§ 8.3
Quadratic Functions and Their Graphs
Graphing Quadratic Functions
The graph of any quadratic function is a parabola.
y  ax2  bx  c
Whether the parabola opens upward or downward depends on the coefficient a
of the leading term of the quadratic.
• If a is positive, the parabola opens upward (like a bowl).
• If a is negative, the parabola opens downward (like an inverted bowl).
Graphs of Quadratic Functions
The graph of the quadratic function
f x   ax2  bx  c, a  0
is called a parabola.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 8.3
Graphing Quadratic Functions
30
30
25
24
20
18
15
Standard
12
10
0
0
-6
-4
-2
Narrow
6
5
0
2
4
-6
6
-4
-2
-5
0
2
4
6
-6
f x   x 2
f x   3x 2
30
24
18
12
Wide
6
0
-6
-4
-2
0
2
-6
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 8.3
4
6
f x  
1 2
x
3
Graphing Quadratic Functions
Graphing Quadratic Functions With
2
Equations in the Form f x  ax  h Tk
2
To graph f x  ax  h  k ,
1) Determine whether the parabola opens upward or downward. If
a > 0, it opens upward. It a < 0, it opens downward.
2) Determine the vertex of the parabola. The vertex is (h, k).
3) Find any x-intercepts by replacing f (x) with 0. Solve the resulting
Quadratic equation for x.
4) Find the y-intercept by replacing x with 0.
5) Plot the intercepts and vertex and additional points as necessary.
Connect these points with a smooth curve that is shaped like a cup.
Page 592
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 8.3
Graphing Quadratic Functions
EXAMPLE
Graph the function f x  2x  42  8.
SOLUTION
We can graph this function by following the steps in the
preceding box. We begin by identifying values for a, h, and k.
f x  ax  h  k
2
f x  2x  4  8
2
1) Determine how the parabola opens. Note that a, the
coefficient of x 2 , is -2. Thus, a < 0; this negative value tells us
that the parabola opens downward.
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 8.3
Graphing Quadratic Functions
CONTINUED
2) Find the vertex. The vertex of the parabola is at (h, k).
Because h = -4 and k = -8, the parabola has its vertex at (-4, -8).
3) Find the x-intercepts. Replace f (x) with 0 in
2
f x  2x  4  8.
2
Find x-intercepts, setting f (x) equal
0  2x  4  8
to 0.
2
Add 2x  42 to both sides.
2x  4  8
x  42  4
x4   4
x  4  2i
x  4  2i
Divide both sides by 2.
Apply the square root property.
Simplify the radical.
Subtract 4 from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 8.3
Graphing Quadratic Functions
CONTINUED
Since no real solutions resulted from this step, there are no xintercepts.
4) Find the y-intercept. Replace x with 0 in f x  2x  4  8.
2
f 0  20  4  8  24  8  216  8  32  8  40
2
2
The y-intercept is -40. The parabola passes through (0,-40).
5) Graph the parabola. With a vertex at (-4,-8), no x-intercepts,
and a y-intercept at -40, the graph of f is shown below. The axis
of symmetry is the vertical line whose equation is x = -4.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 8.3
Graphing Quadratic Functions
CONTINUED
Vertex: (-4,-8)
0
-15
-10
-5
-10
-20
-30
-40
-50
-60
-70
-80
Axis of symmetry:
x = -4
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 8.3
0
5
y-intercept
is: -40
Graphing Quadratic Functions
f x  ax  h  k
h, k 
EXAMPLE Problems from homework
2
Find the vertex
10.
f x  3x  2  12.
12.
f x  2x  4  8
2,12
2
 4,8
2
Find the vertex, intercepts, and sketch
18.
f x  x 1  2
20.
f x  x  3  2
2
Check point 1
2
f ( x)  x 1  4
2
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 8.3
Graphing Quadratic Functions
The Vertex of a Parabola Whose
Equation is f x  ax2  bx Tc
Consider the parabola defined by the quadratic function
f x  ax2  bx  c.
The parabola’s vertex is
 b  b 
  , f    .
 2a  2a  
 b 4ac  b 2 
  ,
.
4a 
 2a
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 8.3
Page 594
Graphing Quadratic Functions
EXAMPLE
Graph the function f x  6  4x  x2 . Use the graph to identify
its domain and its range.
SOLUTION
1) Determine how the parabola opens. Note that a, the
coefficient of x 2, is 1. Thus, a > 0; this positive value tells us
that the parabola opens upward.
2) Find the vertex. We know that the x-coordinate of the vertex
b
is x   . We identify a, b, and c for the given function.
2a
Note that a = 1, b = -4, and c = 6.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 8.3
Graphing Quadratic Functions
CONTINUED
Substitute the values of a and b into the equation for the xcoordinate:
x
b
4
4


