Bell Quiz 2-6 1. Horizontal asymptotes = ? 2. Vertical asymptotes = ? 3. Zeroes = ? 4. What is the “end behavior” of.

Bell Quiz 2-6Bell Quiz 2-61. Horizontal asymptotes = ? 1. Horizontal asymptotes = ?

2. Vertical asymptotes = ?2. Vertical asymptotes = ?

3. Zeroes = ?3. Zeroes = ?

4

12

)(

)(2

2

x

x

xg

xf

4. What is the “end behavior” of4. What is the “end behavior” of

44

13)(

2

3

x

xxf

2.7 2.7 Solving Equations in Solving Equations in oneone variable.variable.

What you’ll learn aboutWhat you’ll learn about Solving Rational EquationsSolving Rational Equations Extraneous SolutionsExtraneous Solutions ApplicationsApplications

… … and whyand whyApplications involving rational functions as Applications involving rational functions as models often require that an equation involving models often require that an equation involving fractions be solved.fractions be solved.

Extraneous Solutions Extraneous Solutions When we divide an equation by an expression containing When we divide an equation by an expression containing variables, the resulting equation may have solutions that variables, the resulting equation may have solutions that are are notnot solutions of the original equation. solutions of the original equation.

These are These are extraneous solutionsextraneous solutions. For this reason we must . For this reason we must check each solution of the resulting equation in the original check each solution of the resulting equation in the original equation.equation.

Fraction ReviewFraction Review6

1

3

2

4

2

x1. Eliminate the denominator by either: 1. Eliminate the denominator by either: (1) eliminate each denomator one at a time (1) eliminate each denomator one at a time (2) Multiply by the LCD (2) Multiply by the LCD (using the property of equality)(using the property of equality)

3. Solve the resulting equation.3. Solve the resulting equation.

1

4

6

1

3

2

4

2

1

4

x

6

4

3

8

1

2

x

1

3

6

4

3

8

1

2

1

3

x

6

12

1

8

1

6

x 286 x

1x

Solving by Clearing FractionsSolving by Clearing Fractions

2Solve 3.x

x

1. Eliminate the denominator.1. Eliminate the denominator. 2. Multiply by the LCD 2. Multiply by the LCD (using the property of equality)(using the property of equality)

3. Solve the resulting quadratic equation.3. Solve the resulting quadratic equation.

4.4. Check for Check for extraneousextraneous solutions (if you multiplied or divided solutions (if you multiplied or divided by an expression with a variable in it).by an expression with a variable in it).

)(3)2

)(( xx

xx

xx 322

0232 xx 0)1)(2( xx 2,1x

Your turn: solve for ‘x’.Your turn: solve for ‘x’.

1.1.3

1

3

5

3

2

xx

2.2.

3

12

3

4

xx

xx

xx

145

3.3.

Eliminating Extraneous Eliminating Extraneous SolutionsSolutions

2

1 2 2Solve the equation .

3 1 4 3

x

x x x x

1.1. These are These are fractionsfractions. Get a common denominator.. Get a common denominator.

34

2

1

2

3

12

xxx

x

x )1)(3(

2

1

2

3

1

xxx

x

x

)1)(3(

2

1

2

3

1

1

1

xxx

x

xx

x

)1)(3(

2

1

2

3

3

)3)(1(

1

xxx

x

x

x

xx

x

Eliminating Extraneous Eliminating Extraneous SolutionsSolutions

)1)(3(

2

1

2

3

3

)3)(1(

1

xxx

x

x

x

xx

x

)1)(3(

2

)1)(3(

62

)3)(1(

1 2

xxxx

xx

xx

x

Now that you have a common denominator Now that you have a common denominator multiple both multiple both sides by the common denominator.sides by the common denominator.

2)62()1( 2 xxx

Eliminating Extraneous Eliminating Extraneous SolutionsSolutions

2. Solve the resulting quadratic equation.2. Solve the resulting quadratic equation.

4. Check for 4. Check for extraneousextraneous solutions (if you multiplied or divided solutions (if you multiplied or divided by an expression with a variable in it).by an expression with a variable in it).

