Quadratic Function

Is a function that can be written in

the standard form:

f(x)=ax²+bx+c

Where a, b, and c are real numbers

and a≠0

Identify a Quadratic Function (1)

To tell if a function is quadratic we must look at the

Differences (subtraction)

in the Domain and Range

Since the first difference is constant we now need to check the range.

Note: In this example we found the differnce by subtracting each element of the domain for example -2-(-1)= -1

[image]

In order to be a quadratic function the range must have a Second difference that is constant.

Note: In this example we found the first difference by subtracting each element of the range for example 8-2= 6. After finding the first difference we need to see if the second difference gives us a constant value. 6-2=4...  Since we have a second difference that is constant, this table of values describes a quadratic function.

Is this a Quadratic Function

{(-4, 8), (-2, 2), (0, 0), (2, 2), (4, 8)}

Lets check the domain's first difference is it constant?

-4-(-2)= -2

-2-0= -2

0-2= -2

2-4= -2 

YEP! Sure is!

Now lets check the range's differences...

First Difference:

8-2=6

2-0=2

0-2 = -2

2-8= -6

Second Difference:

6-2 = 4

2-(-2)= 4

-2-(-6)= 4

YEP! This is a quadratic function since the second difference is a constant value.

If a>0 which way does a

Parabola open?

Since ax²+bx+c is the standard form of a quadratic

If a>0 then the parabola, which is the defining form a quadratic funtion will open upward.

[image]

Think of a smiley face :)

If a<0 which way will a

Parabola open

Since ax²+bx+c is the standard form of a quadratic

If a<0 then the parabola, which is the defining form a quadratic funtion will open downward.

[image]

Think of a frown :(

Define a Maximum

If a<0 then parabola opens downward then the

y value

of the vertex is the Maximum value

Vertex (3, 5)

The vertex you recall is highest or lowest point on the parabola.

In this case the Maximum=5

Define a Minimum

If a>0 then parabola opens upward then the

y value

of the vertex is the Minimum value

Vertex (3, -8)

The vertex you recall is highest or lowest point on the parabola.

In this case the Minimum=-8

Domain of a Quadratic

Unless a specific domain is given, it can be

generalized that the domain of a quadratic function

is All Real Numbers

Recall that we read the domain or the x values from

negative infinity to positive infinity

-∞ to +∞

The Range of a quadratic function

[image]

The range of a quadratic function

is limited by the minimum or maximum value.

Recall that we read the range or the y values from

negative infinity to positive infinity

-∞ to +∞

If the vertex is located at (2, 1) and

a>0 (parabola opens upward)

Then the Range would be written as:

R: y ≥ 1

The Range of a quadratic function

[image]

The range of a quadratic function

is limited by the minimum or maximum value.

Recall that we read the range or the y values from

negative infinity to positive infinity

-∞ to +∞

If the vertex is located at (-2, 6) and

a<0 (parabola opens downward)

Then the Range would be written as:

R: y ≤ 6

Find the Axis of Symmetry

x²-2x-8

Recall that the formula for finding

the axis of symmetry is:

x = -b/2a

Since ax²+bx+c is the standard form of a quadratic we wish to extract the coefficients a and b

from the given problem:

x²-2x+8

a= 1 and b= -2

x= -(-2)/2(1)

x= 2/2 = 1

The axis of symmetry is x=1

This is also the x value of the vertex (x, y)

Find the Vertex of a parabola

f(x)=2x²+x+3

This is a two step process first we must find x. We next substitue the value we found for x back into the equation and solve for f(x) which is our y.

Recall that the formula for finding

the axis of symmetry or x is:

x = -b/2a

Since ax²+bx+c is the standard form of a quadratic we wish to extract the coefficients a and b

from the given problem:

2x²+4x+5

a= 2 and b= 4

x= -(4)/2(2)

x= -4/4 = -1

The axis of symmetry is x=1

This is also the x value of the vertex (-1, y)

Next we simply replace each x with -1

2x²+4x+5

Rewrite the problem using parantheses for x

2(  )²+4(  )+5

Next fill in the parentheses with

the value of x you found

2( -1 )²+4( -1 )+5

Recall your order of operations.

Exponents come before multiplication!

Note:If you distribute the 2 into the parentheses before evaluating the exponent first you will most likely get an incorrect answer!

f(x)=2( -1 )²+4( -1 )+5

f(x)=2( 1 )+4( -1 )+5

f(x)=2-4+5

f(x)=3

or y=3

Or vertex is at (-1, 3)

Finding the Zeros (roots)

of a quadratic function

A Zero of a function is an x-value that makes the function equal to 0

Roots can be found in several ways

1. Looking at a graph
2. Factoring, and then use the Zero Product Property
3. Graphing calculator

• If a parabola crosses the x-axis then it will have two Zeros
•   [image]

• If a parabola's vertex rests on the x-axis then it will have one Zero
• [image]

• If the parabola does not intersect the x-axis then it will have no real roots.
• [image]

Find the Zeros of:

f(x)=x²+8x-9

First see if we can factor:

f(x)=x²+8x-9

(x+9)(x-1)=0

This factors nicely so we set each of the factors equal to zero using the zero product property

x+9=0

x-1=0

Solving for x gives us:

x=-9, and x=1

Therefore we have two roots!

Use Square Roots to Solve

a Qadratic Equation

x² - 16 =0

x² - 16 =0

Solve for x

x² =16

Now take the square root of both sides of the equation

Recall the properties of equality.

x = ±4

Solve using Square Roots

16x² +10 = 131

Isolate x²

16x² +10 = 131

16x² +10 - 10 = 131 - 10

16x² = 121

16   =  16

x = ±11

        4

Distance Formula

Find the distance between:

(-4, 18)     (20, -11)

Meet the distance formula!

To use the distance formula we need two

sets of coordinates:

(x₁, y₁)            (x₂, y₁)

(-4, 18)            (20, -11)