Quadratic Story Problems

The profit from selling local ballet tickets depends on the ticket price. Using past receipts, we find that the profit can be modeled by the function:

p= -15x² +600x +60

What is the maximum profit you can make from selling tickets?

• $6060

• $20

• $600

• $10,250

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.CED.A.1

CCSS.HSA.SSE.A.1

CCSS.HSA.REI.D.10

CCSS.HSF.IF.B.4

CCSS.HSF.IF.C.7

Heather dropped a water balloon over the side of her school building from a height of 80 feet. The approximate height of the balloon at any point during its fall can be represented by the following quadratic equation:

h = -16t² + 80

About how long did it take for the balloon to hit the ground?

• 1.73 seconds

• 2.24 seconds

• 2.45 seconds

• 2.83 seconds

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.CED.A.1

A ball is thrown into the air with an upward velocity of 40ft/s. Its height h in feet after t seconds is given by the function :

h(t) = -16t² + 40t + 10

How many seconds does it take the ball to reach its maximum height?

• 1.25 seconds

• 1.4 seconds

• 2.5 seconds

• 2 seconds

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.SSE.A.1

CCSS.HSF.IF.B.4

If each mark on the x-axis represents one second, when did the object reach the ground?

• 2.5 seconds

• 6 seconds

• 5.5 seconds

• 5 seconds

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.CED.A.1

CCSS.HSA.REI.D.10

CCSS.HSF.IF.B.4

CCSS.HSF.IF.C.7

CCSS.HSA.REI.D.11

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

What is the height of the platform where the object was launched?

• 64 ft.

• 80 ft.

• 128 ft.

• 144 ft.

Tags

CCSS.HSA.SSE.A.1

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

What is the object's maximum height?

• 200 ft.

• 190 ft.

• 128 ft.

• 144 ft.

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.REI.D.10

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

When will the object reach its maximum height?

• 1 second

• 2 seconds

• 5 seconds

• 8 seconds

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.SSE.A.1

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

When will the object reach the ground?

• 1 second

• 2 seconds

• 5 seconds

• 8 seconds

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.CED.A.1

CCSS.HSA.APR.B.3

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

After 1 second, how high is the object?

• 64 ft.

• 80 ft.

• 128 ft.

• 144 ft.

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.SSE.A.1

An object is launched directly upward at 64 feet per second (ft/s) from an elevated platform. The height of the object represented by the equation

h(t) = –16t² + 64t + 80.

What is the domain?

• all real numbers

• -1 ≤ x ≤ 5

• 0 ≤ x ≤ 144

• 0 ≤ x ≤ 5

Tags

CCSS.HSA.REI.B.4