Quadratic Functions and their modeling

Section 2 Quadratic Functions and their modeling

Task 2.1.

Suppose that you are in charge of managing constructing materials at a constructing site. You want to build walls to enclose a rectangular area to store the constructing materials needed for project and you want to have as much area as possible. However, you only have enough materials to build

180

feet of walls.

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Pick your favorite length and width so that you can use all the building materials. Then calculate the area of the region.
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Pick your second favorite length and width and repeat the process of the last question. Which dimension gives larger area?
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Draw a diagram showing the rectangular area, and labeling the length of the region using a variable $ℓ .$ Then find the width of the region in terms of the length $ℓ .$
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Write a function equation describing the area of the rectangular area $A$ in terms of the length $ℓ .$ Use Desmos to graph this function.
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Using the graph of this function, decide how should the wall be built to get the maximum possible area. What is the maximum possible area?

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The function that gives the area of the rectangular region in terms of one side of the region is called a quadratic function. In general, a quadratic function is a function of the form

f (x) = a x^{2} + b x + c

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where

a \neq 0 .

We call this the general form of a quadratic function. Now try to rewrite the equation that you used in Task 2.1 in the form of

f (x) = a x^{2} + b x + c .

What are the values of

a, b

and

c ?

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If we want to find the function equation of a quadratic equation when given its graph, we have to pick three different points to plug into the formula

f (x) = a x^{2} + b x + c

because we have to figure out the values of three unknowns

a, b

and

c .

The computations will be lengthy because it involves solving a system of linear equations of three variables (it can be done pretty easily if you know how to use a computer package). We will not be doing this in this course. Instead, we will manipulate the function equation

f (x) = a x^{2} + b x + c

into a much desired form called the vertex form because a great deal of qualitative information about the function and its graph can simply be read off from this form. We will included the derivation of the algebraic manipulation in the following proof. Feel free to read the proof but it is not required.

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Proof.

Before we start doing the algebra, let us recall that

(X + Y)^{2} = X^{2} + 2 X Y + Y^{2} .

We will need this formula from (2.4) to (2.5). Now we manipulate the quadratic function from the general form into the vertex form.

\begin{aligned} (2.1) & f (x) & = a x^{2} + b x + c \\ (2.2) & f (x) & = a x^{2} + b x + \frac{b^{2}}{4 a} - \frac{b^{2}}{4 a} + c \\ (2.3) & f (x) & = a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}}) - \frac{b^{2}}{4 a} + c \\ (2.4) & f (x) & = a (x^{2} + 2 \frac{b}{2 a} x + (\frac{b}{2 a})^{2}) - \frac{b^{2}}{4 a} + c \\ (2.5) & f (x) & = a (x + \frac{b}{2 a})^{2} - \frac{b^{2}}{4 a} + c \\ (2.6) & f (x) & = a (x + \frac{b}{2 a})^{2} - \frac{b^{2} - 4 a c}{4 a} \end{aligned}

By renaming

h = - \frac{b}{2 a}, k = - \frac{b^{2} - 4 a c}{4 a},

we have

f (x) = a (x - h)^{2} + k

Take Equation (2.6), if we want to solve the quadratic equation

a x^{2} + b x + c = 0,

it is enough to solve the equation

a (x + \frac{b}{2 a})^{2} - \frac{b^{2} - 4 a c}{4 a} = 0

This is easy to solve because we have

\begin{aligned} (2.7) & a (x + \frac{b}{2 a})^{2} - \frac{b^{2} - 4 a c}{4 a} & = 0 \\ (2.8) & a (x + \frac{b}{2 a})^{2} & = \frac{b^{2} - 4 a c}{4 a} \\ (2.9) & (x + \frac{b}{2 a})^{2} & = \frac{b^{2} - 4 a c}{4 a^{2}} \\ (2.10) & x + \frac{b}{2 a} & = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \\ (2.11) & x & = - \frac{b}{2 a} \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \\ (2.12) & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}

Equation (2.12) is the well-known quadratic formula that you have seen before.

