CorrectionKey=NL-D;CA-D Name Class Date 9.5 Solving...

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Name Class Date

Explore Determining the Possible Number of Solutions of a System of Linear and Quadratic Equations

A system of one linear and one quadratic equation may have zero, one, or two solutions.

A The graph of the quadratic function ƒ (x) = x 2 - 2x - 2 is shown. On the same coordinate plane, graph the following linear functions:

g (x) = -x - 2, h (x) = 2x - 6, j (x) = 0.5x - 5

B Look at the graph of the system consisting of the quadratic function, ƒ (x) , and the linear function, g (x) . Based on the intersections of these two graphs, how many solutions exist in a system consisting of these two functions?

C Look at the graph of the system consisting of the quadratic function, ƒ (x) , and the linear function, h (x) . Based on the intersections of these two graphs, how many solutions exist in a system consisting of these two functions?

D Look at the graph of the system consisting of the quadratic function, ƒ (x) , and the linear function, j (x) . Based on the intersections of these two graphs, how many solutions exist in a system consisting of these two functions?

Reflect

1. A system consisting of a quadratic equation and a linear equation can have , , or solutions.

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Module 9 411 Lesson 5

9.5 Solving Nonlinear SystemsEssential Question: How can you solve a system of equations when one equation is linear

and the other is quadratic?

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Explain 1 Solving a System of Linear and Quadratic Equations Graphically

A system of equations consisting of a linear and quadratic equation can be solved graphically by finding the points where the graphs intersect.

Example 1 Solve the system of equations graphically.

A ⎧ ⎨

⎩ y = (x + 1) 2 - 4

y = 2x - 2

Graph the quadratic function. The vertex is the point (–1, –4) . The x-intercepts are the points where y = 0.

(x + 1) 2 - 4 = 0

(x + 1) 2 = 4

x + 1 = ±2

x = 1 or x = -3

Graph the linear function on the same coordinate plane.

The solutions of the system are the points where the graphs intersect. The solutions are (–1, –4) and (1, 0) .

B ⎧ ⎨

⎩ y = 2 (x - 2) 2 - 2

y = -x - 1

Graph the quadratic function. The vertex is the point

⎛

⎜ ⎝ 2, ⎞

⎟ ⎠ .

The x-intercepts are the points where y = 0.

2 (x - 2) 2 - 2 = 0

2 (x - 2) 2 =

(x - 2) 2 =

= ±1

x = or x =

Graph the linear function on the same coordinate plane.

There are intersection points. This system has solution(s).

y

0-2-4

2

4

-242

x

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6

-2

-4

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Your Turn Solve the system of equations graphically.

2. ⎧ ⎨

⎩ y = -2 (x + 2) 2 + 8

y = 4x + 16

3. ⎧

⎨ ⎩ y = (x + 1) 2 - 9

y = 6x - 12

Explain 2 Solving a System of Linear and Quadratic Equations Algebraically

Systems of equations can also be solved algebraically by using the substitution method to eliminate a variable. If the system is one linear and one quadratic equation, the equation resulting after substitution will also be quadratic and can be solved by selecting an appropriate method.

Example 2 Solve the system of equations algebraically.

A ⎧

⎨ ⎩ y = (x + 1) 2 - 4

y = 2x - 2

Set the two the expressions for y equal to each other, and solve for x.

(x + 1) 2 - 4 = 2x - 2

x 2 + 2x - 3 = 2x - 2

x 2 - 1 = 0

x 2 = 1

x = ±1

Substitute 1 and -1 for x to find the corresponding y-values.

y = 2x - 2 y = 2x - 2

y = 2 (1) - 2 = 0 y = 2 (-1) - 2 = -4

The solutions are (1, 0) and (-1, -4) .

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B ⎧

⎨ ⎩ y = (x + 4) (x + 1)

y = -x - 5

Set the two the expressions for y equal to each other, and solve for x.

(x + 4) (x + 1) =

x 2 + x + = -x - 5

x 2 + x + = 0

(x + ) (x + 3) = 0

x =

Substitute -3 for x to find the corresponding y-value.

y = -x - 5

y = - ( ) - 5 =

The solution is .

Reflect

4. Discussion After finding the x-values of the intersection points, why use the linear equation to find the y-values rather than the quadratic? What if the quadratic equation is used instead?

Your Turn Solve the system of equations algebraically.

5. ⎧

⎨ ⎩ y = 2 x 2 + 9x + 5

y = 3x - 3



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Explain 3 Solving a Real-World Problem with a System of Linear and Quadratic Equations

Systems of equations can be solved by graphing both equations on a graphing calculator and using the Intersect feature.

