3. Applications of Quadratic Functions
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Chapter 9
3.
Applications of Quadratic Functions
| Student Learning Objectives: |
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Why
Quadratic functions play a pivotal role in the universe of mathematics. This lesson accentuates the significance of these functions in multiple real-life contexts. From modeling the trajectory of a thrown object to understanding the dynamics of certain economic models, the applications of quadratic functions are vast. The quadratic formula acts as a key tool in solving and interpreting these functions, further broadening the scope of their impact. With a foundation in these principles, one can not only decipher the mathematical intricacies but also develop a profound appreciation for the patterns and behaviors observed in various domains of life.
Lesson Settings & Tools
| | 7 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Applications of Quadratic Functions
Slide of 7
Quadratic functions are often applied in real life. For example, they are used to describe movements of objects, dimensions of regions, profit, and prices. In this lesson, some of these applications will be analyzed and better understood.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Investigating the Price of a Car
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 $11 000.
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b What was the lowest price of the car?
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c What was the price of the car in 2020?
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Example
Analyzing a Swimmer's Dive
Zosia and her classmates were given a half-day from school on one condition — the students try out at a local sports club. Zosia chooses to go to the swim club. She gets to dive from a springboard for the first time!
External credits: @macrovector
a What is the height of the springboard above the water?
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c After how many seconds will Zosia reach the highest point of her jump? Give the exact time.
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Hint
a When x=0, Zosia's height above the water is the same as the springboard's height.
b When Zosia hits the surface of the water, her height above the water is 0 feet.
c The x-coordinate of the vertex of a parabola can be found by the formula x_v=- b2a.
Solution
a When Zosia is standing on the springboard right before taking her dive, her height above the water is the same as the springboard's height. Because x represents the number of seconds since jumping off the board, by substituting x=0 in the given function the height of the springboard can be calculated.
f(x)=- 4x^2+9x+5
Substitute
x= 0
f( 0)=- 4( 0)^2+9( 0)+5
▼
Evaluate right-hand side
CalcPow
Calculate power
f(0)=- 4(0)+9(0)+5
ZeroPropMult
Zero Property of Multiplication
f(0)=0+0+5
IdPropAdd
Identity Property of Addition
f(0)=5
The springboard is 5 feet above the water.
b At the moment of hitting the surface of the water, Zosia is 0 feet above it. That means 0 can be substituted for f(x), then the corresponding time in seconds can be found.
f(x)=- 4x^2+9x+5
Substitute
f(x)= 0
0=- 4x^2+9x+5
▼
Solve for x
UseQuadForm
Use the Quadratic Formula: a = - 4, b= 9, c= 5
x=- 9±sqrt(( - 9)^2-4 ( - 4)( 5))/2( - 4)
CalcPow
Calculate power
x=- 9±sqrt(81-4(- 4)(5))/2(- 4)
MultPosNeg
a(- b)=- a * b
x=- 9±sqrt(81-(- 16)(5))/- 8
MultNegPos
(- a)b = - ab
x=- 9±sqrt(81-(- 80))/- 8
SubNeg
a-(- b)=a+b
x=- 9±sqrt(161)/- 8
StateSol
State solutions
lcx_1= - 9+ sqrt(161)- 8 & (I) x_2= - 9- sqrt(161)- 8 & (II)
(I), (II): Use a calculator
lx_1= - 9+ 12.688577...- 8 x_2= - 9- 12.688577...- 8
(I), (II): Add and subtract terms
lx_1= 3.688577...- 8 x_2= - 21.688577...- 8
(I), (II): Use a calculator
lx_1=- 0.461072... x_2=2.711072...
(I), (II): Round to 1 decimal place(s)
lx_1≈ - 0.5 x_2≈ 2.7
The quadratic equation has two solutions. Since her time in the air cannot be negative, the negative solution does not need to be counted. Zosia reaches the water approximately 2.7 seconds after jumping.
c The highest point Zosia will reach is the vertex of the parabola. The function is given in the standard form, so the x-coordinate of the vertex can be found by the following formula.
x_v=- b/2a In this case, a=- 4 and b=9. By substituting these values, the x-coordinate of the vertex can be calculated.
