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This lesson will cover the key features of quadratic functions, how they can be represented in graphs, and how they can be used to model real-life scenarios.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Exploring Parabolas

The following applet shows a variety of parabolas in the same coordinate plane. Inspect and compare them.

What do these parabolas have in common, and what are their differences?

Extra

Graphs of the Parabolas

The parabolas are shown in the graph below if there is a problem loading the applet.

Discussion

Domain of a Quadratic Function

When given a function, it is not always possible to use any value as an input. Sometimes the input will not make sense in the given context of the function. It could also be that the function is not defined for such value.

The domain of a function is the set of all

x -

values, or inputs, for which the function is defined. For example, consider the following functions.

f (x) g (x) h (x) = 3 x = x = \frac{1}{x}

Their domains can be written by analyzing the definition of each function.

Function	Analysis	Domain
$f (x) = 3 x$	Multiplication by $3$ is defined for all real numbers.	All real numbers
$g (x) = x$	Square roots are not defined for negative numbers.	All non-negative numbers — that is, $x \geq 0$
$h (x) = \frac{1}{x}$	Division by zero is undefined.	All real numbers except $0$ — that is, $x \neq = 0$

Depending on how a function is represented, its domain can be determined by using different methods.

Graph in the coordinate plane, table of values, set of coordinate pairs, and mapping diagram

The domain of a function also depends on what the function describes. For example, let

p (x) = 2 x

be a function representing the price of

x

apples at a market. Although the function is defined for all real numbers, it does not make sense to find the price of a negative number of apples or a fraction of an apple. Here, the domain of

p (x)

is all non-negative integers.

Consider a quadratic function written in standard form.

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

Every operation needed to evaluate this function — addition, multiplication, and exponents — is defined for all real numbers. Therefore, the domain of a quadratic function consists of all real numbers. Furthermore, the graph of a quadratic function extends over the entire

x -

axis, regardless of the graph's shape and position.

However, if a quadratic function is modeling a particular scenario, the domain may have to be modified to fit the context.

Discussion

Range of a Quadratic Function

Given a function and its domain it is possible to study the set of all possible outputs.

Concept

Range

The range of a function is the set of all $y -$ values, or outputs, of the function. The range of a function depends on both the domain and the function itself. For example, consider the following functions and their defined domains.

Function	Domain
$f (x) = 2 x$	All integers
$g (x) = x^{2}$	All real numbers
$h (x) = 4$	All real numbers

By analyzing the definition of each function along with the given domains, the ranges can be determined.

Function	Domain	Analysis	Range
$f (x) = 2 x$	All integers	The function takes any integer input and produces an output that is an even number, as each input is multiplied by $2 .$	All even numbers
$g (x) = x^{2}$	All real numbers	The function takes any real number input and produces an output that is a non-negative number, as each input is squared.	All non-negative numbers — that is, $y \geq 0$
$h (x) = 4$	All real numbers	The function takes any real number input and sends it to $4 .$	Only the number $4$ — that is, the range is ${4}$

Depending on how a function is represented, its range can be determined using different methods.

If two different inputs have the same output, it is not necessary to repeat the output when writing the range.

One of the characteristics of the graph of a quadratic function — a parabola — is the vertex.

Concept

Vertex of a Parabola

Because a parabola either opens upward or downward, there is always one point that is the absolute maximum or absolute minimum of the function. This point is called the vertex.

At the vertex, the function changes from increasing to decreasing, or vice versa.

Knowing the vertex of a parabola is very helpful when looking for the range of a quadratic function. Consider the following quadratic function given in vertex form. It is assumed that this function is defined for all real numbers.

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

The coordinates of the vertex are given by

(h, k) .

Consider the case when

a > 0,

meaning that the parabola opens upward. In this case

k

is the absolute minimum of the function, so the range will start from

k

and extend towards positive infinity.

[k, \infty)

Likewise, if

a < 0

the parabola opens downward, meaning that

k

is the absolute maximum of the function, so the range will come up from negative infinity up to

k .

