5.1 Quadratic Functions

5.1 Quadratic Functions
Chapter 5 Polynomial and Rational Functions

Concepts and Objectives
⚫ Objectives for this section are
⚫ Recognize characteristics of parabolas.
⚫ Understand how the graph of a parabola is related to
its quadratic function.
⚫ Determine a quadratic function’s minimum or
maximum value.
⚫ Solve problems involving a quadratic function’s
minimum or maximum value.

Quadratic Functions
⚫ A function f is a quadratic function if
where a, b, and c are real numbers, and a  0.
⚫ This is called the general form of the quadratic function.
⚫ The graph of a quadratic function is a parabola whose
shape and position are determined by a, b, and c.
⚫ If a > 0, the parabola opens upward.
⚫ If a < 0, the parabola opens downward.
⚫ |a| determines the width of the parabola.
( )= + +
2
f x ax bx c

Vertex Formula
⚫ We can use the general form to find the equation for the
axis of symmetry and the vertex of the parabola:
⚫ If we use the quadratic formula, to
solve for the zeros, we find the value of x
halfway between them is always the equation
for the axis of symmetry.
⚫ To find the y-coordinate of the vertex, plug x into the
general form.
2
4
,
2
b b ac
x
a
−  −
=
2
0
ax bx c
+ + =
,
2
b
x
a
= −

Vertex Form
⚫ The vertex (or standard) form of a quadratic function is
written
⚫ The graph of this function is the same as that of g(x)
translated h units horizontally and k units vertically.
This means that the vertex of f is at (h, k) and the axis of
symmetry is x = h.
⚫ The vertex form is just a vertical and horizontal shift of
the parent function y = x2.
( ) ( )
2
f x a x h k
= − +

Vertex Form (cont.)
⚫ Example: Graph the function and give its domain and
range.
( ) ( )
2
1
4 3
2
f x x
= − − +

Vertex Form (cont.)
⚫ Example: Graph the function and give its domain and
range.
Compare to : h = 4 and k = 3 (Notice
the signs!)
Vertex: (4, 3), axis of symmetry x = 4
We can graph this function by graphing the base
function and then shifting it.
( ) ( )
2
1
4 3
2
f x x
= − − +
( ) ( )
2
k
f x a x h
= − +

Vertex Form (cont.)
⚫ Example, cont.:
Let’s consider the graph of
⚫ Vertex is at (0, 0)
⚫ Passes through (2, ‒2) and
(4, ‒8).
⚫ (I picked 2 and 4 because of
the half.)
( )= − 2
1
2
g x x

Vertex Form (cont.)
⚫ Example, cont.:
To graph f, we just shift everything over 4 units to the
right and 3 units up.
Domain: (‒, )
Range: (‒, 3]

Writing a Function From a Graph
⚫ Given a graph of a quadratic function, to write the
equation of the function in general form:
⚫ Identify the vertex, (h, k). It will always be either the
highest or lowest point of the parabola.
⚫ Substitute the values of h and k in the function
⚫ Substitute the values of any point, other than the
vertex, on the parabola for x and f(x), and solve for a.
⚫ Expand and simplify to write in general form.
( ) ( )
2
f x a x h k
= − +

⚫ Example: Write an equation for the quadratic function g
in vertex form, and then in general form.

First, we identify the vertex,
which is at (–2, –3). Thus,
h = –2 and k= –3, and our
function looks like this:
( ) ( )
( ) ( )
( ) ( )
2
2
2 3
2 3
g x a x
g x a x
= − + −
= + −

Now, we pick a point on the
graph, for example, (0, –1),
to substitute in for x and
g(x).
( )
2
1 0 2 3
4 2
1
2
a
a
a
− = + −
=
=

So, the vertex form for this
graph is
( ) ( )
2
1
2 3
2
g x x
= + −

To write this in general
form, expand the binomial
and simplify:
( ) ( )
( )
2
2
2
2
1
2 3
2
1
4 4 3
2
1
2 2 3
2
1
2 1
2
g x x
x x
x x
x x
= + −
= + + −
= + + −
= + −

General Form to Vertex Form
⚫ Given a quadratic function in general form, to find the
vertex of the parabola (or to write in vertex form):
⚫ Identify a, b, and c from the general form.
⚫ Find h, the x-coordinate of the vertex, by substituting
a and b into .
⚫ Find k, the y-coordinate of the vertex, by evaluating
k .
⚫ Write in vertex form:
2
b
h
a
= −
( )
2
b
f h f
a
 
= = −
 
 
( ) ( )
2
f x a x h k
= − +

⚫ Example: What is the vertex of the function?
( )= − +
2
6 7
f x x x

From the general form, a = 1, b = –6, and c = 7.
The vertex is at (3, ‒2).
Vertex form is
( )= − +
2
6 7
f x x x
( )
6
3
2 1
h
−
= − = ( ) ( )
2
3 3 6 3 7
9 18 7 2
k f
= = − +
= − + = −
( ) ( )
2
3 2
f x x
= − −

Completing the Square
⚫ Another method we can use is “completing the square”
to transform it into vertex form (see section 2.5), but this
time we do everything on the right-hand side.
⚫ For example, the function is not a
binomial square. We can add 0 in the form of 52 – 52
(5 is half of 10), and group the parts that factor to a
binomial square:
( ) + −
= − +
2
2 2
10 30
5 5
f x x x
( )= − +
2
10 30
f x x x
( )
= − + − +
2 2 2
10 5 5 30
x x
( )
= − +
2
5 5
x

Completing the Square (cont.)
⚫ Example: What is the vertex of the function? (same ex.)
( )= − +
2
6 7
f x x x

Completing the Square (cont.)
The vertex is at (3, ‒2).
( )= − +
2
6 7
f x x x
( )− +
+
= − 2
2 2
3 7
3
6
x x
( )
= − − +
2
3 9 7
x
( )
= − −
2
3 2
x
6
3
2
=

⚫ Probably the fastest method to find the vertex is to use
Desmos (surprise!) to graph the function. Desmos will
almost always automatically plot the vertex for you.
⚫ Using our same example:
The graph will show the
vertex as a gray dot.