x_v=- b/2a
SubstituteII
a= - 4, b= 9
x_v=- 9/2( - 4)
▼
Evaluate right-hand side
MultPosNeg
a(- b)=- a * b
x_v=- 9/- 8
RemoveNegFracAndDenom
- a/- b= a/b
x_v=9/8
CalcQuot
Calculate quotient
x_v=1.125
Since x represents the time in seconds since she jumped, Zosia will reach the highest point after 1.125 seconds. To find her height at that moment, 1.125 will be substituted for x into the formula.
f(x)=- 4x^2+9x+5
Substitute
x= 1.125
f( 1.125)=- 4( 1.125)^2+9( 1.125)+5
f(1.125)=10.0625
RoundDec
Round to 1 decimal place(s)
f(1.125)≈ 10.1
Therefore, Zosia will be about 10.1 feet above the water at the highest point of her jump.
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Example
Comparing the Heights of a Football
Zosia's classmates, Mark and Tearrik, take their half-day off from school to try out the football club. Here they are practicing throwing the ball to each other.
External credits: @freepik
a Of Mark and Tearrik, who throws the ball higher?
b If the person who is supposed to catch the ball misses and the ball lands on the floor, how long will the football be airborne? Find the answer for both Mark and Tearrik's throws, then approximate the times to the nearest second.
c In the same coordinate plane that shows the Tearrik's throw, graph the function that describes the height of the ball thrown by Mark.
Answer
a Tearrik
b Mark's ball: about 3 seconds
Tearrik's ball: about 4 seconds
c 
Solution
a When Mark throws the ball, its height is represented by a quadratic function.
M(x)=- 16x^2+48x+5 This function indicates that the ball's path has the shape of a parabola. The given graph of Tearrik's throw is also represented by a parabola. The y-coordinate of the vertex of each parabola is the answer to who throws the ball higher. The given equation will be written into vertex form to determine this coordinate.
M(x)=- 16x^2+48x+5
▼
Rewrite
SplitIntoFactors
Split into factors
M(x)=- 16x^2+16(3)x+5
FactorOut
Factor out - 16
M(x)=- 16(x^2-3x-5/16)
AddDiffZero
a = a+ (3/2)^2- (3/2)^2
M(x)=- 16(x^2-3x+ (3/2)^2- (3/2)^2-5/16)
NumberToFrac
a = 2* a/2
M(x)=- 16(x^2-6/2x+(3/2)^2-(3/2)^2-5/16)
SplitIntoFactors
Split into factors
M(x)=- 16(x^2-2(3)/2x+(3/2)^2-(3/2)^2-5/16)
MovePartNumLeft
a* b/c=a*b/c
M(x)=- 16(x^2-2(3/2)x+(3/2)^2-(3/2)^2-5/16)
CommutativePropMult
Commutative Property of Multiplication
M(x)=- 16(x^2-2x(3/2)+(3/2)^2-(3/2)^2-5/16)
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
M(x)=- 16((x-3/2)^2-(3/2)^2-5/16)
Distr
Distribute - 16
M(x)=- 16(x-3/2)^2+16(3/2)^2+16(5/16)
PowQuot
(a/b)^m=a^m/b^m
M(x)=- 16(x-3/2)^2+16(9/4)+16(5/16)
MoveLeftFacToNum
a*b/c= a* b/c
M(x)=- 16(x-3/2)^2+144/4+16(5/16)
CalcQuot
Calculate quotient
M(x)=- 16(x-3/2)^2+36+16(5/16)
DenomMultFracToNumber
16 * a/16= a
M(x)=- 16(x-3/2)^2+36+5
AddTerms
Add terms
M(x)=- 16(x-3/2)^2+41
FracToDiv
a/b=a÷ b
M(x)=- 16(x-1.5)^2+41
By comparing the obtained equation with a general quadratic function written in vertex form, the coordinates of the vertex can be identified. f(x)&= a(x- h)^2 + k M(x)&=- 16(x- 1.5)^2+ 41 Since (h,k) are the coordinates of the vertex, it can be said that the vertex's coordinates of the given function are (1.5,41). Looking at the y-coordinate, when Mark throws the ball, its maximum height is 41 feet. Mark's football: 41 feet Next, by analyzing the graph, the maximum height of Tearrik's football can be found.
