Describing Quadratic Functions - Quadratic Functions and Equations (Algebra 2) - Mathleaks Online

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Concept

Quadratic Function

A quadratic function is a polynomial function of degree $2.$ That means that the highest exponent of the independent variable is $2.$ The standard form of a quadratic function is expressed as follows.

$y=ax^2+bx+c$

Here, $a, \, b,$ and $c$ are real numbers and $a \neq 0.$ The simplest quadratic function is $y=x^2,$ and the graph of any quadratic function is a parabola.

Concept

Characteristics of Quadratic Functions

The inherent shape of parabolas gives rise to several characteristics that all quadratic functions have in common.

Concept

Direction

A parabola either opens upward or downward. This is called its direction.

Concept

Vertex

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.

Concept

Axis of Symmetry

All parabolas are symmetric, meaning there exists a line that divides the graph into two mirror images. For quadratic functions, that line is always parallel to the $y$ -axis, and is called the axis of symmetry.

The axis of symmetry always intersects the vertex of the parabola, and is written as a vertical line, where $h$ can be any real number.

$x=h$

Concept

Zeros

Depending on its rule, a parabola can intersect the $x$ -axis at $0,$ $1,$ or $2$ points. Since the function's value at an $x$ -intercept is always $0,$ these points are called zeros, or sometimes roots.

Concept

$y$ -intercept

Because all graphs of quadratic functions extend infinitely to the left and right, they each have a

y

-intercept anywhere along the

y

-axis.

Concept

Characteristics of Quadratic Functions in Standard Form

When a quadratic function is written in standard form, it's possible to use $a,$ $b,$ and $c$ to determine characteristics of its graph. $\begin{aligned} \textbf{direction} &: \text{upward when } a>0, \\ &\phantom{:} \text{downward when } a<0 \\ \mathbf{y} \textbf{-intercept} &: (0,c) \\ \textbf{axis of symmetry} &: x=\text{-}\dfrac{b}{2a} \end{aligned}$

Concept

Direction

The direction of the graph is determined by the sign of $a.$ To understand why, consider the quadratic function $y=ax^2.$ Since all squares are positive, $x^2$ will always be positive. When $a$ is positive, then $ax^2$ is also positive. Thus, when moving away from the origin in either direction, the graph extends upward. Similarly, when $a$ is negative, $ax^2$ will be negative. Thus, the graph will extend downward for all $x$ -values.

Concept

$y$ -intercept

The $y$ -intercept of a quadratic function is given by $c,$ specifically at $(0,c).$ This is because substituting $x=0$ into standard form yields the following. $\begin{aligned} y&=ax^2+bx+c\\ y&=a \cdot {\color{#0000FF}{0}}^2 + b \cdot {\color{#0000FF}{0}} + c \\ y&= 0+0+c \\ y&=c \end{aligned}$

Concept

Axis of Symmetry

The equation of the axis of symmetry can be found using the coefficients

a

and

b.

It is derived from the fact that the axis of symmetry divides the parabola in two mirror images. Two points with the same

y

-value are, thus, equidistant from the axis of symmetry. This gives rise to a quadratic equation where the solution is the axis of symmetry.

Rule

Vertex Form

Vertex form is an algebraic format used to express quadratic function rules.

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

In this form, $a$ gives the direction of the parabola. When $a>0,$ the parabola faces upward and when $a<0,$ it faces downward. The vertex of the parabola lies at $(h,k),$ and the axis of symmetry is $x=h.$ Consider the graph of $f(x)=\text{-} \frac{1}{2}(x+4)^2+8.$

From the graph, we can connect the following characterisitcs to the function rule. $\begin{aligned} \textbf{direction} &: \text{downward} &&\rightarrow \quad a<0 \\ \textbf{vertex}&: (\text{-} 4,8) &&\rightarrow \quad h= \text{-} 4, \ k=8\\ \textbf{axis of symmetry}&: x=\text{-} 4 &&\rightarrow \quad h=\text{-} 4 \end{aligned}$

Notice that although the factor in the function rule shows

(x+4)^2,

h

is actually equal to

\text{-} 4.

This coincides with a horizontal translation of a quadratic function.

Rule

Factored Form of a Quadratic Function

Quadratic function rules can be expressed in factored form.

