Section 5: Quadratic Equations and Functions β Part 1
Section 5 β Topic 1 Real-World Examples of Quadratic Functions
Letβs revisit linear functions. Imagine that you are driving down the road at a constant speed of 40mph. This is a linear function.
We can represent the distance traveled versus time on a table (to the right).
Time (in hours)
Distance Traveled (in miles)
1 402 803 1204 160
We can represent the scenario on a graph.
We can represent the distance traveled π(π‘), in terms of time, π‘ hours, with the equation π(π‘) = 40π‘. Linear functions always have a constant rate of change. In this section, we are going to discover a type of non-linear function. Consider the following situation. Liam dropped a watermelon from the top of a 300 ft tall building. He wanted to know if the watermelon was falling at a constant rate over time. He filmed the watermelonβs fall and then recorded his observations in the following table.
Time (in seconds)
Height (in feet)
0 300.01 283.92 235.63 155.14 42.4
What do you notice about the rate of change? It is not constant. Why do you think that the rate of change is not constant? Gravity causes the watermelon to get faster as it gets closer to the ground.
Liam entered the data of the falling watermelon into his graphing calculator. The graph below displays the first quadrant of the graph.
What is the independent variable? Time What is the dependent variable? Height Liam then used his calculator to find the equation of the function, β π‘ = β16π‘3 + 300.
Important facts:
Γ We call this non-linear function a ___________________. Γ The general form of the equation is
_________________________. The graph of π(π₯) = π₯3 is shown below.
Γ This graph is called a ________________________. Why did we only consider the first quadrant of Liamβs graph? You wouldnβt have negative time or height
quadratic
π(π) = πππ + ππ + π
parabola
In Liamβs graph, what was the watermelonβs height when it hit the ground? π feet The time when the watermelonβs height was at zero is called the solution to this quadratic equation. We also call this the ____________ of the equation. There was only one solution to Liamβs equation. Describe a situation where there could be two solutions. Answers vary. Sample answer: A rocketβs height over time after is launched from the ground and lands back on the ground. What about no solutions? Answers vary. Sample answer: A bungee jumpers height over time as they jump from a πππfoot platform.
To solve a quadratic equation using a graph:
Γ Look for the ________________________ of the graph.
Γ The solutions are the values where the graph intercepts the ________________________.
zeros
π-intercepts
π-axis
Letβs Practice! 1. What are the solutions to the quadratic equation graphed
below?
π = π and π = π
Γ
Zeros = π₯-intercepts = Solutions
Try It! 2. Aaron shot a water bottle rocket from the ground. A
graph of height over time is shown below.
a. What type of function best models the rocketβs motion? quadratic
b. After how many seconds did the rocket hit the
ground? π
c. Estimate the maximum height of the rocket.
about π. ππ meters The maximum or minimum point of the parabola is called the ________________. vertex
BEAT THE TEST!
1. Jordan owns an electronics business. During her first year in the business, she collected data and created the following graph showing the relationship between the selling price of an item and the profit.
Part A: Circle the solutions to the quadratic function
graphed above. Part B: What do the solutions represent?
The prices that yielded NO profit.
Part C: Box the vertex of the graph. Part D: What does the vertex represent?
The price that yields the maximum profit.
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Section 5 β Topic 2 Factoring Quadratic Expressions
Letβs review the two methods we used for multiplying polynomials. Area Model:
Distributive Property: 3 π₯ + 2π¦ β 7π§ = ππ + ππ β πππ
We can use these same methods to factor out the greatest common factor of an expression. Area Model:
Distributive Property: 10π₯G β 14π₯3 + 12π₯ = ππ(πππ β ππ + π)
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) 2+ β7-
3
10#$% β14#( 12#
*%+ β,% -
+%
Use the area model to write an equivalent expression for (2π₯ + 5)(π₯ + 3).
πππ + πππ + ππ We can use this same area model to factor a quadratic expression. Look at the resulting trinomial and notice the following four patterns:
Γ The first term of the trinomial can always be found in the ________________ ______________ rectangle.
Γ The last term of the trinomial can always be found in
the ________________ ______________ rectangle. Γ The second term of the trinomial is the ____________ of
the ________________ _____________ and ____________ ______________ rectangles.
Γ The ________________ of the __________________ are
always equal.