First, follow the x-coordinate to about 1.8 seconds after Tearrik throws the ball, and note the y-coordinate. The ball reaches a maximum height of 45 feet. Tearrik's football: 45 feet Tearrik throws the ball higher than Mark.
b Both Mark and Tearrik release the ball at x=0 seconds. To determine how long Mark's ball is airborne, the number of seconds till the ball reaches a height of 0 feet above the ground should be found. To do so, 0 will be substituted for M(x).
M(x)=- 16x^2+48x+5
Substitute
M(x)= 0
0=- 16x^2+48x+5
▼
Solve for x
UseQuadForm
Use the Quadratic Formula: a = - 16, b= 48, c= 5
x=- 48±sqrt(48^2-4( - 16)( 5))/2( - 16)
CalcPow
Calculate power
x=- 48±sqrt(2304-4(- 16)(5))/2(- 16)
MultPosNeg
a(- b)=- a * b
x=- 48±sqrt(2304-(- 64)5)/- 32
MultNegPos
(- a)b = - ab
x=- 48±sqrt(2304-(- 320))/- 32
SubNeg
a-(- b)=a+b
x=- 48±sqrt(2624)/- 32
CalcRoot
Calculate root
x=- 48± 51.224993.../- 32
StateSol
State solutions
lcx_1= - 48+ 51.224993...- 32 & (I) x_2= - 48- 51.224993...- 32 & (II)
(I), (II): Add and subtract terms
lx_1= 3.224993...- 32 x_2= - 99.224993...- 32
MoveNegDenomToFrac
(I): Put minus sign in front of fraction
lx_1=- 3.224993...32 x_2= - 99.224993...- 32
DivNegNeg
(II): - a/- b=a/b
lx_1=- 3.224993...32 x_2= 99.224993...32
(I), (II): Use a calculator
lx_1=- 0.100781... x_2=3.100781...
(I), (II): Round to 1 decimal place(s)
lx_1≈ - 0.1 x_2≈ 3.1
The quadratic equation has two solutions. Since the time the ball is in the air cannot be negative, Mark's ball hits the ground about 3.1 seconds after it is thrown. In other words, it is airborne for about 3.1 seconds. To find how long Tearrik's ball is airborne, analyze the given graph.
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.
c To draw the graph of the function that represents the height of the ball when Mark throws it, the vertex form found in Part A will be used.
M(x)=- 16(x-1.5)^2+41 Here, (1.5,41) are the coordinates of the parabola's vertex. Since the axis of symmetry is a vertical line that passes through the vertex, its equation is x=1.5.
Next, the y-intercept should be determined. For the y-intercept, the x-coordinate is 0. Therefore, it can be found by substituting x=0 into the equation.
M(x)=- 16(x-1.5)^2+41
Substitute
x= 0
M( 0)=- 16( 0-1.5)^2+41
▼
Evaluate right-hand side
SubTerm
Subtract term
M(0)=- 16(- 1.5)^2+41
CalcPow
Calculate power
M(0)=- 16(2.25)+41
MultNegPos
(- a)b = - ab
M(0)=- 36+41
AddTerms
Add terms
M(0)=5
The y-intercept of the parabola occurs at (0,5). This point and its reflection across the axis of symmetry can be added to the graph.
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.
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Example
Making Conclusions About Different Forms of a Quadratic Function
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.
External credits: @freepic
The company's largest profit comes from the sale of baby chairs. It can be described by any of the following equivalent functions. P(x)&=- 3x^2+852x-37 260 P(x)&=- 3(x-54)(x-230) P(x)&=- 3(x-142)^2+23 232 Here, x is the price of a baby chair in dollars, while P is the profit made in dollars. Without changing or rewriting any equation, answer the following questions.
a Which of the given equations reveals the price that produces a profit of zero dollars?
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b Which of the given equations reveals the profit when the price is $0?
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c Which of the given equations reveals the price that produces the highest possible profit?
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Hint
a Look for the quadratic function written in a form that reveals its zeros.
b After substituting x=0, in which equation only a constant will be left on the right-hand side?
c Analyze the value of the coefficient before the x^2-term. What does it say about the direction of the parabola and its highest point?