$y=a(x-s)(x-t)$

As is the case with standard form and vertex form, $a$ gives the direction of the parabola. When $a>0,$ the parabola faces upward, and when $a<0,$ it faces downward. Additionally, the zeros of the parabola lie at $(s,0) \text{ and } (t,0).$ Because the points of a parabola with the same $y$ -coordinate are equidistant from the axis of symmetry, the axis of symmetry lies halfway between the zeros. Consider the graph of $f(x)=(x+4)(x+2).$

From the graph, the following characteristics can be connected to the function rule.

\begin{aligned} \textbf{direction} &: \text{upward} &&\rightarrow \hspace{-6pt} \quad a>0 \\ \\ \textbf{zeros}&: (\text{-} 4,0) \text{ \& } (\text{-} 2, 0 ) \hspace{-6pt} &&\rightarrow \hspace{-6pt} \quad s=\text{-} 4, \ t=\text{-} 2\\ \\ \textbf{axis of symmetry} &: x=\text{-} 3 &&\rightarrow \hspace{-6pt} \quad x=\frac{\text{-}4 + (\text{-}2)}{2}= \text{-} 3 \end{aligned}

Determine the characteristics of the quadratic functions

Exercise

For the following functions, determine the direction, the axis of symmetry, vertex, and zeros. $\begin{aligned} f(x)=(x&+1)(x-5) \quad g(x)=\text{-} 2 (x - 3)^2+2\\[0.3em] &h(x)=0.5x^2+5x+10.5 \end{aligned}$

Solution

We'll focus on each function individually, starting with $f.$

Example

Characteristics of $f(x)$

Notice that the function rule of

f

is written in factored form since it's written as a product two binomials.

f(x)=(x+1)(x-5)

We'll start by identifying

f

's direction. It's determined by the sign of the coefficient in front of the factors. When a function doesn't have a number in front of the factors it can be interpreted as

1,

1\cdot(x+1)(x-5).

Since

1

is positive, the parabola will face upward. Moving on, when a function is in factored form it's straightforward to find its zeros. They can be found by solving

f(x)=0.

Using the Zero Product Property we find

x=\text{-}1

and

x=5.

Therefore, the zeros of the function are

(\text{-}1,0)\text{ and }(5,0).

The axis of symmetry, axis of symmetry, can now be found using the zeros since it lies halfway between them. Add the

x

-coordinates and divide the sum by

2.

x=\dfrac{\text{-}1+5}{2}

Add terms

x=\dfrac{4}{2}

Calculate quotient

x=2

Thus, the axis of symmetry is

x=2.

Lastly, we can find the vertex by determining the function value of the axis of symmetry,

2.

f(2)=(2+1)(2-5)=\text{-}9

The vertex is located at

(2,\text{-}9),

and it is the absolute minimum of the function since

f

opens upward. The characteristics of the function

f

can be concluded as follows.

\begin{aligned} \textbf{direction}&:\text{upward} \\ \textbf{axis of symmetry}&:x=2 \\ \textbf{vertex}&:\text{absolute minimum at }(2,\text{-}9) \\ \textbf{zeros}&:(\text{-}1,0)\text{ and }(5,0) \end{aligned}

Example

Characteristics of $g(x)$

The function rule of

g

is written in vertex form

a(x-h)^2+k.

The direction is given by

a,

the sign of the number in front of the binomial.

g(x)=\text{-} 2 (x - 3)^2+2

Since

a

is negative, the direction is downward. The vertex of the parabola is the point

(h,k).

g(x)=\text{-} 2 (x - {\color{#0000FF}{3}})^2+{\color{#009600}{2}}

Thus, the vertex of the function is

(3,2).

Since the direction of the function is downward, the vertex is an absolute maximum. Because the axis of symmetry intersects the vertex, its equation is

x=3.

What remains is to find the zeros of the function. This is done by solving the equation for

g(x)=0.

\text{-}2(x-3)^2+2=0

Solve for

x

\text{LHS}-2=\text{RHS}-2

\text{-}2(x-3)^2=\text{-}2

\left.\text{LHS}\middle/(\text{-}2)\right.=\left.\text{RHS}\middle/(\text{-}2)\right.

(x-3)^2=1

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x-3=\pm\sqrt{1}

Calculate root

x-3=\pm1

\text{LHS}+3=\text{RHS}+3

x=3\pm1

State solutions

\begin{array}{l}x_1=4 \\ x_2=2 \end{array}

Thus, the zeros of the function are

(2,0)

and

(4,0).

We have now found all of the characteristics of the function

g.

\begin{aligned} \textbf{direction}&:\text{downward} \\ \textbf{axis of symmetry}&:x=3 \\ \textbf{vertex}&:\text{absolute maximum at }(3,2) \\ \textbf{zeros}&:(2,0)\text{ and }(4,0) \end{aligned}

Example

Characteristics of $h(x)$

The function rule of

h

is expressed in standard form. Therefore, the values of

a,

b

and

c

can be used to determine its characteristics. The direction is given by the sign of

a,

the coefficient in front of

x^2.

h(x)=0.5x^2+5x+10.5

Since

a

is positive, the direction is upward. For quadratic functions in standard form, the axis of symmetry can be found by

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

We already stated that

a=0.5,

and

b

is the coefficient of the

x

-term. Thus,

b=5.