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The vertex form of a quadratic equation is of the form

f (x) = a (x - h)^{2} + k .

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The reason this is called the vertex form is because we can read off the vertex of the graph, which is

(h, k) .

The vertex of a quadratic function is the lowest point (if

a > 0

) or the highest point (if

a < 0

). Using the vertex form of a quadratic equation, we can find the function equation of a quadratic equation by first identifying the vertex (so we will know the values of

h

and

k

), and then pick a point on the graph of the quadratic equation other than the vertex to find the value of

a .

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Task 2.2.

Refer to the graph of

k (x)

in Figure Figure 2.3.

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What are the coordinates of the vertex of $k (x) ?$
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Write a function equation for $k (x) .$ Graph $k (x)$ in Desmos.
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Solve the quadratic equation $k (x) = 15$
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Solve the inequality $k (x) \leq 15 .$
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Is it possible that $k (x) = - 11 ?$ What about $200 ?$ What are the possible outputs of the function $k (x) ?$

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Definition 2.4.

The domain of a function is the set of allowable inputs to the function. The range of a function is the set of all possible outputs from the function.

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Definition 2.5.

A function

y = f (x)

is said to be increasing if the graph of

f (x)

is going uphill from left to right. If the graph of

f (x)

is going downhill from left to right, then it is called decreasing.

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Task 2.6.

Refer to the graph of

q (x)

in Figure Figure 2.7.

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Find $q (- 2) .$
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What are the coordinates of the vertex of $q (x) ?$
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Write a function equation for $q (x) .$ Graph $q (x)$ in Desmos.
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Solve the quadratic equation $q (x) = - 6$
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Solve the inequality $q (x) \leq - 6 .$
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Solve the inequality $q (x) > 2 x - 12 .$
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What is the domain and the range of the function $q (x) ?$
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On which interval is the function $q (x)$ increasing? On which interval is the function $q (x)$ decreasing?

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Definition 2.8.

The average rate of change of a function

f (x)

defined on an interval

a \leq x \leq b

\frac{f (b) - f (a)}{b - a} .

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Note that the average rate of change computes the slope of the line between the two points

(a, f (a))

and

(b, f (b)) .

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Task 2.9.

A flower pot falls from the ledge of a balcony on a high-rise building. Any object experiencing the force of gravity can be modeled by the equation

h (t) = - 16 t^{2} + v t + c,

where

t

is the time in seconds,

h (t)

is the height in feet,

c

is the initial height of the object, and

v

is the initial velocity of the object. (This model applies to any object experiencing the force of gravity, where the measurement is in English units (feet). There is another version for metric units (meters).)

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If the pot falls, what is its initial velocity?
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Suppose its takes $3$ seconds for the pot to hit the ground. How high was the balcony?
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Write down the function equation $h (t)$ that gives the height of the flower pot at time $t .$
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What is the domain and the range of the function $h (t) ?$
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When is the flower pot at the height of $100$ feet?
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What is the average velocity of the flower pot from $0$ second to $3$ seconds?

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Task 2.10.

Mike is the owner of a local ready-mix concrete retailer. He finds that if he charges

$ 90

per cubic yard, he can sell

3, 000

cubic yard per day. Thereafter, for every

$ 5

he raises the price,

200

cubic yard fewer will be sold.

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Make a table showing the price of concrete (per cubic yard), the amount of concrete sold per day, and the total income per day, for the prices of $$ 90$ to $$ 120$ per cubic yard, in $$ 5$ increments.
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Write a function equation for the amount of concrete sold per day $A$ as a function of the price of concrete $p .$
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Write an equation describing the total income for one day $I (p)$ in terms of the price of concrete $p$ .
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What is the domain and the range of the function $I (p) ?$
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Suppose Mike will be satisfied if the total income for one day is at least $$ 268, 000 .$ What are the possible price range (per cubic yard) he can charge to earn this income?
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What price will Mike earn the maximum income per day? How many cubic yards of concrete will be sold per day? What is the maximum daily income?

Precalculus

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