2)62()1( 2 xxx

0352 2 xx

35.0 andx

)1)(3(

2

1

2

3

1

xxx

x

xIf x = 3, the result will be division by zero If x = 3, the result will be division by zero x = 3 is an x = 3 is an extraneous solution and must be rejected. extraneous solution and must be rejected.

Your turn: Your turn: solve for ‘x’, check solve for ‘x’, check for extraneous roots.for extraneous roots.

4.4.103

7

2

1

5

32

xxxx

x

5.5.xxxx

x

2

2

1

42

Finding a Minimum Finding a Minimum PerimeterPerimeter

1. Find the dimensions of the rectangle with 1. Find the dimensions of the rectangle with minimumminimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters.

2. Find this least perimeter.2. Find this least perimeter.

Area = 300x

What is the length What is the length of this side ? of this side ?

Can you write an Can you write an expression that usesexpression that uses the length of 1 side the length of 1 side and the given area? and the given area?

Finding a Minimum Finding a Minimum PerimeterPerimeter

1. Find the dimensions of the rectangle with 1. Find the dimensions of the rectangle with minimumminimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters.

2. Find this least perimeter.2. Find this least perimeter.

Area = 300xA= L*wA= L*w

300 = x * w300 = x * w

w= 300/xw= 300/x x

300

Finding a Minimum PerimeterFinding a Minimum Perimeter1. Find the dimensions of the rectangle with 1. Find the dimensions of the rectangle with minimumminimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters.

2. Find this least perimeter.2. Find this least perimeter.

Area = 300xPerimeter = ?Perimeter = ?

Solve the resulting equation graphically.Solve the resulting equation graphically. x

300

xxxxP

300300

xxP

6002

Minimum perimeter = 69.3Minimum perimeter = 69.3

Finding a Minimum PerimeterFinding a Minimum Perimeter

1. Find the dimensions of the rectangle with 1. Find the dimensions of the rectangle with minimumminimum perimeter, if its area is 300 square meters. perimeter, if its area is 300 square meters.

Area = 300x

Solve graphically.Solve graphically.

x

300

xx

60023.69

6003.6920 2 xx

7.17300*7.17300 9.16

300*9.16300

7.179.16300

9.167.17300

17.716.9

Your turn:Your turn:

6. Find the minimum perimeter of this rectangle.6. Find the minimum perimeter of this rectangle.

Area = 2505x

Finding the Dimensions of a Finding the Dimensions of a CanCanA juice company uses 2 liter cans A juice company uses 2 liter cans vol = 2000 cubic cm. vol = 2000 cubic cm.

The can has a surface area of 1000 square cm.The can has a surface area of 1000 square cm.

Find the radius and height of the can.Find the radius and height of the can.

1. Find an equation for volume.1. Find an equation for volume.

2. Find an equation for surface area.2. Find an equation for surface area.

20002 hrV

100022 2 rhrS

Finding the Dimensions of a Finding the Dimensions of a CanCanA juice company uses 2 liter cans A juice company uses 2 liter cans vol = 2000 cubic cm. vol = 2000 cubic cm.

The can has a surface area of 1000 square cm.The can has a surface area of 1000 square cm.

Find the radius and height of the can.Find the radius and height of the can.

20002 hr 100022 2 rhr

3. Use the 3. Use the substitutionsubstitution method to write a method to write a single equation with one variable.single equation with one variable.

4. Solve graphically.4. Solve graphically.

2

2000

rh

10002000

222

2 r

rr

10004000

2 2 r

r

010004000

2 2 r

r

Finding the Dimensions of a Finding the Dimensions of a CanCanA juice company uses 2 liter cans A juice company uses 2 liter cans vol = 2000 cubic cm. vol = 2000 cubic cm.

The can has a surface area of 1000 square cm.The can has a surface area of 1000 square cm.

Find the radius and height of the can.Find the radius and height of the can.

4. Solve graphically.4. Solve graphically.010004000

2 2 r

r

Radius = 4.6 cm or 9.7 cm.Radius = 4.6 cm or 9.7 cm.

2

2000

rh

If radius = 4.6 cm then height = 29.8 cmIf radius = 4.6 cm then height = 29.8 cm

If radius = 9.7 cm then height = 6.83 cmIf radius = 9.7 cm then height = 6.83 cm

HOMEWORKHOMEWORK

Section 2-7Section 2-7