Introduction
A function is a relation in which every element of the
domain is paired with exactly one element of the range;
that is, for every value of x, there is exactly one value of
y. When writing functions based on real-world
information, first we must carefully analyze the words
describing the situation. For example, if we say an
unknown length is 2 more than twice another length, we
must be able to translate the given information into an
algebraic expression. In this case there are two lengths,
with one length dependent on the other. If we let x
represent the second length, then the first length is 2
more (+) than twice (2 ×) x, or 2 + 2x.
5.7.1: Building Functions from Context
1
Introduction, continued
In this lesson, the problems will involve different contexts
that quadratic equations describe. Notice that every
problem will either require you to identify the two distinct
linear expressions whose product will build the
quadratic, or to identify the vertex of the parabola along
with one other point on the parabola (which will allow
you to build the quadratic in the vertex form of a
quadratic equation).
2
5.7.1: Building Functions from Context
Key Concepts
• A quadratic expression is the product of two linear
factors.
• The vertex of a parabola represents either the
maximum or minimum y-value for the equation.
• The vertex of a parabola can be found using
æ -b æ -b ö ö
ç 2a , f ç 2a ÷ ÷ from the general form of the quadratic
è øø
è
equation, where f(x) = ax2 + bx + c.
• The vertex of a parabola can be found using (h, k) in
vertex form, where f(x) = a(x – h)2 + k.
3
5.7.1: Building Functions from Context
Key Concepts, continued
• The curve of a parabola is the graphical
representation of the solution set for the equation of
the parabola.
• A quadratic equation can be built using the vertex and
any other point on the parabola.
• Concavity refers to the direction the parabola faces.
• The vertex of an upward-facing (or concave up)
parabola will be at the bottom or minimum of the
curve.
• Conversely, the vertex of a downward-facing (or
concave down) parabola will be at the top or
maximum of the parabola.
5.7.1: Building Functions from Context
4
Key Concepts, continued
Concave Up
Concave Down
The vertex is a minimum.
The vertex is a maximum.
5
5.7.1: Building Functions from Context
Key Concepts, continued
• The leading coefficient is the coefficient of the term
with the highest power.
• In a quadratic equation, the leading coefficient is the
number that is being multiplied by the x2 term.
• The leading coefficient of the equation of a parabola
determines its concavity.
• If the leading coefficient is positive, the parabola is
concave up and the graph has a minimum.
• If the leading coefficient is negative, the parabola is
concave down and the graph has a maximum.
6
5.7.1: Building Functions from Context
Common Errors/Misconceptions
• not realizing that when given a fixed perimeter, the
corresponding area of the figure is constant
• thinking that a positive leading coefficient yields a
maximum value instead of a minimum value
• mistakenly believing that adding linear expressions will
build quadratics; they must be multiplied
7
5.7.1: Building Functions from Context
Guided Practice
Example 1
A farmer is building a rectangular pen using 100 feet of
electric fencing and the side of a barn. In addition to
fencing, there will be a 4-foot gate also requiring the
electric fencing on either side of the pen. The farmer
wants to maximize the area of the pen. How long should
he make each side of the fence in order to create the
maximum area?
8
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
1. Write the expressions that describe the
length of each side of the pen.
Starting with one of the sides that is perpendicular to
the barn, we can say the length is an unknown
amount, x, plus the 4-foot gate, or x + 4.
The length of the second side of the fence that is
perpendicular to the barn will be the same length as
the first.
9
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
The side that is parallel to the barn is whatever
amount of fence is left over after creating the two
perpendicular sides.
The total amount of fencing is 100 feet, and there
are two sides of length (x + 4), so the length of the
side that is parallel to the barn is 100 – 2(x + 4).
Simplifying the expression, we get 100 – 2x – 8, or
92 – 2x.
10
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
2. Build the equation that describes the area
of the pen.
Remember that area equals length times width, or
A = l • w.
Let l = x + 4 and w = 92 – 2x.
A(x) = (x + 4)(92 – 2x)
A(x) = 92x – 2x2 + 368 – 8x
A(x) = –2x2 + 84x + 368
Substitute values for
length and width.
Multiply.
Reorder and simplify.
11
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
3. To find the maximum area, use the vertex.
Note that the leading coefficient of the equation is
negative. This means the graph of the equation will
be a downward-facing parabola. Therefore, the vertex
of the parabola will describe the greatest amount of
area.
