Quadratic Functions and Their Real-World Applications

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.

Izabella and Kevin want to buy a new car for their business. Because they want to make the best decision, they researched how the value of two example cars will decrease over time. The value of the first car is described by a quadratic function.

V (t) = 40 t^{2} - 1600 t + 18000

Here,

t

is the time in years after the purchase of the car and

V (t)

is the corresponding value of the car in dollars. On the other hand, the value of the second car is modeled by a linear function.

W (t) = 20000 - 1400 t

Similarly,

t

is the time in years since the purchase of the car and

W (t)

is the corresponding value of the car.

How much should Izabella and Kevin pay for the first car?

According to the given models, after how many years will the value of the cars be the same?

The model of the first car's value is a quadratic function. It gives the price V(t) of the car t years after the purchase. V(t)=40t^2-1600t+18 000 When they buy the car, 0 years have passed. This means that we can evaluate the function rule when t= 0 to determine the initial value of the vehicle.

According to the given model, the purchase value of the first car is $18 000.

To find after how many years the value of the cars will be the same, we will equate both functions and find the solutions to the resulting equation.

We have obtained a quadratic equation in standard form. We can solve this equation using the Quadratic Formula. To do so, let's first identify the values of the coefficients a, b, and c in the equation. 40t^2-200t-2000=0 ⇕ 40t^2+( -200)t+( -2000)=0 We have that a = 40, b = -200, and c = -2000. Let's substitute these values into the Quadratic Formula and solve for t.

The solutions to the equation are t= 200± 60080. Let's separate them into the positive and negative cases.

t=200± 600/80
t_1=200 + 600/80	t_2=200 - 600/80
t_1=800/80	t_2=-400/80
t_1=10	t_2=-5

Using the Quadratic Formula, we obtained that the models agree on the value of the cars at t=10 and t=-5. Because we are asked to determine the number of years after the purchase, we only consider t=10. Therefore, after 10 years the value of both cars will be the same.

Kevin's family is planning to build a corral for their horses. They have already bought enough materials to build a corral with a perimeter of

28

meters. What is the maximum area of the corral?

We will determine the maximum area of the rectangular corral. Consider that the area A of a rectangle is given by the product of its length and its width. Let x be the length and y the width of the rectangle. A=xy This means we need to determine the side lengths of the rectangle to find its area. Additionally, we are told that the rectangle has a perimeter of 28 square meters. We can substitute this information into the formula for the perimeter of a rectangle to find an expression for the width y in terms of the length x.

We have obtained that if the length of the corral is x, the width will be 14-x.

We can substitute the expression for y into the formula for the area and simplify to get an expression for the area of the corral.

This function will be modeled with a parabola. Since the leading coefficient is negative, the parabola 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 of the parabola. Consider the formula for the axis of symmetry. Axis of Symmetry [0.5em] x=-b/2a In our quadratic function, a= -1 and b= 14. Let's find the value of x.

We have obtained that the axis of symmetry is x= 7. This means that the length of the corral must be 7 and the width 14- 7= 7 to maximize its area. Let's substitute these values into the formula for the area to determine the maximum area of the rectangle.

Therefore, the maximum area of the corral is 49 square meters.

Kevin plans to build a rectangular paddock for his horses in the meadows bordering Lake of the Ozarks. He has bought enough fence material to build $180$ meters of fence. The materials must be sufficient for the three sides because the fourth side will be the lakeshore, as the figure shows.

Find an expression for the area of the paddock

A (x)

in terms of

x -

variable. Write the expression in standard form

a x^{2} + b x + c .

Find the value of

y

so that the area of the paddock becomes as large as possible.

Since the paddock forms a rectangle, its area is calculated by multiplying its width and its length. A = xy We want to write this expression as a function of x. To do so, we need to rewrite the y-variable in terms of x. We can do this by writing an expression for the perimeter using the given lengths. Keep in mind that we only need to consider three sides of the rectangle. P = 2x + y Recall that the perimeter is 180 meters. Let's substitute P= 180 into the obtained expression and solve it for y.

The dimensions of the rectangle are x meters and (180-2x) meters. We can now write the area as a function of x. Let's also distribute x to rewrite the quadratic function in standard form.

In Part A, we found that the area of the paddock is A(x) = -2x^2+180x. Because the leading coefficient is negative, the quadratic function reaches a maximum value at its vertex. The maximum point lies on the axis of symmetry of the parabola. To find it, consider the area written in the initial form and factor out -2. A(x) = x(180-2x) ⇕ A(x) = -2x(x-90) Notice that the area is now written in the intercept form. This form gives us that the zeros of the function are x_1=0 and x_2=90. The axis of symmetry is the vertical line with equation x= x_1+x_22.

Therefore, the axis of symmetry is given by x=45. This means that the area of the rectangle is the largest when x=45. In Part A, we obtained that y=180-2x. Let's substitute x= 45 into the expression to find the corresponding y.

Finally, we obtained that the area of the rectangle is the largest when y is equal to 90 meters.

A water fountain in the center of New York shoots water whose trajectory can be modeled by the graph of the following quadratic function.

y = - 0.01 x^{2} + x + 5

Here,

x

is the horizontal distance from the fountain and

y

is the height above the lawn. Both

x

and

y

are measured in feet.

Does the water reach a height of

40

feet?

What is the horizontal distance from the fountain at which the water arc is

25

feet above the lawn? Round the distance to

2

decimal places.

Consider the given quadratic function. y = -0.01x^2 + x + 5 We are asked whether the water can reach a height of 40 feet. Since y represents the height above the lawn, start by substituting y= 40 into the given equation. Then, we will rewrite it in standard form to check if there are any solutions.

Now, because the equation is in standard form ax^2+bx+c, we can find the discriminant to check if there are any real solutions to the equation. The discriminant is given by the expression b^2-4ac. Let's substitute a= -0.01, b= 1, and c= -35 into the expression.

Because the discriminant is negative, there are no solutions to the given equation. This means that the water does not reach a height of 40 feet.

We want to find the horizontal distance from the fountain where the water arc is 25 feet above the lawn. Similar to Part A, let's substitute y=25 into the given equation and rewrite it in standard form.

To find the horizontal distance, we will use the Quadratic Formula. Let's now identify the values of a, b, and c in the standard form we just obtained. -0.01x^2+x-20 = 0 ⇕ -0.01x^2+ 1x+( -20) = 0 We will substitute a= -0.01, b= 1, and c = -20 into the Quadratic Formula and solve for x.

Using the Quadratic Formula, we found that the solutions of the equation are x= -1± sqrt(0.2)-0.02. Let's separate the two cases and use a calculator to find the horizontal distances.

x = -1 ± sqrt(0.2)/-0.02
x_1 = -1 + sqrt(0.2)/-0.02	x_2 = -1 - sqrt(0.2)/-0.02
x_1 ≈ 27.64	x_2 ≈ 72.36

Therefore, the water reaches a height of 25 feet twice — at distances of about 27.64 and 72.36 feet from the fountain.

Applications of Quadratic Functions

Catch-Up and Review

Investigating the Price of a Car

Analyzing a Swimmer's Dive

Hint

Solution

Comparing the Heights of a Football

Answer

Hint

Solution

Making Conclusions About Different Forms of a Quadratic Function

Hint

Solution

Calculating the Width of a Sidewalk

Hint

Solution

Analyzing the Price of a Car

Hint

Solution

Applications of Quadratic Functions

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