(- \infty, k]

$a$ -value	Range
$a > 0$	$[k, \infty)$
$a < 0$	$(- \infty, k]$

Example

Optimizing a Fence

A park asks for volunteers to put up a rectangular fence to keep deer from eating tree saplings. Each volunteer will receive a ticket to an amusement park. Dominika and her friend Emily volunteer. They are given

60

feet of wire for the fence, and are told to use all of it. They recall the formula for the area of a rectangle.

A = ℓ w

It is possible to change the area

A

of the fenced region by adjusting the length

ℓ

of the fence. When the length changes, the width

w

also changes.

Help Dominika and Emily analyze this situation so they can do a good job putting up the fence.

a Write a quadratic function

A (ℓ)

for the area of the fenced region in terms of its length

ℓ .

b State the domain of

A (ℓ)

for the fence. Write the answer in interval notation.

c Find the vertex of the quadratic function.

d State the range of

A (ℓ) .

Write the range in interval notation.

e Interpret the coordinates of the vertex.

Answer

A (ℓ) = 30 ℓ - ℓ^{2}

b Domain:

(0, 30)

(15, 225)

d Range:

(0, 225]

e The length of the rectangular fence that makes the area of the fenced region the greatest is

15

feet. The area is

225

square feet.

Hint

a Use the formula for the perimeter of a rectangle to express

w

in terms of

ℓ .

b Is it possible to have negative dimensions?

c Write the function in vertex form.

d Is it possible to have negative areas?

e The vertex is at the absolute maximum or absolute minimum of a quadratic function.

Solution

a The formula for the area of a rectangle is already given.

A = ℓ w

In order to write it in terms of

ℓ,

then

w

needs to be expressed in terms of

ℓ .

That means to find a method to substitute

w

for

ℓ .

Since they want to build a rectangular fence, writing in terms of

ℓ

can be done by using the formula for the perimeter of a rectangle.

P = 2 ℓ + 2 w

Dominika has

60

feet of wiring for the fence. This means that the perimeter of the fence is equal to

60 .

Enter this value into the perimeter of a rectangle formula to solve for

w .

P = 2 ℓ + 2 w

Substitute

$P = 60$

60 = 2 ℓ + 2 w

Solve for

w

SubEqn

$LHS - 2 ℓ = RHS - 2 ℓ$

60 - 2 ℓ = 2 w

DivEqn

$LHS / 2 = RHS / 2$

\frac{6 0 - 2 ℓ}{2} = w

WriteDiffFrac

Write as a difference of fractions

\frac{6 0}{2} - \frac{2 ℓ}{2} = w

CalcQuot

Calculate quotient

30 - ℓ = w

RearrangeEqn

Rearrange equation

w = 30 - ℓ

Now that

w

has been expressed in terms of

ℓ,

the area of the fenced region can be written in terms of

ℓ .

A (ℓ) = ℓ w

Substitute

$w = 30 - ℓ$

A (ℓ) = ℓ (30 - ℓ)

Distr

Distribute $ℓ$

A (ℓ) = 30 ℓ - ℓ^{2}

b The domain of a function depends on the formula of the function and on what the function describes. Since

ℓ

represents the length of the rectangle, it is positive.

ℓ > 0

Furthermore, since there is only

60

feet of wiring, the length of each side must be less than

60 .

Additionally, the perimeter of a rectangle includes twice its length, so that means

2 ℓ

is less than

60 .

2 ℓ < 60 ⇓ ℓ < 30

This is a strict inequality because having a length of exactly

30

feet would imply that the width is

0,

meaning there would be no fenced area! Knowing this, the domain of the function can be written in interval notation.

Domain : (0, 30)

c The quadratic function for the area of the fenced region can be written in vertex form by completing the square.

