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eCourses /

Algebra 1 /

Chapter 9

Applications of Quadratic Functions

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

Image Credits

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.

Quadratic Function
Parabola
Standard Form of a Quadratic Equation
Vertex Form of a Quadratic Equation
Factored Form of a Quadratic Function

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.

The graph of how the price of the car has changed

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 .$

a Interpret the meaning of the point

(25, 41)

that lies on the parabola.

b What was the lowest price of the car?

At which year the car had that price?

c What was the price of the car in

2020 ?

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

Her height above the water in feet can be described by the following quadratic function.

f (x) = - 4 x^{2} + 9 x + 5

Here,

x

represents the number of seconds since the dive is made.

a What is the height of the springboard above the water?

b How many seconds after she jumps from the springboard will Zosia hit the water? Round the time to one decimal place.

c After how many seconds will Zosia reach the highest point of her jump? Give the exact time.

What will her height be at that moment? Round the height to one decimal place.

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} = - \frac{b}{2 a} .

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) = - 4 x^{2} + 9 x + 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) = - 4 x^{2} + 9 x + 5

Substitute

$f (x) = 0$

0 = - 4 x^{2} + 9 x + 5

▼

Solve for

x

UseQuadForm

Use the Quadratic Formula: $a = - 4, b = 9, c = 5$

x = \frac{- 9 \pm ( - 9 ) ^{2} - 4 ( - 4 ) ( 5 )}{2 ( - 4 )}

CalcPow

Calculate power

x = \frac{- 9 \pm 8 1 - 4 ( - 4 ) ( 5 )}{2 ( - 4 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 9 \pm 8 1 - ( - 1 6 ) ( 5 )}{- 8}

MultNegPos

$(- a) b = - a b$

x = \frac{- 9 \pm 8 1 - ( - 8 0 )}{- 8}

SubNeg

$a - (- b) = a + b$

x = \frac{- 9 \pm 1 6 1}{- 8}

StateSol

State solutions

x_{1} = \frac{- 9 + 1 6 1}{- 8} x_{2} = \frac{- 9 - 1 6 1}{- 8} (I) (II)

$(I), (II):$ Use a calculator

x_{1} = \frac{- 9 + 1 2 . 6 8 8 5 7 7 \dots}{- 8} x_{2} = \frac{- 9 - 1 2 . 6 8 8 5 7 7 \dots}{- 8}

$(I), (II):$ Add and subtract terms

x_{1} = \frac{3 . 6 8 8 5 7 7 \dots}{- 8} x_{2} = \frac{- 2 1 . 6 8 8 5 7 7 \dots}{- 8}

$(I), (II):$ Use a calculator

x_{1} = - 0.461072 \dots x_{2} = 2.711072 \dots

$(I), (II):$ Round to $1$ decimal place(s)

x_{1} \approx - 0.5 x_{2} \approx 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} = - \frac{b}{2 a}

In this case,

a = - 4

and

b = 9 .

By substituting these values, the

x -

coordinate of the vertex can be calculated.

x_{v} = - \frac{b}{2 a}

SubstituteII

$a = - 4$ , $b = 9$

x_{v} = - \frac{9}{2 ( - 4 )}

▼

Evaluate right-hand side

MultPosNeg

$a (- b) = - a \cdot b$

x_{v} = - \frac{9}{- 8}

RemoveNegFracAndDenom

$- \frac{a}{- b} = \frac{a}{b}$

x_{v} = \frac{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) = - 4 x^{2} + 9 x + 5

Substitute

$x = 1.125$

f (1.125) = - 4 (1.125)^{2} + 9 (1.125) + 5

▼

Evaluate right-hand side

CalcPow

Calculate power

f (1.125) = - 4 (1.265625) + 9 (1.125) + 5

Multiply

f (1.125) = - 5.0625 + 10.125 + 5

AddTerms

Add terms

f (1.125) = 10.0625

RoundDec

Round to $1$ decimal place(s)

f (1.125) \approx 10.1

Therefore, Zosia will be about

10.1

feet above the water at the highest point of her jump.

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.

Mark and Tearrik throwing football to each other

External credits: @freepik

When Mark throws the ball, its height is represented by the following quadratic function.

