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Quadratic Application Objectives I can solve real life situations represented by quadratic equations.

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Quadratic Application. Objectives I can solve real life situations represented by quadratic equations. - PowerPoint PPT Presentation

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Page 1: Quadratic Application

Quadratic Application

ObjectivesI can solve real life situations

represented by quadratic equations.

Page 2: Quadratic Application

Any object that is thrown or launched into the air, such as a baseball, basketball, or soccer ball, is a projectile. The general function that approximates the height h in feet of a projectile on Earth after t seconds is given.

Note that this model has limitations because it does not account for air resistance, wind, and other real-world factors.

Page 3: Quadratic Application

A golf ball is hit from ground level with an initial vertical velocity of 80 ft/s. After how many seconds will the ball hit the ground?

Example 3: Sports Application

h(t) = –16t2 + v0t + h0

h(t) = –16t2 + 80t + 0

Write the general projectile function.

Substitute 80 for v0 and 0 for h0.

Page 4: Quadratic Application

Example 3 Continued

The ball will hit the ground when its height is zero.

–16t2 + 80t = 0–16t(t – 5) = 0–16t = 0 or (t – 5) = 0t = 0 or t = 5

Set h(t) equal to 0.

Factor: The GCF is –16t.

Apply the Zero Product Property.

Solve each equation.

The golf ball will hit the ground after 5 seconds. Notice that the height is also zero when t = 0, the instant that the golf ball is hit.

Page 5: Quadratic Application

Check It Out! Example 3

A football is kicked from ground level with an initial vertical velocity of 48 ft/s. How long is the ball in the air?

h(t) = –16t2 + v0t + h0

h(t) = –16t2 + 48t + 0

Write the general projectile function.

Substitute 48 for v0 and 0 for h0.

Page 6: Quadratic Application

Check It Out! Example 3 Continued

The ball will hit the ground when its height is zero.

–16t2 + 48t = 0–16t(t – 3) = 0–16t = 0 or (t – 3) = 0t = 0 or t = 3

Set h(t) equal to 0.

Factor: The GCF is –16t.

Apply the Zero Product Property.

Solve each equation.

The football will hit the ground after 3 seconds. Notice that the height is also zero when t = 0, the instant that the football is hit.

Page 7: Quadratic Application

Example 4: Problem-Solving Application

The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x2 + 600x – 4200, where x is the average selling price of a helmet. What range of selling prices will generate a monthly profit of at least $6000?

Page 8: Quadratic Application

1 Understand the Problem

Example 4 Continued

The answer will be the average price of a helmet required for a profit that is greater than or equal to $6000.List the important information:• The profit must be at least $6000.• The function for the business’s profit

is P(x) = –8x2 + 600x – 4200.

Page 9: Quadratic Application

2 Make a Plan

Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.

Example 4 Continued

Page 10: Quadratic Application

Solve3

Write the inequality.–8x2 + 600x – 4200 ≥ 6000

–8x2 + 600x – 4200 = 6000

Find the critical values by solving the related equation.

Write as an equation.

Write in standard form.

Factor out –8 to simplify.

–8x2 + 600x – 10,200 = 0

–8(x2 – 75x + 1275) = 0

Example 4 Continued

Page 11: Quadratic Application

Solve3

Use the Quadratic Formula.

Simplify.

x ≈ 26.04 or x ≈ 48.96

Example 4 Continued

Page 12: Quadratic Application

Solve3

Test an x-value in each of the three regions formed by the critical x-values.

10 20 30 40 50 60 70

Critical values

Test points

Example 4 Continued

Page 13: Quadratic Application

Solve3

–8(25)2 + 600(25) – 4200 ≥ 6000

–8(45)2 + 600(45) – 4200 ≥ 6000

–8(50)2 + 600(50) – 4200 ≥ 6000

5800 ≥ 6000Try x = 25.

Try x = 45.

Try x = 50.

6600 ≥ 6000

5800 ≥ 6000

Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.

x

x

Example 4 Continued

Page 14: Quadratic Application

Solve3

For a profit of $6000, the average price of a helmet needs to be between $26.04 and $48.96, inclusive.

Example 4 Continued

Page 15: Quadratic Application

Look Back4

Enter y = –8x2 + 600x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between 26.04 and 48.96 inclusive result in y-values greater than or equal to 6000.

Example 4 Continued

Page 16: Quadratic Application

A business offers educational tours to Patagonia, a region of South America that includes parts of Chile and Argentina . The profit P for x number of persons is P(x) = –25x2 + 1250x – 5000. The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled?

Check It Out! Example 4

Page 17: Quadratic Application

1 Understand the ProblemThe answer will be the number of people signed up for the trip if the profit is less than $7500.

List the important information:• The profit will be less than $7500.• The function for the profit is P(x) = –25x2 + 1250x – 5000.

Check It Out! Example 4 Continued

Page 18: Quadratic Application

2 Make a Plan

Write an inequality showing profit less than $7500. Then solve the inequality by using algebra.

Check It Out! Example 4 Continued

Page 19: Quadratic Application

Solve3

Write the inequality.

–25x2 + 1250x – 5000 < 7500

–25x2 + 1250x – 5000 = 7500

Find the critical values by solving the related equation.

Write as an equation.

Write in standard form.

Factor out –25 to simplify.

–25x2 + 1250x – 12,500 = 0

–25(x2 – 50x + 500) = 0

Check It Out! Example 4 Continued

Page 20: Quadratic Application

Simplify.

x ≈ 13.82 or x ≈ 36.18

Use the Quadratic Formula.

Solve3

Check It Out! Example 4 Continued

Page 21: Quadratic Application

Test an x-value in each of the three regions formed by the critical x-values.

5 10 15 20 25 30 35

Critical values

Test points

Solve3

Check It Out! Example 4 Continued

Page 22: Quadratic Application

–25(13)2 + 1250(13) – 5000 < 7500

7025 < 7500Try x = 13.

Try x = 30.

Try x = 37.

10,000 < 7500

7025 < 7500

Write the solution as an inequality. The solution is approximately x > 36.18 or x < 13.82. Because you cannot have a fraction of a person, round each critical value to the appropriate whole number.

x

–25(30)2 + 1250(30) – 5000 < 7500

–25(37)2 + 1250(37) – 5000 < 7500

Solve3

Check It Out! Example 4 Continued

Page 23: Quadratic Application

The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people.

Solve3

Check It Out! Example 4 Continued

Page 24: Quadratic Application

Look Back4

Enter y = –25x2 + 1250x – 5000 into a graphing calculator, and create a table of values. The table shows that integer values of x less than 13.81 and greater than 36.18 result in y-values less than 7500.

Check It Out! Example 4 Continued