A (ℓ) = 30 ℓ - ℓ^{2}

Write in vertex form

CommutativePropAdd

Commutative Property of Addition

A (ℓ) = - ℓ^{2} + 30 ℓ

FactorOut

Factor out $- 1$

A (ℓ) = - (ℓ^{2} - 30 ℓ)

AddDiffZero

$a = a + 1 5^{2} - 1 5^{2}$

A (ℓ) = - (ℓ^{2} - 30 ℓ + 1 5^{2} - 1 5^{2})

FacNegPerfectSquare

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

A (ℓ) = - ((ℓ - 15)^{2} - 1 5^{2})

Distr

Distribute $- 1$

A (ℓ) = - (ℓ - 15)^{2} + 1 5^{2}

CalcPow

Calculate power

A (ℓ) = - (ℓ - 15)^{2} + 225

With the quadratic function written in vertex form it is now possible to identify the coordinates of the vertex.

y (x) = a (x - h)^{2} + k ⇓ A (ℓ) = - (ℓ - 15)^{2} + 225

This shows that the vertex is at

(15, 225) .

d In Part C, it was found that the vertex of the quadratic function is at

(15, 225) .

Since the parabola opens downward, this means that

225

is an absolute maximum of the function.

A \leq 225

Since the function describes the area of the fenced region, it has to be positive.

A > 0

This is enough information to write the function's range in interval notation.

Range : (0, 225]

e The vertex of a parabola gives information of the maximum or minimum of the function and the value of the independent variable where this maximum or minimum occurs. In this case, it has already been determined it is the maximum that is being used.

Since the vertex is at $(15, 225),$ the maximum area of the fence is $225,$ and this area is obtained by making a fence $15$ feet long. It is not a coincidence that this is a square fence! Not every seedling can be protected from the deer, but Dominka and Emily did great.

Discussion

Increasing and Decreasing Intervals of Quadratic Functions

It has been stated previously that the vertex of a parabola is either the absolute maximum or absolute minimum of the corresponding quadratic function. There is also an important characteristic of the quadratic function's graph that changes at this point.

Concept

Increasing and Decreasing Intervals

A function is called increasing when as the

x -

values increase, the values of

f (x)

also increase. On the other hand, a function is called decreasing when, as

x

increases,

f (x)

decreases. An increasing interval is an interval of the independent variable for an increasing function. A decreasing interval is an interval of the independent variable for a decreasing function.

A point where a function has a maximum or a minimum is neither included in an increasing or a decreasing interval. The previous applet shows a function that contains two increasing intervals and one decreasing interval. Each can be described in terms of the

x -

values.

From left side to x = - 2 From x = - 2 to x = 0 From x = 0 to right side \to \to \to Increasing Decreasing Increasing

Although the entire graph cannot be seen, it is reasonable to assume that it continues in the same manner. In that case, for all

x -

values less than

x = - 2,

f

will be increasing. For all

x

-values greater than

x = 0,

f

will also be increasing.

Increasing Intervals : - \infty 0 Decreasing Interval : - 2 < x < - 2 < x < \infty < x < 0

A point where the function switches between decreasing and increasing is known as a turning point.

A quadratic function with a positive leading coefficient reaches its absolute minimum at the

vertex .

Therefore, the vertex is the point where the function shifts from decreasing to increasing.

Increasing Decreasing Intervals of a Quadratic Function with positive leading coefficient

Conversely, if the leading coefficient is negative the function reaches its absolute maximum at the

vertex .

Therefore, the vertex is the point where the function shifts from increasing to decreasing.

Increasing Decreasing Intervals of a Quadratic Function with negative leading coefficient

Example

A Roller Coaster Described by Intervals

Dominika and Emily arrive to the amusement park. "Last one to the coaster is crumbs under the toaster!" exclaims Emily. Both of them race to the roller coaster. Dominka takes just a few steps before becoming mesmerized by what she sees.

Dominika sees a section of the roller coaster shinning before the night sky. The cart uses a green light as it goes up, and it changes to a red light as it goes down. Mesmerized, Dominika wonders how this works.