We can substitute

a

and

b

in the formula to find the axis of symmetry.

x=\text{-}\frac{5}{2\cdot0.5}=\text{-}5

Thus, the axis of symmetry is

x=\text{-}5.

We can now find the vertex by calculating the function value

h(\text{-}5).

h(x)=0.5x^2+5x+10.5

x={\color{#0000FF}{\text{-}5}}

h({\color{#0000FF}{\text{-}5}})=0.5({\color{#0000FF}{\text{-}5}})^2+5({\color{#0000FF}{\text{-}5}})+10.5

Calculate power and product

h(\text{-}5)=0.5\cdot25-25+10.5

Multiply

h(\text{-}5)=12.5-25+10.5

Add and subtract terms

h(\text{-}5)=\text{-}2

The vertex of the function is

(\text{-}5,\text{-}2).

Since the direction is upward, it's an absolute minimum. Finally, the zeros of

h

can be found by solving the equation

h(x)=0.

Since

h(x)

is in standard form we'll use the quadratic formula.

0.5x^2+5x+10.5

Solve using the quadratic formula

Use the Quadratic Formula:

a = {\color{#0000FF}{0.5}}, \, b={\color{#009600}{5}}, \, c={\color{#FF0000}{10.5}}

x=\dfrac{\text{-}{\color{#009600}{5}}\pm\sqrt{{\color{#009600}{5}}^2-4\cdot{\color{#0000FF}{0.5}}\cdot{\color{#FF0000}{10.5}}}}{2\cdot{\color{#0000FF}{0.5}}}

Calculate power and product

x=\dfrac{\text{-}5\pm\sqrt{25-21}}{1}

\dfrac{a}{1}=a

x=\text{-}5\pm\sqrt{25-21}

Subtract term

x=\text{-}5\pm\sqrt{4}

Calculate root

x=\text{-}5\pm2

State solutions

\begin{array}{l}x_1=\text{-}3 \\ x_2=\text{-}7 \end{array}

Thus, the zeros are

(\text{-}7,0)

and

(\text{-}3,0).

We have now determined the characteristics for the function

h.

\begin{aligned} \textbf{direction}&:\text{downward} \\ \textbf{axis of symmetry}&:x=3 \\ \textbf{vertex}&:\text{absolute maximum at }(3,2) \\ \textbf{zeros}&:(2,0)\text{ and }(4,0) \end{aligned}

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Rule

Average Rate of Change

A function's rate of change gives an indication of how its outputs ( $y$ ) change with respect to its inputs ( $x$ ).

$\text{Rate of Change} = \dfrac{\Delta y}{\Delta x}$

When the rate of change is constant, the function is linear. However, when the function is not linear, it's possible to determine an average rate of change over an arbitrary interval. This is an increase or decrease between the endpoints of the interval. If the $x$ -coordinates of the endpoints are $x_1$ and $x_2,$ and the function is $f,$ the average rate of change is defined as follows.

$\text{Average Rate of Change} = \dfrac{f(x_2)-f(x_1)}{x_2-x_1}$

The definition is similar to the slope formula. In fact, the average rate of change can be interpreted as the slope of the line going through the endpoints of the interval.

Determine the average rate of change

Exercise

The function $f(x)$ is quadratic.

Determine the average rate of change over the interval $[\text{-} 3,2].$

Solution

To determine the average rate of change, we must know the $x$ -values of the endpoints. Since the interval is $[\text{-} 3, 2],$ the $x$ -values are $x=\text{-} 3$ and $x=2.$ Let's mark these points on the graph to find their corresponding function values.

The

y

-values are

\text{-}2

and

\text{-}7.

Now, we can determine the average rate of change using the formula.

\dfrac{f(x_2)-f(x_1)}{x_2-x_1}

x_1={\color{#0000FF}{\text{-}3}}

,

x_2={\color{#009600}{2}}

\dfrac{f({\color{#009600}{2}})-f({\color{#0000FF}{\text{-}3}})}{{\color{#009600}{2}}-({\color{#0000FF}{\text{-}3}})}

f(2)={\color{#0000FF}{\text{-}7}}

,

f(\text{-}3)={\color{#009600}{\text{-}2}}

\dfrac{{\color{#0000FF}{\text{-}7}}-({\color{#009600}{\text{-}2}})}{2-(\text{-}3)}

a-(\text{-} b)=a+b

\dfrac{\text{-}5}{5}

Calculate quotient

\text{-}1

The average rate of change over the interval

[\text{-} 3, 2]

is

\text{-}1.

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Exercises

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