M (x) = - 16 x^{2} + 48 x + 5

For the function representing Mark's throw,

x

is the number of seconds since Mark throws the ball, and

M

is the height of the ball in feet. Tearrik also throws the ball back to Mark. The height of the football when Tearrik throws it is modeled by the following graph.

The graph of the height of Tearrik's football

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

Mark's and Tearrik's ball's heights on one coordinate plane

Hint

a Find the maximum of each function.

b Solve the equation for

M (x) = 0 .

Find the point on the graph when the ball is

0

feet above the ground.

c Graph the equation in vertex form on the coordinate plane by plotting the vertex, the

y -

intercept and the axis of symmetry.

Solution

a When Mark throws the ball, its height is represented by a quadratic function.

M (x) = - 16 x^{2} + 48 x + 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) = - 16 x^{2} + 48 x + 5

▼

Rewrite

SplitIntoFactors

Split into factors

M (x) = - 16 x^{2} + 16 (3) x + 5

FactorOut

Factor out $- 16$

M (x) = - 16 (x^{2} - 3 x - \frac{5}{1 6})

AddDiffZero

$a = a + (\frac{3}{2})^{2} - (\frac{3}{2})^{2}$

M (x) = - 16 (x^{2} - 3 x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

NumberToFrac

$a = \frac{2 \cdot a}{2}$

M (x) = - 16 (x^{2} - \frac{6}{2} x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

SplitIntoFactors

Split into factors

M (x) = - 16 (x^{2} - \frac{2 ( 3 )}{2} x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

M (x) = - 16 (x^{2} - 2 (\frac{3}{2}) x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

CommutativePropMult

Commutative Property of Multiplication

M (x) = - 16 (x^{2} - 2 x (\frac{3}{2}) + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

M (x) = - 16 ((x - \frac{3}{2})^{2} - (\frac{3}{2})^{2} - \frac{5}{1 6})

Distr

Distribute $- 16$

M (x) = - 16 (x - \frac{3}{2})^{2} + 16 (\frac{3}{2})^{2} + 16 (\frac{5}{1 6})

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

M (x) = - 16 (x - \frac{3}{2})^{2} + 16 (\frac{9}{4}) + 16 (\frac{5}{1 6})

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

M (x) = - 16 (x - \frac{3}{2})^{2} + \frac{1 4 4}{4} + 16 (\frac{5}{1 6})

CalcQuot

Calculate quotient

M (x) = - 16 (x - \frac{3}{2})^{2} + 36 + 16 (\frac{5}{1 6})

DenomMultFracToNumber

$16 \cdot \frac{a}{1 6} = a$

M (x) = - 16 (x - \frac{3}{2})^{2} + 36 + 5

AddTerms

Add terms

M (x) = - 16 (x - \frac{3}{2})^{2} + 41

FracToDiv

$\frac{a}{b} = a \div b$

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) M (x) = - 1 a (x - h)^{2} + k = - 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) = - 16 x^{2} + 48 x + 5

Substitute

$M (x) = 0$

0 = - 16 x^{2} + 48 x + 5

▼

Solve for

x

UseQuadForm

Use the Quadratic Formula: $a = - 16, b = 48, c = 5$

x = \frac{- 4 8 \pm 4 8 ^{2} - 4 ( - 1 6 ) ( 5 )}{2 ( - 1 6 )}

CalcPow

Calculate power

x = \frac{- 4 8 \pm 2 3 0 4 - 4 ( - 1 6 ) ( 5 )}{2 ( - 1 6 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 4 8 \pm 2 3 0 4 - ( - 6 4 ) 5}{- 3 2}

MultNegPos

$(- a) b = - a b$

x = \frac{- 4 8 \pm 2 3 0 4 - ( - 3 2 0 )}{- 3 2}

SubNeg

$a - (- b) = a + b$

x = \frac{- 4 8 \pm 2 6 2 4}{- 3 2}

CalcRoot

Calculate root

x = \frac{- 4 8 \pm 5 1 . 2 2 4 9 9 3 \dots}{- 3 2}

StateSol

State solutions

x_{1} = \frac{- 4 8 + 5 1 . 2 2 4 9 9 3 \dots}{- 3 2} x_{2} = \frac{- 4 8 - 5 1 . 2 2 4 9 9 3 \dots}{- 3 2} (I) (II)

