algebra 2 section 2-7

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Section 2-7 Solving Equations by Graphing

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Page 1: Algebra 2 Section 2-7

Section 2-7Solving Equations by Graphing

Page 2: Algebra 2 Section 2-7

Essential Questions

• How do you find x- and y-intercepts of functions?

• How do you solve equations by examining graphs of the related functions?

Page 3: Algebra 2 Section 2-7

Vocabulary1. Intercepts:

2. y-intercept:

3. x-intercept:

4. Root of the Equation:

5. Zero of a Function:

Page 4: Algebra 2 Section 2-7

Vocabulary1. Intercepts: Points where a graph crosses an axis

2. y-intercept:

3. x-intercept:

4. Root of the Equation:

5. Zero of a Function:

Page 5: Algebra 2 Section 2-7

Vocabulary1. Intercepts: Points where a graph crosses an axis

2. y-intercept: Point where a graph crosses the y-axis

3. x-intercept:

4. Root of the Equation:

5. Zero of a Function:

Page 6: Algebra 2 Section 2-7

Vocabulary1. Intercepts: Points where a graph crosses an axis

2. y-intercept: Point where a graph crosses the y-axis

3. x-intercept: Point where a graph crosses the x-axis

4. Root of the Equation:

5. Zero of a Function:

Page 7: Algebra 2 Section 2-7

Vocabulary1. Intercepts: Points where a graph crosses an axis

2. y-intercept: Point where a graph crosses the y-axis

3. x-intercept: Point where a graph crosses the x-axis

4. Root of the Equation: The solution to an equation

5. Zero of a Function:

Page 8: Algebra 2 Section 2-7

Vocabulary1. Intercepts: Points where a graph crosses an axis

2. y-intercept: Point where a graph crosses the y-axis

3. x-intercept: Point where a graph crosses the x-axis

4. Root of the Equation: The solution to an equation

5. Zero of a Function: Another name for the root of the equation; The value of x for which f(x) = 0

Page 9: Algebra 2 Section 2-7

Example 1Find the x- and y-intercepts of the graph.

y = 45x 2 + 12

5x − 8

Page 10: Algebra 2 Section 2-7

Example 1Find the x- and y-intercepts of the graph.

y = 45x 2 + 12

5x − 8

x-intercepts:

Page 11: Algebra 2 Section 2-7

Example 1Find the x- and y-intercepts of the graph.

y = 45x 2 + 12

5x − 8

x-intercepts:(-5, 0), (2, 0)

Page 12: Algebra 2 Section 2-7

Example 1Find the x- and y-intercepts of the graph.

y = 45x 2 + 12

5x − 8

x-intercepts:

y-intercept:(-5, 0), (2, 0)

Page 13: Algebra 2 Section 2-7

Example 1Find the x- and y-intercepts of the graph.

y = 45x 2 + 12

5x − 8

x-intercepts:

y-intercept:(-5, 0), (2, 0)

(0, -8)

Page 14: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

Page 15: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 16: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 17: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 18: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 19: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 20: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 21: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 22: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 23: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

Page 24: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

The root is at x = -6

Page 25: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.a. 0 = 1

3x + 2

x

y

The root is at x = -6

The x-intercept is at (-6, 0)

Page 26: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

Page 27: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 28: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 29: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 30: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 31: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 32: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

Page 33: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

The root is at x = 3

Page 34: Algebra 2 Section 2-7

Example 2Find the root of each equation by graphing the related

function.b. 0 = 5x −15

xy

The root is at x = 3

The x-intercept is at (3, 0)

Page 35: Algebra 2 Section 2-7

Example 3Matt Mitarnowski found a service that allows him to download albums to his smart phone.

Each album costs $1.25 to download. He also paid a subscription fee of $8 to access or just stream the music. Someone needs to inform

him of Spotify Premium, where there is no extra fee on top of the subscription fee to sync music.

If the total cost for the download and subscription fee was $15.50, how many albums did he download? Solve by graphing the related

function.

Page 36: Algebra 2 Section 2-7

Example 3

Page 37: Algebra 2 Section 2-7

Example 3a = albums

Page 38: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

Page 39: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.50

Page 40: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 0

Page 41: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

Page 42: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Page 43: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Page 44: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Page 45: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Page 46: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Page 47: Algebra 2 Section 2-7

Example 3

8 +1.25a = 15.50

a = albums

−15.50−15.501.25a − 7.5 = 054a − 15

2= 0

a

c

Matt bought 6 albums