Dominika and Emily wondering how the light knows when to change

The two of them look around for the answer. They find a signboard that explains that the cart uses software that controls the lights. By knowing the function and stating the increasing and decreasing intervals, the lights can switch accordingly. The signboard shows the blueprints.

Roller Coaster Blueprints Parabola Section

The blueprints also show the quadratic function that describes the parabolic section of the roller coaster.

f (x) = - (x - 2)^{2} + 4, if 0 \leq x \leq 4

What is the increasing interval of the function?

What is the decreasing interval of the function?

Hint

The $x -$ value for which a function reaches its maximum or minimum is not included in the increasing or decreasing interval.

Solution

The function that describes the parabolic section of the roller coaster is given in vertex form.

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

The vertex is at

(2, 4) .

Since the parabola opens downward, this point is an absolute maximum.

It is also given that the domain of this function is

[0, 4] .

This means that the function increases starting from

x = 0

until it reaches its maximum at

x = 2,

which is not included in the increasing interval.

[0, 2)

From there, the function decreases until it reaches its end at

x = 4 .

As a reminder, the maximum at

x = 2

is not included in the decreasing interval.

(2, 4]

Discussion

Intercepts of Quadratic Functions

A quadratic function might cross the $x -$ and $y -$ axes. These points are known as intercepts.

Concept

Intercept

The $x -$ intercept of a line is the $x -$ coordinate of the point where the line crosses the $x -$ axis. The $y -$ intercept of a line is the $y -$ coordinate of the point where the line crosses the $y -$ axis. The $y -$ intercept of an equation is also known as its initial value.

When talking about functions, the

x -

intercepts are the zeros of the function. Sometimes, only one coordinate of these points is referenced. For example, if the

x -

intercept lies at

(a, 0),

it can be said that the

x -

intercept is at

x = a .

The same is true for the

y -

intercept. A relation can have several intercepts. A function can have multiple

x -

intercepts, but it can only have one

y -

intercept.

To find the

x -

intercepts of a quadratic function begin by setting the function equal to

0 .

f (x) = 0

This results in a quadratic equation that can be solved by a method of choice. The solutions to this equation, if any, are the

x -

intercepts. Since a quadratic equation has at most two solutions, a quadratic function can have either one, two, or no

x -

intercepts.

The

y -

intercept of a quadratic function is found by evaluating the function at

x = 0 .

f (0)

This corresponds to the

c

value of a quadratic equation given in standard form. A quadratic function whose domain includes

x = 0

always has one

y -

intercept.

Example

The Motion of a Ball Described by a Quadratic Function

Dominika and Emily, enjoying the theme park, want to try Shoot 'N' Hoop. If they make their one shot attempt, they will win a teddy bear. As the ball goes up, Dominika imagines her shot being modeled by a quadratic function.

The function representing Dominika's shot is graphed on the following coordinate plane. Take note that the drawn path of the ball in the air in the applet versus when mapped on the graph looks slightly different.

Vertical throw as function of time graph

The height of the ball is represented by $y,$ and time the ball is in the air after Dominika shoots the ball is $x .$ The worker hosting the game tells Dominika and Emily that if they can answer the following math questions correctly, he will give them a second teddy bear.

a Find the

y -

intercept of the function.

b Find the

x -

intercept of the function. Round the answer to two decimal places.

c Interpret both intercepts.

Answer

6

b About

2.17

c The ball was released from

6

feet above the ground. It remained in the air for about

2.17

seconds.

Hint

a Evaluate the function at

x = 0 .

b Set the function equal to

0

and solve using the Quadratic Formula.

c What does

x

and

y

represent in this scenario?

Solution

a The

y -

intercept of the function can be found by evaluating it at

x = 0 .

y = - 16 x^{2} + 32 x + 6

Substitute

$x = 0$

y = - (0)^{2} + 32 (0) + 6

Simplify right-hand side

CalcPow

Calculate power

y = - (0) + 32 (0) + 6

Multiply

y = 0 + 0 + 6

AddTerms

Add terms

y = 6

The

y -

intercept of the function is

6 .