Example 3 Create and solve a system of equations to solve the problem.

A A rock climber is pulling his pack up the side of a cliff that is 175.5 feet tall at a rate of 2 feet per second. The height of the pack in feet after t seconds is given by h = 2t. The climber drops a coil of rope from directly above the pack. The height of the coil in feet after t seconds is given by h = -16 t 2 + 175.5. At what time does the coil of rope hit the pack?

Create the system of equations to solve.

⎧

⎨ ⎩ h = -16t 2 + 175.5

h = 2t

Graph the functions together and find any points of intersection.

The intersection is at (-3.375, -6.75) . The intersection is at (3.25, 6.5) .

The x-value represents time, so this This solution indicates that the coil hits solution is not reasonable. the pack after 3.25 seconds.

B A window washer is ascending the side of a building that is 520 feet tall at a rate of 3 feet per second. The elevation of the window washer after t seconds is given by h = 3t. The supplies are lowered to the window washer from the top of the building at the same time that he begins to ascend the building. The height of the supplies in feet after t seconds is given by h = -2 t 2 + 520. At what time do the supplies reach the window washer?

Create the system of equations to solve.

⎧

⎨

⎩

h = t 2 +

h = t



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Graph the functions together and find any points of intersection.

The intersection is at about ⎛

⎜ ⎝ ,

⎞

⎟ ⎠ . The intersection is at about

⎛

⎜ ⎝ ,

⎞

⎟ ⎠ .

The x-value represents , so this This solution indicates that

solution is

.

Reflect

6. How did you know which intersection to use in the example problems?

Your Turn Write and solve a system of equations to solve the problem.

7. A billboard painter is using a pulley system to hoist a can of paint up to a scaffold at a rate of half a meter per second. The height of the can of paint as a function of time is given by h (t) = 0.5t. Five seconds after he starts raising the can of paint, his partner accidentally kicks a paint brush off of the scaffolding, which falls to the ground. The height of the falling paint brush can be represented by h (t) = -4.9 (t - 5) 2 + 30. When does the brush pass the paint can?



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Elaborate

8. Discussion When solving a system of equations consisting of a quadratic equation and a linear equation by graphing, why is it difficult to be sure there is one solution as opposed to 0 or 2?

9. How can you use the discriminant to determine how many solutions a linear-quadratic system has?

10. Essential Question Check-in How can the graphs of two functions be used to solve a system of a quadratic and a linear equation?

1. The graph of the function ƒ (x) = - 1 _ 4 (x - 3) 2 + 4 is shown. Graph the functions g (x) = x + 1, h (x) = x + 2, and j (x) = x + 3 with the graph of ƒ (x) , and determine how many solutions each system has.

ƒ (x) and g (x) :

ƒ (x) and h (x) :

ƒ (x) and j (x) :

Solve each system of equations graphically.

2. ⎧

⎨ ⎩ y = (x + 3) 2 - 4

y = 2x + 2

3. ⎧

⎨ ⎩ y = x 2 - 1

y = x - 2

• Online Homework• Hints and Help• Extra Practice

Evaluate: Homework and Practice

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4. ⎧

⎨ ⎩ y = (x - 4) 2 - 2

y = -2

5. ⎧

⎨ ⎩ y = - x 2 + 4

y = -3x + 6

6. ⎧

⎨ ⎩ y = - (x - 2) 2 + 9

y = 3x + 3

7. ⎧

⎨ ⎩ y = 3 (x + 1) 2 - 1

y = x - 4

Solve the system of equations algebraically.

8. ⎧

⎨ ⎩ y = x 2 + 1

y = 5

9. ⎧

⎨ ⎩ y = x 2 - 3x + 2

y = 4x - 8

10. ⎧

⎨ ⎩ y = (x - 3) 2

y = 4

11. ⎧

⎨ ⎩ y = - x 2 + 4x

y = x + 2

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12. ⎧

⎨ ⎩ y = 2 x 2 - 5x + 6

y = 5x - 6

13. ⎧

⎨ ⎩ y = x 2 + 7

y = -9x + 29

14. ⎧

⎨ ⎩ y = 4 x 2 + 45x + 83

y = 5x - 17

15. ⎧

⎨ ⎩ y = (x + 2) (x + 4)

y = 3x + 2

Create and solve a linear quadratic system to solve the problem.