A(x) = –2x2 + 84x + 368
Given that the general form of a quadratic function is
y = ax2 + bx + c, we can determine that a = –2,
b = 84, and c = 368.
12
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
Find the x-coordinate of the vertex by substituting in a
and b values from the quadratic function into the
-b
:
expression
2a
-b -(84) -84
=
=
= 21
2a 2(-2) -4
The x-coordinate of the vertex is 21.
13
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
Find the y-coordinate of the vertex by substituting the
x-coordinate from the vertex into the quadratic
function.
A(x) =
–2x2
+ 84x + 368
Quadratic function
A(21) = –2(21)2 + 84(21) + 368 Substitute 21 for x.
A(21) = –882 + 1764 + 368
Simplify, then
solve.
A(21)
= 1250 area of the pen is 1,250 ft2.
The
maximum
14
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
4. Finally, use the x-value from the vertex to
find the lengths of each side of the pen.
x = 21
Each side that is perpendicular to the barn is equal to
x + 4.
(x + 4) = (21 + 4) = 25 feet
The side that is parallel to the barn is equal to
92 – 2x.
(92 – 2x) = [92 – 2(21)] = 50 feet
✔
15
5.7.1: Building Functions from Context
Guided Practice: Example 1, continued
16
5.7.1: Building Functions from Context
Guided Practice
Example 2
An amusement park has commissioned the design of a
steel roller coaster with a drop section that is modeled
by a parabola. Part of the roller coaster’s track will go
through an underground tunnel. In this section, the roller
coaster will dip 12 feet below ground level. The roller
coaster will dip below ground level at a horizontal
distance of 24 feet from the peak just before the drop
and reemerge to ground level at a horizontal distance of
36 feet from the peak just before the drop. Find an
equation of the parabola that describes the drop and the
height of the roller coaster at the peak.
5.7.1: Building Functions from Context
17
Guided Practice: Example 2, continued
1. Derive the coordinates of the points on
the parabola.
Let the x-axis represent ground level and the yintercept represent the peak of this section of the
roller coaster.
Since the horizontal distances are given as 24 feet
and 36 feet, it can be determined that the roller
coaster dips below ground level at (24, 0) and
reemerges at (36, 0).
18
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
2. Establish the linear factors and the
equation of the parabola.
From the x-intercepts, the linear factors of the
equation are (x – 24) and (x – 36). Therefore, the
equation of the parabola is y = a(x – 24)(x – 36).
19
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
3. Find the vertex of the parabola.
Expand the factored form of the equation of the
parabola into the general form, y = ax2 + bx + c.
y = a(x – 24)(x – 36)
y = a(x2 – 36x – 24x + 864)
y = a(x2 – 60x + 864)
y = ax2 – 60ax + 864a
Equation of the parabola
Multiply the factors.
Simplify.
Distribute a.
20
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
The x-coordinate for the vertex of a quadratic
-b
equation is
. Substitute the values from the
2a
quadratic function to determine the x-coordinate.
-b
2a
=
-(-60a)
2(a)
=
30a
a
= 30
The x-coordinate is 30.
Combine this information with the given minimum of
the drop, 12 feet below ground level, to find the vertex
of the equation: (30, –12).
21
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
4. Find the value of a.
Substitute the point (30, –12) into the factored form of
the quadratic equation and solve for a.
Factored form of the
y = a(x – 24)(x – 36)
quadratic equation
Substitute 30 and –12 for
–12 = a(30 – 24)(30 – 36)
x and y.
22
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
–12 = a(6)(–6)
–12 = –36a
a=
Simplify, then solve for a.
1
3
After solving for a, the equation of the parabola is
1
y = (x - 24)(x - 36).
3
23
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
5. Find the height of the peak of this section
of the roller coaster.
The height of the peak is the roller coaster’s vertical
distance above ground level. Since the peak occurs
at the y-intercept, where the x-coordinate equals 0,
set x = 0 in the quadratic equation and solve for y.
24
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
1
y = (0 - 24)(0 - 36)
3
1
y = (-24)(-36)
3
1
y = (864)
3
y = 288
The height of the roller coaster’s peak
just before the drop is 288 feet.
✔
25
5.7.1: Building Functions from Context
Guided Practice: Example 2, continued
26
5.7.1: Building Functions from Context