There is a an alternative method to finding the same result. Since the function is given in standard form, identify the constant term of the function.

y = - 16 x^{2} + 32 x + 6

b To find the

x -

intercepts of the function begin by setting it equal to

0 .

- 16 x^{2} + 32 x + 6 = 0

This expression is a quadratic equation which can be solved using the Quadratic Formula. Determine what are the coefficients in the standard form

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

- 16 x^{2} + 32 x + 6 = 0

In this case,

a = - 16,

b = 32,

and

c = 6 .

The Quadratic Formula can now be applied.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The values of

a,

b,

and

c

will be substituted into the formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 3 2 \pm 3 2 ^{2} - 4 ( - 1 6 ) ( 6 )}{2 ( - 1 6 )}

Simplify right-hand side

CalcPow

Calculate power

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

MultNegNegOnePar

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

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

Multiply

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

AddTerms

Add terms

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

MoveNegDenomToNum

Put minus sign in numerator

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

Distr

Distribute $- 1$

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

CalcRoot

Calculate root

x = \frac{3 2 \pm 3 7 . 5 2 3 3 2 6 \dots}{3 2}

Since negative values of time do not make sense in this scenario, only the positive solution will be considered. The time the ball was in the air will be rounded to two decimal places.

x = \frac{3 2 + 3 7 . 5 2 3 3 2 6 \dots}{3 2}

UseCalc

Use a calculator

x = 2.172603 \dots

RoundDec

Round to $2$ decimal place(s)

x \approx 2.17

c The

y -

intercept is the

y -

value at which

x

equals zero. Since

x

represents time in this scenario, the

y -

intercept represents the starting height of the ball. The

x -

intercept is the

x -

value at which

y

equals zero. Since

y

represents height in this scenario, the

x -

intercept represents the time the ball remains in the air.

Discussion

Positive and Negative Intervals of Quadratic Functions

Parts of the graph of a quadratic function can be either above or below the $x -$ axis. The intervals of $x -$ values where this happens receive a special name.

Concept

Positive and Negative Intervals

A function is said to be $positive$ where its graph is above the $x -$ axis and is said to be $negative$ where its graph is below the $x -$ axis.

A positive interval is an interval for which the function is positive. Likewise, a negative interval is an interval for which the function is negative. The graph above shows two negative intervals and two positive intervals. Each interval is described in terms of the

x -

values.

From left side to x = - 2 From x = - 2 to x = 0 From x = 0 to x = 2 From x = 2 to right side \to \to \to \to negative positive negative positive

Although the entire graph cannot be shown, it is reasonable to assume that it continues in the same manner. Thus, for all

x -

values less than

x = - 2,

f

will be negative. Likewise, for all

x

-values greater than

x = 2,

f

will be positive. A point where the function is equal to zero is neither included in a positive nor a negative interval.

Positive Intervals : - 2 2 Negative Intervals : - \infty 0 < x < 0 < x < \infty < x < - 2 < x < 2

Depending on the opening of the parabola and the position of its vertex, a quadratic function may have different positive and negative intervals. If the parabola opens upward, it always has at least one positive interval. A negative interval appears when the vertex is below the

x -

axis.

Positive and Negative intervals of upward parabola moving the vertex

In contrast, if the parabola opens downward, it always has at least one negative interval. A positive interval appears when the vertex is above the

x -

axis.

Positive and Negative intervals of downward parabola moving the vertex

Notice that in either case, the ends of the intervals correspond to the

x -

intercepts of the function.

Example

Filling a Birdbath

Emily and Dominika are relaxing on a bench at the theme park. Next to them, a bluebird appears in a bird bath! The bath seems to be a bit low on water, however. They look up the blueprints of similar birdbaths online and find a cross section of one that includes its quadratic function.