⚫ Probably the fastest method to find the vertex is to use
Desmos (surprise!) to graph the function. Desmos will
almost always automatically plot the vertex for you.
⚫ Using our same example:
The graph will show the
vertex as a gray dot. Click
to get the coordinates.

Practice
⚫ Find the vertex using the method of your choice. Then
give the axis of symmetry and the domain and range.
1. f(x) = x2 +8x + 5
2. g(x) = x2 – 5x + 8
3. h(x) = 3x2 + 12x – 5

Practice (cont.)
1. f(x) = x2 +8x + 5
The vertex is at (‒4, ‒11).
The axis of symmetry is x = –4.
The domain is (–∞, ∞), and the range is [–11, ∞)
( ) ( )
+
= + − +
2
2 2
8 4 4 5
f x x x
( )
= + − +
2
4 16 5
x
( )
= + −
2
4 11
x
( )
( ) ( )
2
8
4
2 1
4 8 4 5
16 32 5 11
h
k
= − = −
= − + − +
= − + = −

Practice (cont.)
2. g(x) = x2 – 5x + 8
The vertex is at , and the axis is .
The domain is (–∞, ∞), and the range is .
 
−
 
 
5 7
,
2 4
5
2
x = −
7
,
4
 


 
( )
2
5 5
2 1 2
5 5
5 8
2 2
25 25 25 50 32 7
8
4 2 4 4 4 4
h
k
−
= − =
   
= − +
   
   
= − + = − + =

Practice (cont.)
3. h(x) = –3x2 – 12x – 5
The vertex is at (‒2, 7), and the axis is x = –2.
The domain is (–∞, ∞), and the range is (–∞, 7].
( )
( ) ( )
2
12
2
2 3
3 2 12 2 5
12 24 5 7
h
k
−
= − = −
−
= − − − − −
= − + − =
Remember, the –3 flips the graph, so
the vertex is at the top.

Practice (cont.)
3. h(x) = –3x2 – 12x – 5 (Completing the Square method)
( ) ( )
2
4 5
3
h x x x
= + −
−
( ) ( )
2 2 2
3
4 2 2
3 5
x x
= +
+
− + −
( )
2
3 2 12 5
x
= − + + −
( )
3 2 7
x
= − + +
Notice that I factored
the –3 out of the first
two expressions. This means that I’m
adding and
subtracting –3(22).

Practice (cont.)
3. h(x) = –3x2 – 12x – 5 (Desmos)

Maximum and Minimum Values
⚫ The output of the quadratic function is the maximum or
minimum value of the function, depending on the
orientation of the parabola.
⚫ There are many real-world scenarios that involve
finding the maximum or minimum value of a quadratic
function.
⚫ If you are solving a problem involving quadratics,
and they ask for the maximum/minimum value, find
the vertex.
⚫ Depending on the problem, you will need either the
x-coordinate or the y-coordinate to solve the
problem.

Maximum and Minimum Values
⚫ Example: A backyard farmer wants to enclose a
rectangular space for a new garden within her fenced
backyard. She has purchased 80 feet of wire fencing to
enclose three sides, and she will use a section of the
backyard fence as the fourth side.
⚫ Find a formula for the area enclosed by the fence if
the sides of the fencing perpendicular to the existing
fence have length L.
⚫ What dimensions should she make her garden to
maximize the enclose area?

Maximum and Minimum Value
⚫ Example (cont.)
(We want to do this because the area will be A = LW.)
She only has 80 feet of
fencing available, so the
three sides would be
L + W + L = 80, or
2L + W = 80.
This means we can write W
in terms of L:
W = 80 – 2L

⚫ Example (cont.)
This is our quadratic function.
To find the maximum area, we find the vertex.
The area of the garden will
thus be
A = LW = L(80 – 2L)
or
2
2
80 2
2 80
A L L
L L
= −
= − +

⚫ Example (cont.):
( )
( ) ( )
2
80
20
2 2
2 20 80 20
800 1600 800
h
k
= − =
−
= − +
= − + =
( ) ( )( )
( )
2 2 2
2
2 40 20 2 20
2 20 800
20, 800
A L L
L
h k
= − − + − −
= − − +
= =

If you don’t see the parabola immediately, you may
have to zoom out and/or move the graph. If you still
don’t see it, make sure you typed the equation
correctly!

⚫ Whichever method you used, we have our vertex at
(20, 800). So what does it mean?
⚫ Remember our function was a function of L, which
was the length. So 20 represents the L at the
maximum area, which is 800.
⚫ To finish this out, the two perpendicular sides should
be 20 feet, the side between them will be
80 – 2(20) = 40 feet, and the maximum area for the
garden will thus be 800 square feet.

Classwork
⚫ College Algebra 2e
⚫ 5.1: 8-24 (×4); 4.3: 22-30 (even); 4.2: 46-58 (even)
⚫ 5.1 Classwork Check
⚫ Quiz 4.3

5.1 Quadratic Functions

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5.1 Quadratic Functions