πππ β ππ = ππππ and ππ β ππ = ππππ
2"
5
" 3
%&% '&
(& )(
top left
bottom right
bottom left top
right
sum
product diagonals
Use the distributive property to write an equivalent expression for (2π₯ + 5)(π₯ + 3).
ππ + π π + π = πππ + ππ + ππ + ππ = πππ + πππ + ππ We can also use the distributive property to factor a quadratic expression. What are the two middle terms of the expanded form? ππand ππ
Consider the resulting trinomial. 2π₯3 + 11π₯ + 15 Notice that the product of the two middle terms of expanded form are equal to the product of the first and last term of the trinomial. The middle terms also sum to the middle term of the trinomial. ππ β ππ = ππππ and ππ + ππ = πππ Letβs consider how we can use this and the distributive property to factor a quadratic expression.
Factor 2π₯3 + 3π₯ β 5 using the distributive property. Γ Multiply the first term by the last term.
πππ β βπ = βππππ
Γ Find two factors whose product is equal to β10π₯3 and whose sum is equal to 3π₯.
ππ and β ππ
Γ Replace the middle term with these two factors. πππ + ππ β ππ β π
Γ Factor the polynomial by grouping the first 2 terms and the last 2 terms.
πππ + ππ + (βππ β π) π ππ + π β π(ππ + π)
(ππ + π)(π β π)
Letβs Practice! 1. Consider the quadratic expression 3π₯3 + 4π₯ β 4.
a. Factor using the area model.
(π + π)(ππ β π)
b. Factor using the distributive property.
πππ + ππ β π = πππ β ππ + ππ β π
= πππ β ππ + ππ β π = π ππ β π + π ππ β π = (ππ β π)(π + π)
!
"
#! β"
#!" β"!
%! β&
Γ
You can check your answer by using the distributive property. The product of the factors should always result in the original trinomial.
Try It! 2. Consider the quadratic expression 4π€3 β 21π€ + 20.
a. Factor using the area model.
(ππ β π)(π β π)
b. Factor using the distributive property. πππ β πππ + ππ
= πππ β ππ β πππ + ππ = πππ β ππ + βπππ + ππ
= π ππβ π β π ππ β π = (ππ β π)(π β π)
!"
β$
" β!
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β$" %(
BEAT THE TEST!
1. Identify all factors of the expression 18π₯3 β 9π₯ β 5. Select all that apply.
Β¨ 2π₯ + 5 Γ½ 6π₯ β 5 Β¨ 18π₯ β 5 Β¨ 3π₯ + 5 Γ½ 3π₯ + 1
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Section 5 β Topic 3 Solving Quadratic Equations by Factoring
Once a quadratic equation is factored, we can use the zero product property to solve the equation.
The zero product property states that if the product of two factors is zero, then one (or both) of the factors must be _________________.
Γ If ππ = 0, then either π = 0, π = 0,or π = π = 0.
To solve a quadratic equation by factoring:
Step 1: Set the equation equal to zero. Step 2: Factor the quadratic. Step 3: Set each factor equal to zero and solve. Step 4: Write the solution set.
zero
Letβs Practice! 1. Solve for π by factoring π3 + 8π + 15 = 0.
ππ + ππ + ππ + ππ = π π π + π + π π + π = π π + π π + π = π π + π = π or π + π = π π = βπ or π = βπ
{βπ,βπ} 2. Solve for π by factoring 10π3 + 17π + 3 = 0.
ππ + π ππ + π = π ππ + π = π or ππ + π = π π = βπ
π or π = βπ
π
βππ ,β
ππ
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$" %
Try It! 3. Solve for π by factoring 6π3 β 19π + 14 = 0.
ππ β π π β π = π ππ β π = π or π β π = π π = π
π or π = π
ππ , π
!"
β$
" β%
!"% β&%"
β$" &'
BEAT THE TEST!
1. Tyra solved the quadratic equation π₯3 β 10π₯ β 24 = 0 by factoring. Her work is shown below:
Step 1: π₯3 β 10π₯ β 24 = 0 Step 2: π₯3 β 4π₯ β 6π₯ β 24 = 0 Step 3: π₯3 β 4π₯ + (β6π₯ β 24) = 0 Step 4: π₯ π₯ β 4 β 6(π₯ β 4) = 0Step 5: (π₯ β 4)(π₯ β 6) = 0 Step 6: π₯ β 4 = 0, π₯ β 6 = 0 Step 7: π₯ = 4 or π₯ = 6 Step 8: 4, 6
Tyra did not find the correct solutions. Investigate the steps, decipher her mistakes, and explain how to correct Tyraβs work.