$(I), (II):$ Add and subtract terms

x_{1} = \frac{3 . 2 2 4 9 9 3 \dots}{- 3 2} x_{2} = \frac{- 9 9 . 2 2 4 9 9 3 \dots}{- 3 2}

MoveNegDenomToFrac

$(I):$ Put minus sign in front of fraction

x_{1} = - \frac{3 . 2 2 4 9 9 3 \dots}{3 2} x_{2} = \frac{- 9 9 . 2 2 4 9 9 3 \dots}{- 3 2}

DivNegNeg

$(II):$ $\frac{- a}{- b} = \frac{a}{b}$

x_{1} = - \frac{3 . 2 2 4 9 9 3 \dots}{3 2} x_{2} = \frac{9 9 . 2 2 4 9 9 3 \dots}{3 2}

$(I), (II):$ Use a calculator

x_{1} = - 0.100781 \dots x_{2} = 3.100781 \dots

$(I), (II):$ Round to $1$ decimal place(s)

x_{1} \approx - 0.1 x_{2} \approx 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

$(- a) b = - a b$

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.

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) P (x) P (x) = - 3 x^{2} + 852 x - 37260 = - 3 (x - 54) (x - 230) = - 3 (x - 142)^{2} + 23232

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?

What is a price that produces a profit of zero dollars? If there are more than one, write any of them.

b Which of the given equations reveals the profit when the price is

$ 0 ?

What is the value of that profit?

c Which of the given equations reveals the price that produces the highest possible profit?

What is that price?

What is the value of the corresponding profit?

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

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) P (0) P (0) = - 3 (0)^{2} + 852 (0) - 37260 = - 3 (0 - 54) (0 - 230) = - 3 (0 - 142)^{2} + 23232

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) P (0) P (0) = - 3 (0)^{2} + 852 (0) - 37260 = 0 + 0 - 37260 = - 37260

Therefore, the first equation — written in standard form — reveals that if the price of the item is

$ 0,

the profit will be

- $ 37260 .

This means that if the company gives away the product for free, they will lose

$ 37260 .

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) P (x) = - a (x - p)^{2} + q = - 3 (x - 142)^{2} + 23232

In the vertex form,

(p, q)

are the coordinates of the vertex, and the given situation has its vertex at

(142, 23232) .

This means that by setting the baby chair's price at

$ 142,

the company makes its highest possible profit of

$ 23232 .

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?

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

2 x

less than the dimensions of the place found.

h = 40 - 2 x w = 50 - 2 x

Because the rink has a rectangular shape, its area is the product of its width and its height.

A = h w ⇓ A = (40 - 2 x) (50 - 2 x)

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 - 2 x) (50 - 2 x)

Substitute

$A = 1575$

1575 = (40 - 2 x) (50 - 2 x)

▼

Multiply parentheses

Distr

Distribute $(40 - 2 x)$

1575 = (40 - 2 x) (50) - (40 - 2 x) (2 x)

Distr

Distribute $50 & 2 x$

1575 = 2000 - 100 x - 80 x + 4 x^{2}

SubTerm

Subtract term

1575 = 2000 - 180 x + 4 x^{2}

CommutativePropAdd

Commutative Property of Addition

1575 = 4 x^{2} - 180 x + 2000

SubEqn

$LHS - 1575 = RHS - 1575$

0 = 4 x^{2} - 180 x + 425

Now, using the Quadratic Formula, the solutions to the equation can be found.