Solution
a The equation that reveals the price that produces a profit of $ 0 is the equation that gives the value of x for which P(x)=0. In other words, it is the equation written in a form that its zeros can be immediately identified. Recall the meaning of the constants p and q in the factored form of a quadratic function.
y=a(x- p)(x- q) Here, because p and q are the zeros of the parabola, the second equation — written in factored form — reveals the zeros of the parabola. P(x)=- 3(x- 54)(x- 230) For x=54 and x=230, the value of P is 0. These values mean that the prices of $54 and $230 result in a profit of zero dollars.
b To determine which equation reveals the profit when the price is $0, substitute 0 for x in each equation and see if the value of P can be found without any calculations.
P( 0)&=- 3( 0)^2+852( 0)-37 260 P( 0)&=- 3( 0-54)( 0-230) P( 0)&=- 3( 0-142)^2+23 232 The second and third equations require further calculations. Conversely, in the first equation after substituting 0 for x, the first two terms are equal to 0, according to the Zero Property of Multiplication. Because of this, the value of P is equal to the constant term of the equation's right-hand side. P(0)&=- 3(0)^2+852(0)-37 260 P(0)&=0+0-37 260 P(0)&=- 37 260 Therefore, the first equation — written in standard form — reveals that if the price of the item is $0, the profit will be - $37 260. This means that if the company gives away the product for free, they will lose $37 260.
c Since the company's profit is given by a quadratic function, its graph is a parabola. Additionally, the coefficient a has a negative value of - 3, which indicates that the parabola opens downward. In this case, the highest point of the graph is its vertex.
Out of the three given equations, only the third equation is written in vertex form. Therefore, it reveals the coordinates of the vertex without having to do any calculations. f(x)&= a(x - p)^2 + q P(x)&=- 3(x- 142)^2+ 23 232 In the vertex form, (p,q) are the coordinates of the vertex, and the given situation has its vertex at (142,23 232). This means that by setting the baby chair's price at $142, the company makes its highest possible profit of $23 232.
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Example
Calculating the Width of a Sidewalk
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.
External credits: @pikisuperstar
If the icerink must have an area of 1575 square meters, what is the width of the sidewalk?
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Hint
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.
Solution
Let x be the width of the sidewalk around the ice skating rink.
External credits: @pikisuperstar
Since there is a sidewalk on each side of the rink, its dimensions are by 2x less than the dimensions of the place found. h=40-2x w=50-2x Because the rink has a rectangular shape, its area is the product of its width and its height. A= h w ⇓ A= (40-2x) (50-2x) It is known that the rink should have an area of 1575 square meters. By substituting this value for A, a quadratic equation can be formed. Finally, it can be solved for x to find the width of the sidewalk.
A=(40-2x)(50-2x)
Substitute
A= 1575
1575=(40-2x)(50-2x)
▼
Multiply parentheses
Distr
Distribute (40-2x)
1575=(40-2x)(50)-(40-2x)(2x)
Distr
Distribute 50 & 2x
1575=2000-100x-80x+4x^2
SubTerm
Subtract term
1575=2000-180x+4x^2
CommutativePropAdd
Commutative Property of Addition
1575=4x^2-180x+2000
SubEqn
LHS-1575=RHS-1575
0=4x^2-180x+425
Now, using the Quadratic Formula, the solutions to the equation can be found.
0=4x^2-180x+425
▼
Solve for x
UseQuadForm
Use the Quadratic Formula: a = 4, b= - 180, c= 425
x=- ( - 180)±sqrt(( - 180)^2-4( 4)( 425))/2( 4)
NegNeg
- (- a)=a
x=180±sqrt((- 180)^2-4(4)(425))/2(4)
CalcPow
Calculate power
x=180±sqrt(32 400-4(4)(425))/2(4)
Multiply
Multiply
x=180±sqrt(32 400-6800)/8
SubTerm
Subtract term
x=180±sqrt(25600)/8
CalcRoot
Calculate root
x=180± 160/8
Therefore, the quadratic equation has two solutions. ccc x_1=180+160/8& & x_2=180-160/8 ⇓ & & ⇓ x_1=40& & x_2=2.5 If the sidewalk is 40 meters wide, then there will be no space for the rink. Therefore, the value of x_1 can be disregarded. This means that the sidewalk around the rink is 2.5 meters wide. What a great chance for Tadeo to practice some math when he initially thought he would be skating!
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Closure
Analyzing the Price of a Car
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 $11 000.
a Interpret the meaning of the point (25,41) that lies on the parabola.
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b What was the lowest price of the car?
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c What was the price of the car in 2020?
