Download Quadratic Application

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.
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) = 0
t = 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.
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?
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.
Example 4 Continued
3
Solve
Write the inequality.
–8x2 + 600x – 4200 ≥ 6000
Find the critical values by solving the related equation.
–8x2 + 600x – 4200 = 6000
–8x2 + 600x – 10,200 = 0
–8(x2 – 75x + 1275) = 0
Write as an equation.
Write in standard form.
Factor out –8 to simplify.
Example 4 Continued
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 26.04 or x ≈ 48.96
Example 4 Continued
3
Solve
Test an x-value in each of the three regions
formed by the critical x-values.
Critical values
10
20
30
40
Test points
50
60
70
Example 4 Continued
3
Solve
–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
3
Solve
For a profit of $6000, the average price of a helmet
needs to be between $26.04 and $48.96, inclusive.
Example 4 Continued
4
Look Back
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?
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
3
Solve
Write the inequality.
–25x2 + 1250x – 5000 < 7500
Find the critical values by solving the related equation.
–25x2 + 1250x – 5000 = 7500
–25x2 + 1250x – 12,500 = 0
–25(x2 – 50x + 500) = 0
Write as an equation.
Write in standard form.
Factor out –25 to simplify.
Check It Out! Example 4 Continued
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 13.82 or x ≈ 36.18
Check It Out! Example 4 Continued
3
Solve
Test an x-value in each of the three regions
formed by the critical x-values.
Critical values
5
10
15
20
25
30
Test points
35
Check It Out! Example 4 Continued
3
Solve
–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
3
Solve
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
4
Look Back
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.