Warm UpTuesday 11/9/10 We will graph linear equations using

intercepts.

Simplify each expression:

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a. 8x(x − 3) −10(x + 2)

b. 2x(x − 5) − 3(x + 4)

c. 3x(x − 2) − 5(x − 2)

d. 7x(x − 2) − 6(x + 3)

Warm UpWednesday 11/10/10 We will graph linear equations

using intercepts.

Solve each equation:

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a. 2

9x − 3 = 4

b. 3

5x − 8 =10

c. 2

7x −10 =12

Today’s Objective

•To be able to find the x and y intercepts of an equation and use them to draw a quick graph.

The Intercepts

y-Intercept = 6

x-Intercept = 2

The intercepts are

where the line crosses

the axis.

Intercepts: Find the x and y intercepts then graph the line

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4x + 5y = −20x intercept y = 0Substitute 0 in for y then solve for x

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4x + 5(0) = −20

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4x + 0 = −20

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4x = −20

x = −5

(−5, 0)

y intercept x = 0Substitute 0 in for x then solve for y

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4(0) + 5y = −20

0 + 5y = −20

5y = −20

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y = −4

(0, − 4)

Graph of equation

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4x + 5y = −20

Next graph the x intercept x = -5 at the point (-5,0)

First graph the y intercept y = -4 at the point (0,-4)

Finally connect the points to form a line

Intercepts: Find the x and y intercepts then graph the line

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3y = 2x − 24x intercept y = 0Substitute 0 in for y then solve for x

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3(0) = 2x − 24

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0 = 2x − 24

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24 = 2x

x =12

(12, 0)

y intercept x = 0Substitute 0 in for x then solve for y

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3y = 2(0) − 24

3y = 0 − 24

3y = −24

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y = −8

(0,−8)

Graph of equation

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3y = 2x − 24

Next graph the x intercept x = 12 at the point (12,0) (change scale count by 2s)

First graph the y intercept y = -8 at the point (0,-8)

Finally connect the points to form a line

Compare the 2 graphs3x+ y = 6

(1,3)

(0,6) (2,0)

(-1,9)

3x+ y = 6

x-intercept = 2y-intercept = 6

Find the intercepts and graph

3x + 4y = 12

Finding the x-intercept 3x + 4y = 12 3x + 4(0) = 12 3x + 0 = 12 3x = 12 x = 4

Finding the y-intercept 3x + 4y = 12 3(0) + 4y = 12 0 + 4y = 12 4y = 12 y = 3

The graph of 3x + 4y = 12

x-intercept = 4y-intercept = 3

Find the intercepts and graph

y = 4x - 4

You try this one.

Finding the x-intercept y = 4x - 4 0 = 4x - 4 0 + 4 = 4x -4 + 4 4 = 4x 1 = x

Finding the y-intercept y = 4x - 4 y = 4(0) - 4 y = -4

The graph of y = 4x - 4

x-intercept = 1

y-intercept = -4

What is the x-intercept of3x – 4y = 24?

1. (3, 0)

2. (8, 0)

3. (0, -4)

4. (0, -6)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

What is the y-intercept of-x + 2y = 8?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

1. (-1, 0)

2. (-8, 0)

3. (0, 2)

4. (0, 4)

What is the y-intercept ofx = 3?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

1. (3, 0)

2. (-3, 0)

3. (0, 3)

4. None

Word Problem• You make and sell decorative bows. You sell

small bows for $3 and large bows for $5. You want to earn $60 per week. This situation can be modeled by 3x + 5y = 60 where x is the number of small bows and y is the number of large bows.

• Find the intercepts of the graph.• Graph the equation• Give 3 possible solutions

Finding the y-intercept 3x + 5y = 60 3(0) + 5y = 60 5y = 60 y = 12 (0, 12)

Finding the x-intercept 3x + 5y = 60 3x + 5(0) = 60 3x = 60 x = 20

(20, 0)

Graph of the Equation

Small Bows Sold

Three Possible Solutions

Small Bows Sold

0 small, 12 large0 small, 12 large

10 small, 6 large10 small, 6 large

20 small, 0 large20 small, 0 large

Word Problem• You borrow $1800 from your parents. To repay your debt, you give them $150 per month.

Your debt d (in dollars) is given by the function: d = 150t -1800 where t represents time in (months) • Find the intercepts of the graph of the function and

state what they represent.

• Graph the equation and identify its domain and range.

Finding the t-intercept d = 150t - 1800

Set d = 0 0 = 150t – 1800 1800 = 150t

12 = t

In 12 months the debt = 0

Finding the d-intercept d = 150t - 1800

Set t = 0 d = 150(0) - 1800 d = - 1800

The debt is -$1800 when time is 0.

Graph the function d = 150t -1800

Put time (t) on the x axisPut time (t) on the x axis

Put

deb

t (d)

on

the

y ax

isP

ut d

ebt (

d) o

n th

e y

axis

Graph of the function d = 150t -1800

deb

t (d)

in d

olla

rs d

ebt (

d) in

dol

lars

time (t) in months time (t) in months

(12, 0) (12, 0)

Identify the domain and range

deb

t (d)

in d

olla

rs d

ebt (

d) in

dol

lars

time (t) in months time (t) in months

Domain:

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0 ≤ t ≤12

Range:

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−1800 ≤ d ≤ 0

Guided Practice Pgs. 225 – 228

Examples 1 – 5