Step 2: She replaced βπππ with βππ and β ππ. While their sum is β πππ, their product is not β ππππ. Step 4: When she factored out β π, she should have gotten β π π + π . ππ β πππ + ππ β ππ = π π π β ππ + π π β ππ = π π β ππ π + π = π π β ππ = π and π + π = π π = ππ and π = βπ
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Section 5 β Topic 4 Solving Other Quadratic Equations by Factoring
Many quadratic equations will not be in standard form. Γ The equation wonβt always equal zero. Γ There may be a greatest common factor (GCF) within
all of the terms. Letβs Practice! 1. Solve for π: 3π3 + 30π β 168 = 0.
π ππ + πππβ ππ = π π π+ ππ πβ π
π =ππ
π+ ππ = π or πβ π = π π = βππ or π = π
2. Solve for π₯: π₯ + 4 π₯ β 5 = β8.
ππ β ππ + ππ β ππ = βπ ππ β π β ππ + π = π ππ β π β ππ = π π β π π + π = π π = π or π = βπ
Try It! 3. Solve for π: 6π3 + 5π = 1. ππ π + ππ β π = π
ππ β π π + π = π π = π
π or π = βπ
4. Solve for π: π3 + 36 = 13π.
ππ β πππ + ππ = π π β π π β π = π π β π = π or π β π = π
π = π or π = π
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β$" $
BEAT THE TEST!
1. What are the solutions to 40π₯3 β 30π₯ = 135? Select all that apply.
Β¨ β ]
3 Β¨
G^
Β¨ β ]^ Β¨
G3
Γ½ β32 Γ½ ]
^
Β¨ βG^
ππππ β πππ β πππ = π π πππ β ππ β ππ
π =ππ
πππ β ππ β ππ = π
ππ + π ππ β π = π ππ + π = π or ππ β π = π π = βπ
π or π = π
π
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Section 5 β Topic 5 Solving Quadratic Equations by Factoring β
Special Cases
There are a few special cases when solving quadratic equations by factoring. Perfect Square Trinomials:
Γ π₯3 + 6π₯ + 9 is an example of perfect square trinomial. We see this when we factor.
Γ A perfect square
trinomial is created when you square a ________________________.
Recognizing a Perfect Square Trinomial: A quadratic expression can be factored as a perfect square trinomial if it can be re-written in the form π3 + 2ππ + π3. Factoring a Perfect Square Trinomial:
Γ If π3 + 2ππ + π3 is a perfect square trinomial, then π3 + 2ππ + π3 = π + π 3.
Γ If π3 β 2ππ + π3 is a perfect square trinomial, then
π3 β 2ππ + π3 = π β π 3.
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binomial
Letβs Practice! 1. Determine whether 16π₯3 + 88π₯ + 121 is a perfect square
trinomial. Justify your answer.
It is a perfect square trinomial. It can be written in the form ππ + πππ + ππ.
ππ π + π ππ ππ + ππ π 2. Solve for π: π3 β 10π + 25 = 0.
π π β π π π + π π = π π β π π = π (π β π) = π or (π β π) = π π = π
π Try It! 3. Determine whether π₯3 β 8π₯ + 64 is a perfect square
trinomial. Justify your answer. It is NOT a perfect square trinomial. It cannot be written in the form ππ + πππ + ππ.
ππ β ππ + ππ β (π)π β π π π + π π
4. Solve for π€: 4π€3 + 49 = β28π€. πππ + πππ + ππ = π ππ π + π(ππ)(π) + π π = π
ππ + π π = π π = βπ
π βπ
π
5. What do you notice about the number of solutions to perfect square quadratic equations? There is always one solution.
6. Sketch the graph of a quadratic equation that is a perfect
square trinomial. Sample Answer:
Difference of Squares Use the distributive property to multiply the following binomials. (π₯ + 5)(π₯ β 5) ππ β ππ + ππ β ππ ππ β ππ (5π₯ + 3)(5π₯ β 3) ππππ β πππ + πππ β π ππππ β π Describe any patterns you notice. Answers vary. Sample answer: The middle terms always canceled out.