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Hint
a Recall what the axes of the coordinate plane represent.
b Use the given information about the car price in different years to find the values of a, b, and c and write the quadratic function of the given parabola.
c Calculate the number of years that have passed between 1970 and 2020 and substitute that value for x into the equation found in Part B.
Solution
a Start by recalling what the axes of the coordinate plane represent. The horizontal axis denotes the number of years since 1970, while the vertical axis represents the price of the car at a particular year.
This means that 25 years after 1970 — in the year 1995 — the car had a price of $ 41 000.
b Since the function's graph is a parabola facing upwards, its minimum is located at the vertex of the parabola. Analyzing the graph, the first coordinate of the vertex appears to be 5. However, it is hard to determine the exact value of its second coordinate.
To find its exact value, the equation of the parabola should be found. Since the price of the car in 1970 was $ 3500, the graph intersects the vertical axis at ( 0, 3.5). By substituting this point into the general equation of a quadratic function written in standard form, the value of the constant c can be determined. 3.5=a( 0)^2+b( 0)+c ⇓ c=3.5 Next, from Part A, it is known that (25,41) lies on the parabola. Also, in 1985, which is 15 years since 1970, the price of the car reached $11 000. Substitute ( 25, 41) and ( 15, 11) into the equation to form a system of equations in terms of a and b. 41=a( 25)^2+b( 25)+3.5 & (I) 11=a( 15)^2+b( 15)+3.5 & (II) The values of a and b can be found by solving this system.
41=a(25)^2+b(25)+3.5 11=a(15)^2+b(15)+3.5
▼
(I), (II): Simplify
(I), (II): Calculate power
41=a(625)+b(25)+3.5 11=a(225)+b(15)+3.5
(I), (II): LHS-3.5=RHS-3.5
37.5=a(625)+b(25) 7.5=a(225)+b(15)
(I), (II): Multiply
37.5=625a+25b 7.5=225a+15b
▼
(I): Solve for b
SubEqn
(I): LHS-625a=RHS-625a
37.5-625a=25b 7.5=225a+15b
DivEqn
(I): .LHS /25.=.RHS /25.
37.5-625a25=b 7.5=225a+15b
WriteDiffFrac
(I): Write as a difference of fractions
37.525- 625a25=b 7.5=225a+15b
MovePartNumRight
(I): a* b/c=a/c* b
37.525- 62525a=b 7.5=225a+15b
CalcQuot
(I): Calculate quotient
1.5-25a=b 7.5=225a+15b
RearrangeEqn
(I): Rearrange equation
b=1.5-25a 7.5=225a+15b
Substitute
(II): b= 1.5-25a
b=1.5-25a 7.5=225a+15( 1.5-25a)
▼
(II): Solve for a
Distr
(II): Distribute 15
b=1.5-25a 7.5=225a+22.5-375a
SubTerm
(II): Subtract term
b=1.5-25a 7.5=- 150a+22.5
SubEqn
(II): LHS-22.5=RHS-22.5
b=1.5-25a - 15=- 150a
DivEqn
(II): .LHS /(- 150).=.RHS /(- 150).
b=1.5-25a 0.1=a
RearrangeEqn
(II): Rearrange equation
b=1.5-25a a=0.1
Substitute
(I): a= 0.1
b=1.5-25( 0.1) a=0.1
b= - 1 a= 0.1
Now, the quadratic equation that describes the parabola can be completed. y= 0.1x^2 -1x+3.5 ⇕ y=0.1x^2-x+3.5 Finally, by substituting 5 for x, the lowest price of the car can be calculated.
The lowest price of the car was $1000. It was 5 years since 1970, which is 1975.
c In order to determine the price of the car in 2020, first the number of years since 1970 should be found.
2020-1970=50 years Next, substitute x=50 into the equation found in Part B to determine the corresponding value of y.
y=0.1x^2-x+3.5
Substitute
x= 50
y=0.1( 50)^2- 50+3.5
▼
Evaluate right-hand side
CalcPow
Calculate power
y=0.1(2500)-50+3.5
Multiply
Multiply
y=250-50+3.5
AddSubTerms
Add and subtract terms
y=203.5
Therefore, in 2020 the car had a price of 203.5 thousands of dollars or $ 2 035 000.
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Applications of Quadratic Functions
Exercise 2.1
The figure shows two rectangles that have the side lengths of x and (8-x) centimeters.
