Sign In
| 7 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
When Fantastic Car 1
was released in 1970, the price of the car was about $3500. Since then, the way its market value changed over the years can be described by the following graph.
Here, C represents the price of the car in thousands of dollars and t the number of years since 1970. In 1985, it has been reported that the Fantastic Car 1 was being sold for $11000.
x=0
Calculate power
Zero Property of Multiplication
Identity Property of Addition
f(x)=0
Use the Quadratic Formula: a=-4,b=9,c=5
Calculate power
a(-b)=-a⋅b
(-a)b=-ab
a−(-b)=a+b
State solutions
(I), (II): Use a calculator
(I), (II): Add and subtract terms
(I), (II): Use a calculator
(I), (II): Round to 1 decimal place(s)
x=1.125
Round to 1 decimal place(s)
Tearrik's ball: about 4 seconds
Split into factors
Factor out -16
a=a+(23)2−(23)2
a=22⋅a
Split into factors
ca⋅b=a⋅cb
Commutative Property of Multiplication
(a−b)2=a2−2ab+b2
Distribute -16
(ba)m=bmam
a⋅cb=ca⋅b
Calculate quotient
16⋅16a=a
Add terms
ba=a÷b
M(x)=0
Use the Quadratic Formula: a=-16,b=48,c=5
Calculate power
a(-b)=-a⋅b
(-a)b=-ab
a−(-b)=a+b
Calculate root
State solutions
(I), (II): Add and subtract terms
(I): Put minus sign in front of fraction
(II): -b-a=ba
(I), (II): Use a calculator
(I), (II): Round to 1 decimal place(s)
The ball will reach a height of 0 feet after around 3.8. Rounded to the nearest integer, this value approximates to 4. Therefore, Tearrik's ball is airborne for about 4 seconds, while Mark's ball is airborne for about 3 seconds.
x=0
Subtract term
Calculate power
(-a)b=-ab
Add terms
Finally, by connecting the three plotted points with a smooth curve, the parabola can be drawn.
As stated in Part A, it can be seen that Tearrik throws the ball higher than Mark.
Zosia's friend Emily takes her half-day off from school to visit the furniture club. A speaker from a local furniture company comes to teach the club about their business.
Tadeo takes a half-day off from school to visit the ice skating club. The rink is under construction! Tadeo, interested, reads the plan posted on the closed entrance doors. It will be a rectangular ice rink with dimensions 40 by 50 meters. According to the plan, there should be a sidewalk around the rink.
Use a variable to denote the width of the sidewalk and write the expressions for the dimensions of the rink. Then solve the quadratic equation for the area of the rink.
Let x be the width of the sidewalk around the ice skating rink.
A=1575
Distribute (40−2x)
Distribute 50&2x
Subtract term
Commutative Property of Addition
LHS−1575=RHS−1575
Finally, the challenge presented at the beginning of the lesson can be solved. When the Fantastic Car 1
was released in 1970, the price of the car was about $3500. Since then the way its market value changed over the years can be described by the following graph.
Here, C represents the price of the car in thousands of dollars and t the number of years since 1970. In 1985, it has been reported that the Fantastic Car 1 was being sold for $11000.
This means that 25 years after 1970 — in the year 1995 — the car had a price of $41000.
(I), (II): Calculate power
(I), (II): LHS−3.5=RHS−3.5
(I), (II): Multiply
(I): LHS−625a=RHS−625a
(I): LHS/25=RHS/25
(I): Write as a difference of fractions
(I): ca⋅b=ca⋅b
(I): Calculate quotient
(I): Rearrange equation
(II): b=1.5−25a
(II): Distribute 15
(II): Subtract term
(II): LHS−22.5=RHS−22.5
(II): LHS/(-150)=RHS/(-150)
(II): Rearrange equation
(I): a=0.1
x=50
Calculate power
Multiply
Add and subtract terms
We are asked about the profit of the company. The profit is defined as the company's revenue minus the total costs. P(x) = R(x) - C(x) Let's substitute the given expressions for R(x) and C(x) to write the profit in terms of x-variable.
We need to determine for what values of x the profit is greater than 0. Since the leading coefficient in the profit function is negative, the function reaches a maximum value. Depending on this value, the function may or may not reach 0. We will now determine whether there are any zeros by setting the function equal to zero.
Because the obtained quadratic equation is written in standard form, we will use the Quadratic Formula to solve it for x.
Using the Quadratic Formula, we found that the solutions of the equation are x= -2.9 ± sqrt(5.41)-0.1. Let's separate the two cases and use a calculator to find the values of x for which the profit is 0.
x = -2.9 ± sqrt(5.41)/-0.1 | |
---|---|
x_1 = -2.9 + sqrt(5.41)/-0.1 | x_2 = -2.9 - sqrt(5.41)/-0.1 |
x_1 ≈ 6 | x_2 ≈ 52 |
We obtained that there are two boundary values of x, for which the profit is 0. Now, consider the values of the function for x between 6 and 52. Remember that the function reaches a maximum value and there are two zeros.
The values of the profit function are positive for the values of x between about 6 and 52. This means that the company should make more than 6 and less than 52 million of cinnamon rolls to profit.
The maximum profit for the company is reached at the maximum of the function P(x). The maximum point of a quadratic function is its vertex, which lies on the axis of symmetry.
In Part A, we found that the zeros of the functions are at x_1≈ 6 and x_2≈ 52. By using these values, we can find the equation of the axis of symmetry x = x_1+x_22.
Finally, we will calculate the maximum value of the profit function by substituting x=29 into the function rule.
Therefore, the maximum profit of the company is 27 million dollars.
The company Jump and Bounce NY sells rectangular trampolines. For each trampoline, its longer side is twice as long as its shorter side. The company recommends that there is a 2-meter wide security zone around the trampoline and that the security zone area must be at least three times larger than the area of the trampoline.
The length of the trampoline is two times greater than the width. This means that if we let the width be x, the length must equal 2x. We also know that the security zone around the trampoline is 2-meter wide. Let's write the dimensions of the area with the security zone on the diagram.
Because the area under the trampoline has the shape of a rectangle, we can calculate its area of the rectangle A_t by multiplying the lengths of its sides. A_t = 2x(x) = 2x^2 The area of the trampoline is 2x^2 meters. The surface with the security zone is also rectangular. Because the security zone does not include the area under the trampoline, we need to subtract the area of the trampoline when we calculate the area of the security zone A_s.
We are given that the area of the security zone is 3 times larger than the trampoline. This gives us the following equation. A_s = 3A_t ⇓ 12x+16 = 3( 2x^2 ) Let's rewrite the equation in standard form and solve it for x using the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions to the equation are x= -12 ± sqrt(528)-12. Let's separate the two cases and use a calculator to find the solutions.
x = -12 ± sqrt(528)/-12 | |
---|---|
x_1 = -12 + sqrt(528)/-12 | x_2 = -12 - sqrt(528)/-12 |
x_1 ≈ -0.9 | x_2 ≈ 2.9 |
Because the length of the side cannot be negative, we obtained that the width of the trampoline is about 2.9 meters. This means that the length is equal to 2( 2.9) = 5.8 meters.