Γ When we have a binomial in the form π3 β π3, it is called the difference of two squares. We can factor this as π + π π β π .
Letβs Practice! 7. Solve the equation 49π3 = 64 by factoring. ππππ β ππ = π ππ β π ππ + π = π
ππ β π = π or ππ + π = π π = π
π or π = βππ
βππ, ππ
Try It! 8. Solve the equation 0 = 121π3 β 100.
πππππ β πππ = π (πππ β ππ)(πππ + ππ) = π
πππ β ππ = π or πππ + ππ = π π = ππ
ππ or π = βππππ
βππππ, ππππ
BEAT THE TEST!
1. Which of the following expressions are equivalent to 8πG β 98π? Select all that apply. Γ½ 2(4πG β 49π) Γ½ 2π(4π3 β 49) Β¨ 2π(4πG β 49π) Β¨ 2π β 7 (2π + 7) Β¨ 2 2π β 7 (2π + 7) Γ½ 2π 2π β 7 (2π + 7)
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Section 5 β Topic 6 Solving Quadratic Equations by Taking Square Roots
Consider the following quadratic equation.
2π₯3 β 36 = 0
When quadratic equations are in the form ππ₯3 + π = 0, solve by taking the square root.
Step 1: Get the variable on the left and the constant on the right.
Step 2: Take the square root of both sides of the equation. (Donβt forget the negative root!)
Solve for π₯ by taking the square root. 2π₯3 β 36 = 0 πππ
π =πππ
ππ = ππ π = Β± ππ π = Β±π π or π β Β±π. ππ
Letβs Practice! 1. Solve π₯3 β 121 = 0. ππ = πππ π = Β± πππ
π = Β±ππ Try It! 2. Solve β5π₯3 + 80 = 0.
βπππ
βπ =βππβπ
ππ = ππ π = Β± ππ π = Β±π
BEAT THE TEST!
1. What is the smallest solution to the equation 2π₯3 + 17 = 179?
A βπ B β3 C 3 D 9 Answer is A.
2. A rescuer on a helicopter that is 50 feet above the sea drops a lifebelt. The distance from the lifebelt to the sea can be modeled by the equation β(π‘) = β16π‘3 + π , where π‘ is the time, in seconds, after the lifebelt is dropped, and π is the initial height, in feet, of the lifebelt above the sea.
How long will it take for the lifebelt to reach the sea? Round your answer to the nearest tenth of a second. βππππ + ππ = π βππππ
βππ =βππβππ
π = Β± ππππ
π = Β± πππ or π = Β±π. π
π. πseconds
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Section 5 β Topic 7 Solving Quadratic Equations by Completing the Square
Sometimes you wonβt be able to solve a quadratic equation by factoring. However, you can rewrite the quadratic equation so that you can complete the square to factor and solve. Letβs start by determining what number we can add to a quadratic expression to make it a perfect square trinomial.
What value could be added to the quadratic expression to make it a perfect square trinomial? π₯3 + 6π₯ + π π₯3 + 8π₯ + 3 + ππ π₯3 β 22π₯ β 71 +192 Letβs see how this can be used to solve quadratic equations.
Recall how we factored perfect square trinomials. If π3 + 2ππ + π3 is a perfect square trinomial, then π3 + 2ππ + π3 = π + π 3 and π3 β 2ππ + π3 = π β π 3. Solve π π₯ = ππ₯3 + ππ₯ + π by completing the square. Step 1: Group ππ₯3and ππ₯ together.
π π₯ = (ππ₯3 + ππ₯ +___) + π Step 2: If π β 1, then factor out π.
π π₯ = π(π₯3 +ππ π₯ + ______) + π
Step 3: Divide j
k by two and square the result. Add that
number to the grouped terms. Subtract the product of that number and π from π so that you have not changed the equation.
π π₯ = π(π₯3 + j
kπ₯ + j
l
^kl) + π β jl
^k
Step 4: Factor the trinomial.
π π₯ = π π₯ +π2π
3
+4ππ β π3
4π Step 5: This is vertex form. Now we can solve the equation by
setting the function equal to zero, moving the constant to the opposite side, and taking the square root of both sides.
Letβs Practice! 1. Consider the following quadratic expression 2π₯3 β 8π₯ + 5.
a. Complete the square to write the quadratic expression in vertex form. π(π β π)π β π
b. If the expression represents a function, find the solutions to the quadratic function.