That birdbath's manual suggests to fill the bath with water up to the

x -

axis. This way, the birdbath will be filled in its negative interval. Knowing that the

x -

intercepts are at

x = 0.5

and

x = 5.5,

state the the positive and negative intervals.

Answer

Positive Intervals: $[0, 0.5)$ and $(5.5, 6]$
Negative Interval: $(0.5, 5.5)$

Hint

The positive and negative intervals do not include the points where the function is equal to zero.

Solution

The domain of the function describing the cross section of the birdbath consists of the numbers between

0

and

6 .

[0, 6]

Note that the function graph is

above

the

x -

axis between

x = 0

and

x = 0.5,

meaning that it is

positive

in this interval. Since the positive and negative intervals do not include the points where the function equals zero,

x = 0.5

is not included in this interval.

[0, 0.5)

The function then goes

below

the

x -

axis between

x = 0.5

and

x = 5.5,

meaning that it is

negative

in this interval.

(0.5, 5.5)

In the last bit of the birdbath the function graph goes

above

the

x -

axis once again, meaning that the function is

positive

here.

(5.5, 6]

The intervals can now be summarized.

Positive Intervals : Negative Interval : [0, 0.5) (5.5, 6] (0.5, 5.5)

Dominika and Emily can explain the meaning of this information to the park employees to make sure the birdbath has enough water.

Discussion

End Behavior of Quadratic Functions

In general, the domain of quadratic functions consists of all real numbers. Investigating what happens to the function as the $x -$ values increase or decrease infinitely is valuable in better understanding quadratic functions.

Concept

End Behavior

The end behavior of a function is the value to which

f (x)

tends as

x

extends to the left or the right infinitely. If

f (x)

keeps increasing without bound, it is said to tend to positive infinity. The end behavior of this case is stated as up.

f (x) \to + \infty

Conversely, if

f (x)

keeps decreasing without bound, it is said to tend to negative infinity. In this case, the end behavior is stated as down.

f (x) \to - \infty

For example, consider the graph of a function

g (x) .

From the arrows on the graph, it can be seen that the left end of the graph extends downward, while the right end extends upward. The end behavior of

g

can then be expressed as follows.

As x \to - \infty, As x \to + \infty, g (x) \to - \infty g (x) \to + \infty

To state the end behavior of a function in words, begin by stating the left-end behavior, then state the right-end behavior. A dash can also be used to separate the words. For instance, the end behavior of the graph of

g (x)

can be written as down and up or as down-up.

Since quadratic functions are polynomial functions of degree $2$ their end behavior is either up and up or down and down, all depending on the sign of the leading coefficient.

End Behavior of Quadratic Functions by lead coefficient

Closure

Analyzing a Quadratic Function

Given any quadratic function, all of its key features can be found using the tools provided in this lesson. Consider the following function.

f (x) = x^{2} - 6 x + 5

The graph of the function is shown below.

a Find its vertex. Is it a maximum or a minimum?

b State the domain and range of the function.

c Find the

y -

intercept and the

x -

intercepts, if any.

d State the decreasing and increasing intervals.

e State the positive and negative intervals.

f What is the end behavior of this function?

Answer

(3, - 4),

minimum

b Domain: All real numbers
Range:

[- 4, \infty)

y -

intercept:

5

x -

intercepts:

1,

5

d Decreasing Interval:

(- \infty, 3)

Increasing Interval:

(3, \infty)

e Positive Intervals:

(- \infty, 1)

and

(5, \infty)

Negative Interval:

(1, 5)

f Up and up

Hint

a Write the function in vertex form by completing the square.

b Use the result from Part A.

c Find the

y -

intercept by evaluating the function at

x = 0 .

Find the

x -

intercepts by setting the function equal to

0

and solving for

x .

d The function switches from decreasing to increasing at the vertex.

e Which part of the graph is above the

x -

axis? Which part is below?

f Look at both ends of the graph.

