizontal Axis is the X – Axis. (y = 0)

D. Vertical Axis is the Y- Axis (x = 0)

Plot the following points:

a) (3,7) b) (-4,5) c) (-6,-1) d) (6,-7)

e) (5,0) f) (0,5) g) (-5,0) f) (0, -5)

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)

6) Connect the dots.

Do Now:

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 (1,2) 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= 1 y= 2

 

m = -7 – 2 . 2 = 3(1) + b

-2 – 1 2 = 3 + b

-3 -3

m = -9 . -1 = b

-3

m= 3 y = 3x – 1

 

 

III. Special Slopes

A. Zero Slope B. No Slope (undefined slope)

* No change in Y * No change in X

* Equation will be Y = * Equation will be X =

* Horizontal Line * 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)

Graph both of the lines on the same set of axis:

y = -2x + 6 y = -2x – 5

IV. Parallel and Perpendicular Lines:

A. Parallel Lines

* Do not intersect

* Have same slopes

For the given line, find a line that is parallel and passes through the given point.

Given Line: Parallel: Given Line: Parallel:

7) y = ⅓ x + 4 (6,1) 8) y = 4x – 5 (2,13)

Given Line: Parallel: Given Line: Parallel:

9) y = -⅔ x + 2 (-9,2) 10) –5x + 6 (4,-27)

 

 

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) Graph both lines on the same set of axis. (Use your own graph paper.)

 

Given Line: Parallel:

1) (-5, 13) (3, -3) (4,-10)

2) (-6,0) (3,6) (6,3)

3) (2,6)(-3,-19) (5,30)

4) (-4,3) (-8,6) (-4, 10)

5) (2,-5) (-2, -5) (8,-2)

6) (-9,-11)(6,9) (-3,-9)

7) (8,-3) (-4,9) (-2, 1)

8) (3,6)(3,-6) (7,-3)

9) (4,-3)(-6,-8) (6,7)

10) (2,4)(-6,-12) (-3,-5)

11) Find the equation of the line parallel to y = 3x – 2, passing through (-2, 1).

12) Find the equation of the line parallel to y = -½x – 5, passing through (-2, 7)

13) Find the equation of the line parallel to y = -¼ x + 2, passing through (-8, 4)

14) Find the equation of the line parallel to y = (3/2)x + 6, passing through (-6, -11)

15) Find the equation of the line parallel to y = -5, passing through (2,7)

16) Find the equation of the line parallel to x = 5, passing through (6, -4).

Solving Systems of Equations

Graphically

A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. When solving a system containing two linear equations there will be one ordered pair (x,y) that will work in both equations.

To solve such a system graphically, we will graph both lines on the same set of axis and look for the point of intersection. The point of intersection will be the one ordered pair that works in both equations. We must then CHECK the solution by substituting the x and y coordinates in BOTH ORIGINAL EQUATIONS.

1) Solve the following system graphically:

y = 2x – 5

y = - ⅓x + 2

Solve each of the systems of equations graphically:

2) y + 1 = -3(x – 1)

7x + 7y = 42

3) y – 9 = ¾ (x – 12)

6x + 12y = -60

4) 12x – 8y = 48

y – 4 = -2(x – 2)

5) y + 5 = 2(x + 4)

y – 10 = - ½ (x + 4)

Solve each system graphically and check:

6) y = -4x -5 10) y-2= (3/5)(x-10)

y = 2x -7 y+11 =2(x+7)

7) 6x + 3y =21 11) 6x + 9y = 45

12x + 16y = -48 9x +15y = 75

8) 12x – 6y = -6 12) x = 5

16x -8y = 40 y-12 = -3(x+2)

9) y= -4 13) 9x – 18y = 126

x = 7 y = -4

Graphing Inequalities

            When we solved and graphed inequalities with only one variable (ex: x > 3), we moved on to compound inequalities (AND/OR).  We would graph both inequalities on the same number line and decide what to keep based on whether it was an AND or an OR problem.  When we graphed linear equations on the coordinate plane we moved on to solving systems of equations graphically.

When we graph inequalities in two variables on the coordinate plane, we do not graph compound inequalities.  We move on to solving systems of inequalities.  It takes a little from both inequalities with one variable and solving systems graphically.

Graph the Inequality:

                        y > ¼ x + 3

Step 1:  Graph the line.     

    y > ¼ x + 3                                                                 

m = ¼ = ▲y = up 1

▲x r 4                                                                   

y-int= (0,3)  

Step 2: Test a point one up from the from

the y-int and one down from the y-int):                                                                

 (0, 2)                          (0, 4)                                

2 > ¼ (0)+3          4 > ¼ (0) + 3                    

2  >  3                         4 > 3                                   

 FALSE                     TRUE                                    

Step 3: Shade towards the “true” point (0,4)

When you “test”, you must do it in the original equation!

1) 6x - 9y > 36

2) y - 3 > -2(x + 1)

3) 12x + 9y < 27

4) y + 4 > -3(x - 3)

5) y > 4

6) x < -6

Graphing Systems of Inequalities

Solve the system of inequalities graphically:           y > ¼ x + 3

y < 3x – 5

Step 1:  Graph the 1st inequality (graph              Step 2:  Graph the 2nd inequality (graph the line and test a point one up from the          the line and test a point one up from the y-int and one down from the y-int):                  y-int and one down from the y-int):

y > ¼ x + 3                                                                  y < 3x – 5

m = ¼ =                                                                       m = 3/1 =

y-int= (0,3)                                                                   y-int.= (0,-5)

 (0, 2)           TEST               (0, 4)                               (0-6)          TEST             (0,-4)

2 > ¼ (0)+3          -3 > ¼ (0) + 3                    -6 < 3(0) - 5                -4 < 3(0) - 5    

2  >  3                         4 > 3                                    -6 < -5                               -4 < -5

 FALSE                     TRUE                                     TRUE                                FALSE

Step 3: Label the area where the shading intersects with an “S”

2) y - 3 < - ⅓(x – 6)

12x – 6y > -12

3) x > 4

y < -5

4) 24x + 6x > -6

y > 2

5) y – 6 < ⅔(x - 9)

x < -3

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