π β π. ππ or π β π. ππ
Try It! 2. Consider the quadratic expression 3π₯3 + 12π₯ + 31.
a. Complete the square to write the quadratic expression in vertex form. π(π + π)π + ππ
b. If the expression represents a function, find the solutions to the quadratic function. No Solution. This graph would not cross the π β axis.
BEAT THE TEST! 1. The equations shown below are steps to solve the function
π π₯ = 2π₯3 + 24π₯ β 29 by completing the square.
A. π π₯ = 2 π₯3 + 12π₯ + 36 β 29 β 72
B. π₯ + 6 = Β± 50.5
C. π π₯ = 2(π₯3 + 12π₯ + ____) β 29
D. π₯ = β6 Β± 50.5
E. 2 π₯ + 6 3 β 101 = 0
F. π₯ + 6 3 = 50.5
G. π₯ + 6 3 = Β± 50.5
H. π π₯ = 2 π₯ + 6 3 β 101
Place the equations in the correct order by writing the letter corresponding to each step in the boxes below.
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Step 1 Step 2 Step 3 Step 4
Step 8 Step 7 Step 6 Step 5
C A H E
D B G F
Section 5 β Topic 8 Deriving the Quadratic Formula
We can use the process of completing the square to derive a formula to solve any quadratic equation. Consider the quadratic equation, π π₯ = ππ₯3 + ππ₯ + π, where π β 0. Recall our steps for completing the square as a method for solving π(π₯). Step 1: Group ππ₯3and ππ₯ together. π π = πππ + ππ + + π Step 2: If π β 1, then factor out π.
π π = π ππ +πππ +
ππ
πππ + π βππ
πππ
Step 3: Divide j
k by two and square the result. Add that
number to the grouped terms. Subtract the product of that number and π from π so that you have not changed the equation.
πππ β ππ
ππ Step 4: Factor the trinomial.
π π = π π +πππ
π
+πππ β ππ
ππ
Step 5: Solve the equation.
π +πππ
π
=ππ β ππππππ
π +πππ
π
= Β±ππ β ππππππ
π +πππ =
Β± ππ β πππππ
π = βπππ Β±
ππ β πππππ
π =βπ Β± ππ β πππ
ππ
BEAT THE TEST! 1. Complete the missing steps in the derivation of the
quadratic formula:
π π₯ = ππ₯3 + ππ₯ + π
π π₯ = π π₯3 + ππ π₯ +
π2
4π2 + π β π24π
π π₯ = π π₯ + π2π
3+ 4ππβπ2
4π
π π₯ +π2π
3
+4ππ β π2
4π = 0
π₯ + π2π
3= π2β4ππ
4π2
π₯ + π2π = Β± π2β4ππ
2π
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π(π) = nπππ + ππ +o + π+ π
pqπ + πππrπ= Β±ππsπππ
πππ
π =βπ Β± βππ β πππ
ππ
Section 5 β Topic 9 Solving Quadratic Equations Using the Quadratic
Formula For any quadratic equation ππ₯3 + ππ₯ + π = 0, where π β 0,
π₯ =βπ Β± π3 β 4ππ
2π To use the quadratic formula:
Step 1: Set the quadratic equation equal to zero.
Step 2: Identify π, π, and π.
Step 3: Substitute π, π, and π into the quadratic formula and evaluate to find the zeros.
Letβs Practice! 1. Use the quadratic formula to solve π₯3 β 4π₯ + 3 = 0.
π = π
π = βπ π = π
π =πΒ± βπ πβπ(π)(π)
π(π)
π = πΒ± ππβπππ(π)
π =π Β± π
π
π =π Β± ππ
π = π or π = π π, π
2. Consider the graph of the quadratic equation π¦ = π₯3 β 4π₯ + 3.
Does the graph verify the solutions we found using the quadratic formula? Yes, the π-intercepts are π and π.
3. Use the quadratic formula to solve 2π€3 + π€ = 5.
π = π π = π π = βπ
π =βπΒ± π πβπ(π)(βπ)
π(π)
π = βπΒ± π+πππ
π =βπ Β± ππ
π π β π. ππ or π β βπ. ππ {βπ. ππ, π. ππ}
Try It! 4. Use the quadratic formula to solve 3π3 β 11 = 20π.