0 = 4 x^{2} - 180 x + 425

▼

Solve for

x

UseQuadForm

Use the Quadratic Formula: $a = 4, b = - 180, c = 425$

x = \frac{- ( - 1 8 0 ) \pm ( - 1 8 0 ) ^{2} - 4 ( 4 ) ( 4 2 5 )}{2 ( 4 )}

NegNeg

$- (- a) = a$

x = \frac{1 8 0 \pm ( - 1 8 0 ) ^{2} - 4 ( 4 ) ( 4 2 5 )}{2 ( 4 )}

CalcPow

Calculate power

x = \frac{1 8 0 \pm 3 2 4 0 0 - 4 ( 4 ) ( 4 2 5 )}{2 ( 4 )}

Multiply

x = \frac{1 8 0 \pm 3 2 4 0 0 - 6 8 0 0}{8}

SubTerm

Subtract term

x = \frac{1 8 0 \pm 2 5 6 0 0}{8}

CalcRoot

Calculate root

x = \frac{1 8 0 \pm 1 6 0}{8}

Therefore, the quadratic equation has two solutions.

x_{1} = \frac{1 8 0 + 1 6 0}{8} ⇓ x_{1} = 40 x_{2} = \frac{1 8 0 - 1 6 0}{8} ⇓ 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!

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 $$ 11000 .$

a Interpret the meaning of the point

(25, 41)

that lies on the parabola.

b What was the lowest price of the car?

At which year the car had that price?

c What was the price of the car in

2020 ?

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 $$ 41000 .$

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

$ 11000 .

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 11 = a (15)^{2} + b (15) + 3.5 (I) (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 = 625 a + 25 b 7.5 = 225 a + 15 b

▼

(I):

Solve for

b

SubEqn

$(I):$ $LHS - 625 a = RHS - 625 a$

{37.5 - 625 a = 25 b 7.5 = 225 a + 15 b

DivEqn

$(I):$ $LHS / 25 = RHS / 25$

{\frac{3 7 . 5 - 6 2 5 a}{2 5} = b 7.5 = 225 a + 15 b

WriteDiffFrac

$(I):$ Write as a difference of fractions

{\frac{3 7 . 5}{2 5} - \frac{6 2 5 a}{2 5} = b 7.5 = 225 a + 15 b

MovePartNumRight

$(I):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{\frac{3 7 . 5}{2 5} - \frac{6 2 5}{2 5} a = b 7.5 = 225 a + 15 b

CalcQuot

$(I):$ Calculate quotient

{1.5 - 25 a = b 7.5 = 225 a + 15 b

RearrangeEqn

$(I):$ Rearrange equation

{b = 1.5 - 25 a 7.5 = 225 a + 15 b

Substitute

$(II):$ $b = 1.5 - 25 a$

{b = 1.5 - 25 a 7.5 = 225 a + 15 (1.5 - 25 a)

▼

(II):

Solve for

a

Distr

$(II):$ Distribute $15$

{b = 1.5 - 25 a 7.5 = 225 a + 22.5 - 375 a

SubTerm

$(II):$ Subtract term

{b = 1.5 - 25 a 7.5 = - 150 a + 22.5

SubEqn

$(II):$ $LHS - 22.5 = RHS - 22.5$

{b = 1.5 - 25 a - 15 = - 150 a

DivEqn

$(II):$ $LHS / (- 150) = RHS / (- 150)$

{b = 1.5 - 25 a 0.1 = a

RearrangeEqn

$(II):$ Rearrange equation

{b = 1.5 - 25 a a = 0.1

Substitute

$(I):$ $a = 0.1$

{b = 1.5 - 25 (0.1) a = 0.1

▼

(I):

Solve for

b

Multiply

$(I):$ Multiply

{b = 1.5 - 2.5 a = 0.1

SubTerm

$(I):$ Subtract term

{b = - 1 a = 0.1

Now, the quadratic equation that describes the parabola can be completed.

y = 0.1 x^{2} - 1 x + 3.5 ⇕ y = 0.1 x^{2} - x + 3.5

Finally, by substituting

5

for

x,

the lowest price of the car can be calculated.

y = 0.1 x^{2} - x + 3.5

Substitute

$x = 5$

y = 0.1 (5)^{2} - 5 + 3.5

▼

Evaluate right-hand side

CalcPow

Calculate power

y = 0.1 (25) - 5 + 3.5

Multiply

y = 2.5 - 5 + 3.5

AddSubTerms

Add and subtract terms

y = 1

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 = 0.1 x^{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

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

$ 2035000 .

Applications of Quadratic Functions

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