Download 3.1 Quadratic Functions A quadratic function is a

3.1 Quadratic Functions
A quadratic function is a second degree polynomial function that can written in the general form
The graph of a quadratic function is a parabola that opens upward if
if
0.
0 and downward
The turning point on the graph of a parabola is called the parabola’s vertex. If the graph of the
function opens upward then the function has a minimum value. If the graph of the function
opens downward, then the function has a maximum value.
The equation of a quadratic function in standard form is
The graph is a parabola with vertex
,
.
Example 1. State the vertex for the graph of each quadratic function.
Function
Vetrex
3, ,4
a)
2
3
4
1, 5
b)
3
1
5
2,0
2
c)
0,1
d)
1
2
To graph a quadratic function, plot at least two points to the left of the vertex and two points to
the right of the vertex.
Examples: Find the vertex, axis of symmetry, the function’s extreme value, and the domain and
range. Graph the function.
2.
1
2
Solution: The vertex of the parabola is 1,2
1
2
1
0
2
3
2
1
1
2
Axis of Symmetry x = 1
Maximum value is 2
Domain ∞, ∞ Range
∞, 2
3.
2
2
3
Solution: The vertex of the parabola is
2
2
3
4
3
5
1
1
0
1
5
Axis of symmetry x =-2
Minimum Value is -3
Domain ∞, ∞
Range 3, ∞
2, 3
The vertex of the graph of the quadratic function
2
2
4.
8
,
2
5
2
Solution: The vertex is
8 2
2 2
8 16 5
3
The Vertex is 2,3
2
8
5
0
1
3
4
is given by
5
1
3
1
5
Axis of symmetry x = 2
Maximum of function is 3
Domain ∞, ∞
Range ∞, 3
5
Math 1314
3.1 Applications of Quadratic Functions
(Solutions are on the next page)
1. Determine the minimum or maximum values of each function.
72
90
b)
2
8
a)
6
1
2. Find two numbers whose difference is 80 and whose product is the minimum possible value.
3. Find two numbers whose sum is 16 and whose product is the maximum possible value.
4. The revenue of a charter bus company depends on the number of unsold seats. If the revenue
in dollars is given by
50
5000, where x is the number of unsold seats, find
the number of unsold seats that produce maximum revenue. What is the maximum revenue?
5. A fourth grade class decides to enclose a rectangular garden, using the side of the school as
one side of the rectangle. What is the maximum area that the class can enclose with 32 feet of
fence? What should the dimensions of the garden be in order to yield this area?
6. A farmer wishes to put a fence around a rectangular field and then divide the field into three
rectangular plots by placing two fences parallel to one of the sides. If the farmer can afford only
1000 yards of fencing, what dimensions will give the maximum rectangular area?
Solutions:
1 a) Since 6 > 0, the function has a minimum value.
6
6
6 6
72 6
90
6
216 432 90 738
The function’s minimum value is 738.
The minimum value is
b) Since -2 < 0, the function has a maximum value.
2
8 2
1
2
2 2
2
8 16 1 9
The Maximum value of the function is 9
The maximum value is
80
2.
80
variable.
substitute this value for x in the product to make the product a function of one
80
80
80
40
2
2
80
80
40 40
The two numbers are 40 and -40 and the smallest product is -1600
16
3.
16
variable.
substitute this value for x in the product to make the product a function of one
16
16
16
8
2
2
16
16 8 8
The two numbers are 8 and 8 and the largest product is 64.
4. The function has a maximum value since -1 < 0.
The number of unsold seats which maximizes revenue is given by
.
25 . The number of unsold seats which maximizes revenue is 25.
The maximum revenue is
25
50 25
5000
625 1250 5000 $5625
5.
W
W
L
Find the maximum area A = LW, where 2W + L = 32.
L = 32 – 2W substitute this expression for L in the area function in order to rewrite the area as a
function of a single variable.
A = LW
32 2
2
32 . The value of W which maximizes area is
8
L = 32 – 2W = 32 – 2(8) =16. The dimensions should be 8 ft wide and 16 ft long to maximize
area.
6.
W
L
Find the maximum area A = LW, where 2L + 4W = 1000.
2L = 1000 – 4W or L = 500 – 2W substitute this expression for L in the area function in order to
rewrite the area as a function of a single variable.
A = LW
500 2
2
500 . The value of W which maximizes area is
125
L = 500 – 2W = 500 – 2(125) = 250. The dimensions should be 125 yards wide and 250 yards
long to maximize area.