π = π π = βππ π = βππ
π =ππΒ± βππ πβπ(π)(βππ)
π(π)
π = ππΒ± πππ+ππππ
π =ππ Β± πππ
π π β π. ππ or π β βπ. π {βπ. π, π. ππ}
BEAT THE TEST!
1. Your neighborβs garden measures 12 meters by 16 meters. He plans to install a pedestrian pathway all around it, increasing the total area to 285 square meters. The new area can be represented by 4π€3 + 56π€ + 192. Use the quadratic formula to find the width, π€, of the pathway.
Part A: Write an equation that can be used to solve for
the width of the pathway. πππ + πππ + πππ = πππ
πππ + πππ β ππ = π
Part B: Use the quadratic formula to solve for the width of the pathway.
π = π π = ππ π = βππ
π =βππΒ± ππ πβπ(π)(βππ)
π(π)
π = βππΒ± ππππ+πππππ
π =βππ Β± ππππ
π π = π. π or π = βππ. π
The width of the path is approximately π. π meters.
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Section 5 β Topic 10 Quadratic Functions in Action
Letβs consider solving real-world situations that involve quadratic functions. Consider an object being launched into the air. We compare the height versus time elapsed. From what height is the object launched?
πmeters Once the object is launched, how long does it take to reach its maximum height?
π. π seconds
What is the maximum height? πmeters Once the object is launched, how long does it take for it to hit the ground? π. π seconds Once the object is launched, when does it return to a height of three meters? πseconds
Question How to Answer it
1. From what height is the object launched?
This is the π¦-intercept. In the standard form, ππ₯3 + ππ₯ + π, πis the π¦-intercept.
2. How long does it take for the object to reach its maximum height?
This is the π₯-coordinate of the vertex, π₯ = sj
3k, where values of π
and π come from the standard form of a quadratic equation. π₯ = sj3k is also the equation that
represents the axis of symmetry.
3. What is the maximum height?
This is the π¦-coordinate of the vertex. Substitute the π₯-coordinate from the step above and evaluate to find π¦. In vertex form, the height is π and the vertex is (β, π).
4. How long does it take for the object to hit the ground?
The π₯-intercept(s) are the solution(s), or zero(s), of the quadratic function. Solve by factoring, using the quadratic formula, or by completing the square. In a graph, look at the π₯-intercept(s).
5. When does the object return to a height of three meters?
In function π» π‘ = ππ‘3 + ππ‘ + π, if height is given, then substitute the value for π»(π‘). If time is given, then substitute for π‘.
Letβs Practice! 1. Ferdinand is playing golf. He hits a shot off the tee box that
has a height modeled by the function β π‘ = β16π‘3 + 80π‘, where β(π‘) is the height of the ball, in feet, and π‘ is the time in seconds it has been in the air. The graph that models the golf ballβs height over time is shown below.
a. When does the ball reach its maximum height?
π. π seconds
b. What is the maximum height of the ball?
πππfeet
c. What is the height of the ball at 3 seconds? When will the ball reach that same height again?
About ππ feet, About π seconds
Try It! 2. Recall exercise 1.
a. When is the ball 65 feet in the air? Explain.
After about π second and again after about π seconds
b. How long does it take until the golf ball hits the ground? π seconds
BEAT THE TEST!
1. A neighborhood threw a fireworks celebration for the 4th of July. A bottle rocket was launched upward from the ground with an initial velocity of 160 feet per second. The formula for vertical motion of an object is
β π‘ = 0.5ππ‘3 + π£π‘ + π , where the gravitational constant, π, is β32 feet per square second, π£ is the initial velocity, π is the initial height, and β(π‘) is the height in feet modeled as a function of time, π‘.
Part A: What function describes the height, β, of the
bottle rocket after π‘ seconds have elapsed?
π π = βππππ + ππππ Part B: What was the maximum height of the bottle
rocket? Time to reach max. height: π = β πππ
π βππ = π seconds
Maximum height: π π = βππ π π + πππ π = πππ feet
Test Yourself! Practice Tool
Great job! You have reached the end of this section. Now itβs time to try the βTest Yourself! Practice Tool,β where you can practice all the skills and concepts you learned in this section. Log in to Algebra Nation and try out the βTest Yourself! Practice Toolβ so you can see how well you know these topics!
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