Find the largest total area that the two rectangles can have together.
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The area A of a rectangle is calculated by multiplying its length l by its width w. A_R=l w The given figure consists of two congruent rectangles with a length of x centimeters and a with of 8-x centimeters. We can multiply these expressions to get the area A_R of one rectangle. A_R=l w Substitute A_R= x( x-8) Multiplying A_R by 2 will give us the total area A_T of both rectangles as function of x. A_T(x)= 2A_R ⇒ A_T(x)= 2x(x-8) We can expand this function to identify its type.
This is a quadratic function whose graph is a parabola. Since its leading coefficient is negative, it opens downward and reaches its maximum at its vertex. This means that we can find the value of x that maximizes the area by determining the axis of symmetry. Consider the equation for the axis of symmetry. x=x_1+x_2/2 In this equation, x_1 and x_2 are the zeros of the parabola. Let's use the factored form of the function to find the zeros of the parabola.
The function has zeros at x_1= 0 and x_2= 8. Now, we can find the equation of the axis of symmetry. x = 0+ 8/2 = 4 Finally, by substituting x= 4 into A_T(x), we can calculate the maximum area of the rectangles.
The maximum total area of the rectangles is 32 square centimeters.
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Izabella plans to build a rectangular enclosure to protect her strawberries from hungry rabbits. She wants to build it alongside one barn wall so that only three sides of the fence are needed.
Izabella wants the area of the enclosure to be as large as possible. However, she only has 9 meters of fence materials and a gate with a width of 1 meter. What is the largest possible area of the enclosure?
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To find the maximum area of the enclosure, we will need to first write an expression for the area. Let's mark one of the sides of the rectangle as x and the other one as y.
We are told that Izabella has 9 meters of fence material and a gate with a width of 1 meter. Including the gate width, this means that the sum of the lengths of all three sides should be 9+ 1=10 meters. This information leads us to write the following equation, where x and y represent the fence sides. 2x + y = 10 [0.4em] ⇕ [0.4em] y = 10 - 2x Since the enclosure will have a rectangular shape, we can calculate its area by multiplying one side length by the other. This will result in the following function. A(x) = x(10-2x) We want to determine the value of x that maximizes this function. We will expand the product on the right-hand side to identify the main parts of the function.
We have a quadratic function whose graph is a parabola. Additionally, because its leading coefficient is negative, its maximum value will be reached at the vertex of the parabola. We can find this value by determining the axis of symmetry. To do so, we will use the equation x=- b2a. Let's substitute a= -2 and b= 10 into the equation.
We will substitute x= 2.5 into the factored form of A(x) to find the maximum area of the enclosure.
Therefore, the maximum area of the enclosure is 12.5 square meters.
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We are given a diagram and some information about the area and the length of the fencing. To write an expression for x in terms of y, we will first redraw the diagram and label all sides of the fence.
From the diagram, we can see that the total length of vertical fencing is 3y and the total length of horizontal fencing is 4x. Because we are given that Kevin wants to use 320 meters of fencing, we will equate the sum of 4x and 3y to 320 to write an expression for the total length of the fencing. 4x+3y=320 Next, we can solve this equation for x to find the expression for x in terms of y.
To find the possible values of y, we will use the information that Kevin plans for the fence to be 1400 square meters. From the diagram, we have noted that each pasture is rectangular with a width of x and a length of y.
The area of a rectangle is given by the product of its width and its length. Previously, we found that x = 80 - 34y. Let's use this result to write the area A of the rectangle.
Since the area of each pasture is planned to be 1400 square meters, we can substitute this value into our equation. Let's substitute it and rewrite the equation in standard form.
Now that the equation is in standard form, we can use the Quadratic formula to find the possible values for y.
Using the Quadratic Formula, we found that the solutions of the equation are y= -80 ± sqrt(2200)-1.5. Let's separate the two cases and use a calculator to find the values of y.
| y = -80 ± sqrt(2200)/-1.5 | |
|---|---|
| y_1 = -80 + sqrt(2200)/-1.5 | y_2 = -80 - sqrt(2200)/-1.5 |
| y_1 ≈ 22 | y_2 ≈ 85 |
Thanks to our calculations, we can say that the possible values of y are about 22 or 85 meters.
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Applications of Quadratic Functions
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