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

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.

Example 3: Sports 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? h(t) = –16t2 + v0t + h0 Write the general projectile function. h(t) = –16t2 + 80t + 0 Substitute 80 for v0 and 0 for h0.

The ball will hit the ground when its height is zero. Example 3 Continued The ball will hit the ground when its height is zero. –16t2 + 80t = 0 Set h(t) equal to 0. –16t(t – 5) = 0 Factor: The GCF is –16t. –16t = 0 or (t – 5) = 0 Apply the Zero Product Property. t = 0 or t = 5 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.

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 Write the general projectile function. h(t) = –16t2 + 48t + 0 Substitute 48 for v0 and 0 for h0.

Check It Out! Example 3 Continued The ball will hit the ground when its height is zero. –16t2 + 48t = 0 Set h(t) equal to 0. –16t(t – 3) = 0 Factor: The GCF is –16t. –16t = 0 or (t – 3) = 0 Apply the Zero Product Property. t = 0 or t = 3 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.

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?

Understand the Problem Example 4 Continued 1 Understand the Problem 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.

Example 4 Continued 2 Make a Plan Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.

Find the critical values by solving the related equation. Example 4 Continued Solve 3 Write the inequality. –8x2 + 600x – 4200 ≥ 6000 Find the critical values by solving the related equation. –8x2 + 600x – 4200 = 6000 Write as an equation. –8x2 + 600x – 10,200 = 0 Write in standard form. –8(x2 – 75x + 1275) = 0 Factor out –8 to simplify.

Solve x ≈ 26.04 or x ≈ 48.96 Example 4 Continued 3 Use the Quadratic Formula. Simplify. x ≈ 26.04 or x ≈ 48.96

Example 4 Continued Solve 3 Test an x-value in each of the three regions formed by the critical x-values. Critical values 10 20 30 40 50 60 70 Test points

Example 4 Continued Solve 3 –8(25)2 + 600(25) – 4200 ≥ 6000 Try x = 25. 5800 ≥ 6000 x –8(45)2 + 600(45) – 4200 ≥ 6000 Try x = 45. 6600 ≥ 6000  –8(50)2 + 600(50) – 4200 ≥ 6000 Try x = 50. 5800 ≥ 6000 x Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.

Example 4 Continued Solve 3 For a profit of $6000, the average price of a helmet needs to be between $26.04 and $48.96, inclusive.

Example 4 Continued Look Back 4 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.

Check It Out! Example 4 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?

Understand the Problem Check It Out! Example 4 Continued 1 Understand the Problem The 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 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 Solve 3 Write the inequality. –25x2 + 1250x – 5000 < 7500 Find the critical values by solving the related equation. –25x2 + 1250x – 5000 = 7500 Write as an equation. –25x2 + 1250x – 12,500 = 0 Write in standard form. –25(x2 – 50x + 500) = 0 Factor out –25 to simplify.

Check It Out! Example 4 Continued Solve 3 Use the Quadratic Formula. Simplify. x ≈ 13.82 or x ≈ 36.18

Check It Out! Example 4 Continued Solve 3 Test an x-value in each of the three regions formed by the critical x-values. Critical values 5 10 15 20 25 30 35 Test points

Check It Out! Example 4 Continued Solve 3 –25(13)2 + 1250(13) – 5000 < 7500 Try x = 13. 7025 < 7500  –25(30)2 + 1250(30) – 5000 < 7500 Try x = 30. 10,000 < 7500 x –25(37)2 + 1250(37) – 5000 < 7500 Try x = 37. 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.

Check It Out! Example 4 Continued Solve 3 The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people.

Check It Out! Example 4 Continued Look Back 4 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.