16. The height in feet of a skydiver t seconds after deploying her parachute is given by h (t) = -300t + 1000. A ball is thrown up toward the skydiver, and after t seconds, the height of the ball in feet is given by h (t) = -16 t 2 + 100t. When does the ball reach the skydiver?

17. A wildebeest fails to notice a lion that is charging from behind at 65 feet per second until the lion is 40 feet away. The lion’s position as a function of time is given by p (t) = 65t - 40. The wildebeest has to begin accelerating from a standstill until it is captured or reaches a top speed fast enough to stay ahead of the lion. The wildebeest’s position as a function of time is given by d (t) = 35 t 2 . Does the wildebeest escape?



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18. An elevator in a hotel moves at 20 feet per second. Leaving from the ground floor, its height in feet after t seconds is given by the formula h (t) = 20t. A bolt comes loose in the elevator shaft above, and its height in feet after falling for t seconds is given by h (t) = -16 t 2 + 200. At what time and at what height does the bolt hit the elevator?

19. A bungee jumper leaps from a bridge 100 meters over a gorge. Before the 40-meter-long bungee begins to slow him down, his height is characterized by h (t) = -4.9 t 2 + 100. Two seconds after he jumps, a car on the bridge blows out a tire. The sound of the tire blow-out moves down from the top of the bridge at the speed of sound and has a height given by h (t) = -340 (t - 2) + 100. How high will the bungee jumper be when he hears the sound of the blowout?

20. Explain the Error A student is asked to solve the system of equations y = x 2 + 2x - 7 and y - 2 = x + 1. For the first step, the student sets the right hand sides equal to each other to get the equation x 2 + 2x - 7 = x + 1. Why does this not give the correct solution?

21. Explain the Error After solving the system of equations in Exercise 18 (the elevator and the bolt), a student concludes that there are two different times that the bolt hits the elevator. What is the error in the student’s reasoning?

22. Multi-part ClassificationThe functions listed are graphed here. f 1 (x) = 2 (x + 3) 2 + 1 and f 2 (x) = - 3 _ 4 (x - 2) 2 + 3

g 1 (x) = x + 3 and g 2 (x) = 3 and g 3 (x) = - 1 _ 2 x + 1

Use the graph to classify each system as having 0, 1, or 2 solutions.

a. ⎧ ⎨ ⎩ y = ƒ 1 (x)

y = g 1 (x)

b.

⎧ ⎨ ⎩ y = ƒ 1 (x)

y = g 2 (x)

c.

⎧ ⎨ ⎩ y = ƒ 1 (x)

y = g 3 (x)

d. ⎧ ⎨ ⎩ y = ƒ 2 (x)

y = g 1 (x)

e.

⎧ ⎨ ⎩ y = ƒ 2 (x)

y = g 2 (x)

f.

⎧ ⎨ ⎩ y = ƒ 2 (x)

y = g 3 (x)

y

0-2

4

2

x



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H.O.T. Focus on Higher Order Thinking

23. Explain the Error After solving the system of equations in Exercise 16 (the skydiver and the ball), a student concludes there are two valid solutions because they both have positive times. The ball must pass by the skydiver twice. What is the error in the student’s reasoning?

24. Multi-Part Problem The path of a baseball hit for a home run can be modeled by y = - x 2 _ 484 + x + 3, where x and y are in feet and home plate is at the origin.

The ball lands in the stands, which are modeled by 4y - x = -352 for x ≥ 400. Use a graphing calculator to graph the system.

a. What do the variables x and y represent?

b. About how far is the baseball from home plate when it lands?

c. About how high up in the stands does the baseball land?

25. Draw Conclusions A certain system of a linear and a quadratic equation has two solutions, (2, 7) and (5, 10) . The quadratic equation is y = x 2 - 6x + 15. What is the linear equation? Justify your answer.

26. Justify Reasoning It is possible for a system of two linear equations to have infinitely many solutions. Explain why this is not possible for a system with one linear and one quadratic equation.



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Lesson Performance TaskA race car leaves pit row at a speed of 40 feet per second and accelerates at a constant rate of 44 feet per second squared. Its distance from the pit exit is given by the function d r (t) = 22 t 2 + 40t. The race car leaves ahead of an approaching pace car traveling at a constant speed of 120 feet per second. In each case, find out if the pace car will catch up to the race car, and if so, how far down the track it will catch up. If there is more than one solution, explain how you know which one to select.

a. The pace car passes by the exit to pit row 1 second after the race car exits.

b. The pace car passes the exit half a second after the race car exits.



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