  2  2.
2a
2 1
2
The x-coordinate for the vertex is 2. We substitute 2 for x in the
equation of the function to find the corresponding y-coordinate.
f 2  6  4  2  22  6  8  4  2.
The vertex is (2,2).
3) Find the x-intercepts. Replace f (x) with 0 in the original
function. We obtain 0  6  4 x  x 2. This equation cannot be
solved by factoring. We will use the quadratic formula instead.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 8.3
Graphing Quadratic Functions
CONTINUED
 b  b 2  4ac   4   4  4 1 6 4  16  24
x


2a
2 1
2
2
Clearly, the discriminant is going to be negative, 16 – 24 = -8.
Therefore, there will be no x-intercepts for the graph of the
function.
4) Find the y-intercept. Replace x with 0 in the original
function.
f 0  6  4  0  02  6  0  0  6
The y-intercept is 6. The parabola passes through (0,6).
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 8.3
Graphing Quadratic Functions
CONTINUED
5) Graph the parabola. With a vertex of (2,2), no x-intercepts,
and a y-intercept at 6, the graph of f is shown below. The axis
of symmetry is the vertical line whose equation is x = 2.
y-intercept
is: 6
30
30
25
25
20
20
15
15
Axis of
symmetry: x = 2
10
10
5
5
0
-4
-2
Range: All
real numbers
greater than
or equal to 2.
0
0
Vertex: (2,2)
2
4
6
8
-4
-2
Domain: All
real numbers
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 8.3
0
2
4
6
8
Graphing Quadratic Functions
CONTINUED
Now we are ready to determine the domain and range of the
original function. We can use the second parabola from the
preceding page to do so. To find the domain, look for all inputs
on the x-axis that correspond to points on the graph.
Domainof f is x | x is a real number or  ,  .
To find the range, look for all the outputs on the y-axis that
correspond to points on the graph. Looking at the first parabola
from the preceding page, we see the parabola’s vertex is (2,2).
This is the lowest point on the graph. Because the y-coordinate
of the vertex is 2, outputs on the y-axis fall at or above 2.
Range of f is y | y  2 or 2, .
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 8.3
Graphing Quadratic Functions
EXAMPLE Problems
f x  ax2  bx  c.
14.
Find the vertex
 b  b 
  , f    .
 2a  2a  
 b 4ac  b 2 
  ,
.
4a 
 2a
f x  3x 2 12x  1
4  3 1  (12)
43