Solution

a The quadratic function is given in standard form.

f (x) = x^{2} - 6 x + 5

To find its vertex the function will be rewritten in the vertex form by completing the square.

f (x) = x^{2} - 6 x + 5

Write in vertex form

Rewrite

Rewrite $6 x$ as $2 \cdot 3 x$

f (x) = x^{2} - 2 \cdot 3 x + 5

AddDiffZero

$a = a + 3^{2} - 3^{2}$

f (x) = x^{2} - 2 \cdot 3 x + 3^{2} - 3^{2} + 5

FacNegPerfectSquare

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

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

CalcPow

Calculate power

f (x) = (x - 3)^{2} - 9 + 5

SubTerm

Subtract term

f (x) = (x - 3)^{2} - 4

With the quadratic function written in vertex form it is now possible to find its vertex.

f (x) = (x - 3)^{2} - 4

Therefore, the vertex is at

(3, - 4) .

Since the parabola opens upward this is a minimum.

b There are no given restrictions on the function, so it can be assumed that its domain consists of all real numbers.

Domain : all real numbers

From Part A it is known that the vertex of the parabola at

(3, - 4)

is a minimum. Therefore, all the values of the function are greater than or equal to

- 4 .

Range : [- 4, \infty)

c The

y -

intercept can be found by evaluating the function at

x = 0 .

f (x) = x^{2} - 6 x + 5

Substitute

$x = 0$

f (0) = 0^{2} - 6 (0) + 5

Simplify right-hand side

CalcPow

Calculate power

f (0) = 0 - 6 (0) + 5

Multiply

f (0) = 0 - 0 + 5

AddSubTerms

Add and subtract terms

f (0) = 5

Therefore, the

y -

intercept is

5 .

To find the

x -

intercepts begin by setting the function equal to

0 .

x^{2} - 6 x + 5 = 0

The resulting expression is a quadratic equation. By looking at the graph it seems that

x = 1

and

x = 5

are solutions to the equation. This can be verified by solving the equation by factoring.

x^{2} - 6 x + 5 = 0

Factor

Rewrite

Rewrite $- 6 x$ as $- 5 x - x$

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

FactorOut

Factor out $x$

x (x - 5) - x + 5 = 0

FactorOut

Factor out $- 1$

x (x - 5) - (x - 5) = 0

FactorOut

Factor out $(x - 5)$

(x - 1) (x - 5) = 0

ZeroProdProp

Use the Zero Product Property

x - 1 = 0 x - 5 = 0 (I) (II)

(I), (II):

Solve for

x

AddEqn

$(I):$ $LHS + 1 = RHS + 1$

x = 1 x - 5 = 0

AddEqn

$(II):$ $LHS + 5 = RHS + 5$

x = 1 x = 5

The solutions to the quadratic equation are

x = 1

and

x = 5 .

Therefore, the

x -

intercepts are

1

and

5 .

The coordinate plane needs to be extended in order to see the intercepts.

d It was found in Part A that the vertex is at

(3, - 4)

and that it is a minimum. Since the domain of the function is all real numbers the function will be decreasing from negative infinity until it reaches the vertex. Then it will start increasing. Remember that the

vertex

is not included in decreasing or increasing intervals.

Knowing this, the decreasing and increasing intervals can be stated.

Decreasing Interval : (- \infty, 3) Increasing Interval : (3, \infty)

e The

x -

intercepts of the function were found in Part C. It can be noted that the graph is

below

the

x -

axis between these intercepts. Knowing that the points at which the function is equal to zero are not to be included, the negative interval can be stated.

Negative Interval : (1, 5)

The graph is

above

the

x -

axis for

x -

values less than

1

and greater than

5 .

Once again, these values are not included in the positive intervals.

Positive Intervals : (- \infty, 1) and (5, \infty)

The graph is divided into a negative interval and two positive intervals.

f It can be noted by taking a look at the graph that as

x

extends to the left or to the right infinitely the function increases in either case. Therefore, its end behavior is up and up.

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