2,11
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 8.3
2
12  144
 11
12
2,11
Graphing Quadratic Functions
EXAMPLE Problems
f x  ax  bx  c.
2
 b  b 
  , f    .
 2a  2a  
Find the vertex, intercepts, and sketch
28.
f x  x 2  2x 15
32.
f x   5  4 x  x 2
34.
f x   x 2  4 x  1
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 8.3
 b 4ac  b 2 
  ,
.
4a 
 2a
Minimums & Maximums
Minimum and Maximum: Quadratic Functions
2
Consider f x  ax  bx  c.
1) If a > 0, then f has a minimum that occurs at x  
b
.
2a
 b 
.
 2a 
This minimum value is f  
b
.
2) If a < 0, then f has a maximum that occurs at x  
2a
 b 
This maximum value is f  
.
 2a 
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 8.3
Minimums & Maximums
EXAMPLE (similar to number 59 and Example 6)
A person standing close to the edge on the top of a 200-foot
building throws a baseball vertically upward. The quadratic
function
st   16t 2  64t  200
models the ball’s height above the ground, s (t), in feet, t seconds
after it was thrown.
How many seconds does it take until the ball finally hits the
ground? Round to the nearest tenth of a second.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 8.3
Minimums & Maximums
CONTINUED
SOLUTION
It might first be useful to have some sort of picture representing
the situation. Below is some sort of picture.
Heigth of Ball (feet)
300
250
200
st   16t 2  64t  200
150
100
Point of
interest
50
0
0
2
4
Time (seconds)
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 8.3
6
8
Minimums & Maximums
CONTINUED
When the ball is released, it is at a height of 200 feet. That is,
when the ball is released, the value of s (t) = 200. By the same
token, when the ball finally hits the ground, it will of course be
0 feet above the ground. That is, when the ball hits the ground,
the value of s (t) = 0. Therefore, to determine for what value of t
the ball hits the ground, we replace s (t) with 0 in the original
function.
st   16t 2  64t  200
0  16t 2  64t  200


0  8 2t 2  8t  25
0  2t 2  8t  25
This is the given function.
Replace s (t) with 0.
Factor -8 out of all terms.
Divide both sides by -8.
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 8.3
Minimums & Maximums
CONTINUED
We will use the quadratic formula to solve this equation.
 b  b2  4ac   8 
t

2a
 82  4  2 25
2 2
8  64  200

4
8  264 8  16.2 8  16.2
8  16.2



or
4
4
4
4
24.2
 8.2

or
4
4
 6.05 or  2.05
Since time cannot be a negative quantity, the answer cannot be
-2.05 seconds. Therefore, the ball hits the ground after 6.05
seconds (to the nearest tenth of a second).
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 8.3
Minimums & Maximums
(handout included)
EXAMPLE (Number 59 similar to problem above and Example 6)
A person standing close to the edge on the top of a 160-foot building throws a
baseball vertically upward. The quadratic function
2
s t   16t  64t  160
models the ball’s height above the ground, s (t), in feet, t seconds after it was
thrown.
a.
After how many seconds does the ball reach its maximum height? What is
the maximum height.
b.
How many seconds does it take until he ball finally hits the ground? Round
to the nearest tenth of a second.
c.
Find s(0) and describe what this means.
d.
Use your results from parts(a) through (c) to graph the quadratic function.
Begin the graph with t=0 and end with the value of t for which the ball hits
the ground.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 8.3
Minimums & Maximums
Strategy for Solving Problems Involving Maximizing
or Minimizing Quadratic Functions
1) Read the problem carefully and decide which quantity is to be maximized or
minimized.
2) Use the conditions of the problem to express the quantity as a function in
one variable.
2
3) Rewrite the function in the form f x  ax  bx  c.
4) Calculate  b . If a > 0, f has a minimum at x   b . This minimum
2a
value is f  
2a
b
b 
x


.
If
a
<
0,
f
has
a
maximum
at
. This maximum

2
a
 2a 
 b 
value is f    .
 2a 
5) Answer the question posed in the problem.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 8.3
Minimums & Maximums
EXAMPLE (similar to 61 and Example 7)
Among all pairs of numbers whose sum is 20, find a pair whose
product is as large as possible. What is the maximum product?
SOLUTION
1) Decide what must be maximized or minimized. We must
maximize the product of two numbers. Calling the numbers x
and y, and calling the product P, we must maximize
P = xy.
2) Express this quantity as a function in one variable. In the
formula P = xy, P is expressed in terms of two variables, x and y.
However, because the sum of the numbers is 20, we can write
x + y = 20.
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 8.3
Minimums & Maximums
CONTINUED
We can solve this equation for y in terms of x, substitute the
result into P = xy, and obtain P as a function of one variable.
y = 20 - x
Subtract x from both sides of
the equation: x + y = 20.
Now we substitute 20 – x for y in P = xy.
P = xy = x(20 – x).
Because P is now a function of x, we can write
P (x) = x(20 – x).
3) Write the function in the form f x  ax2  bx  c . We
apply the distributive property to obtain
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 8.3
Minimums & Maximums
CONTINUED
P (x) = (20 – x)x = 20x - x 2 .
b = 20
a = -1
4) Calculate  b . If a < 0, the function has a maximum at
2a
this value. The voice balloons show that a = -1 and b = 20.
b
20
20
x


  10  10
2a
2 1
2
This means that the product, P, of two numbers who sum is 20 is
a maximum when one of the numbers, x, is 10.
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 8.3
Minimums & Maximums
CONTINUED
5) Answer the question posed by the problem. The problem
asks for the two numbers and the maximum product. We found
that one of the numbers, x, is 10. Now we must find the second
number, y.
y = 20 – x = 20 – 10 = 10.
The number pair whose sum is 20 and whose product is as large
as possible is 10, 10. The maximum product is 10 x 10 = 100.
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 8.3
Minimums & Maximums
EXAMPLE (number 65 and example 8)
You have 200 feet of fencing to enclose a rectangular plot that
borders on a river. If you do not fence the side along the river,
find the length and width of the plot that will maximize the area.
What is the largest area that can be enclosed?
SOLUTION
1) Decide what must be maximized or minimized. We must
maximize area. What we do not know are the rectangle’s
dimensions, x and y.
x
x
y
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 8.3
Minimums & Maximums
CONTINUED
2) Express this quantity as a function in one variable.
Because we must maximize area, we have A = xy. We need to
transform this into a function in which A is represented by one
variable. Because you have 200 feet of fencing, the sum of the
lengths of the three sides of the rectangle that need to be fenced
is 200 feet. This means that
2x + y = 200.
We can solve this equation for y in terms of x, substitute the
result into A = xy, and obtain A as a function in one variable. We
begin by solving for y.
y = 200 – 2x
Subtract 2x from both sides.
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 8.3
Minimums & Maximums
CONTINUED
Now we substitute 200 – 2x for y in A = xy.
A = xy = x(200 – 2x)
The rectangle and its dimensions are illustrated in the
picture at the beginning of this exercise. Because A is
now a function of x, we can write
A (x) = x(200 – 2x).
This function models the area, A (x), of any rectangle whose
perimeter is 200 feet (and one side is not counted) in terms of
one of its dimensions, x.
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 8.3
Minimums & Maximums
CONTINUED
3) Write the function in the form f x  ax2  bx  c . We
apply the distributive property to obtain
Ax  x200 2x  200x  2x 2  2x2  200x.
a = -2
b = 200
4) Calculate  b . If a < 0, the function has a maximum at
2a
this value. The voice balloons show that a = -2 and b = 200.
b
200
200
x


  50  50
2a
2 2
4
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 8.3
Minimums & Maximums
CONTINUED
This means that the area, A(x), of a rectangle with a “3-sided”
perimeter 200 feet is a maximum when the lengths of the two
sides that are the same, x, are 50 feet.
5) Answer the question posed in the problem. We found that x
= 50. The picture at the beginning of this exercise shows that the
rectangle’s other dimension is 200 – 2x = 200 – 2(50) = 200 –
100 = 100 feet. The dimensions of the rectangle that maximize
the enclosed area are 50 feet by 100 feet. The rectangle that
gives the maximum area has an area of (50 feet) x (100 feet) =
5,000 square feet.
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 8.3
In summary…
We consider two standard forms for the quadratic function. In either
form, it is easy to see whether the parabola opens upward or
downward. We just consider the sign of a in either equation. In the
first form that we considered, we could easily see the vertex of the
parabola. In the second form, we could easily see the y intercept.
You must decide which form is easiest for you to use in a given
situation.
The vertex of the parabola is important. For a parabola opening
upward, at the vertex we obtain a minimum function value. For a
parabola opening downward, at the vertex we obtain a maximum
function value.
Have you met Mini and Maxi? Mini is always smiling. Maxi is always frowning.
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 8.3
DONE
Graphing Quadratic Functions
Graphs of Quadratic Functions
The graph of the quadratic function
f x  ax2  bx  c, a  0
is called a parabola.
Blitzer, Intermediate Algebra, 5e – Slide #36 Section 8.3