For a quadratic given in standard form, the vertex (h, k) is located at:
[latex]\displaystyle{h}=-\frac{{b}}{{{2}{a}}},{k}={f{{({h})}}}={f{{(\frac{{-{b}}}{{{2}{a}}})}}}[/latex]
Example 1
Find the vertex of the quadratic
[latex]\displaystyle{f{{({x})}}}={2}{x}^{{2}}-{6}{x}+{7}[/latex]. Rewrite the quadratic into transformation form (vertex form).
The horizontal coordinate of the vertex will be at
[latex-display]\displaystyle{h}=-\frac{{b}}{{{2}{a}}}=-\frac{{-{6}}}{{{2}{({2})}}}=\frac{{6}}{{4}}=\frac{{3}}{{2}}[/latex-display]
The vertical coordinate of the vertex will be at
[latex-display]\displaystyle{f{{(\frac{{3}}{{2}})}}}={2}{(\frac{{3}}{{2}})}^{{2}}-{6}{(\frac{{3}}{{2}})}={7}=\frac{{5}}{{2}}[/latex-display]
Rewriting into transformation form, the stretch factor will be the same as the
a in the original quadratic. Using the vertex to determine the shifts, [latex]\displaystyle{f{{({x})}}}={2}{({x}-\frac{{3}}{{2}})}^{{2}}+\frac{{5}}{{2}}[/latex]
Try it Now 1
Given the equation
[latex]\displaystyle{g{{({x})}}}={13}+{x}^{{2}}[/latex] write the equation in standard form and then in transformation/vertex form.
In addition to enabling us to more easily graph a quadratic written in standard form, finding the vertex serves another important purpose—it allows us to determine the maximum or minimum value of the function, depending on which way the graph opens.
Example 2
Returning to our backyard farmer from the beginning of the section, what dimensions should she make her garden to maximize the enclosed area?
Earlier we determined the area she could enclose with 80 feet of fencing on three sides was given by the equation
[latex]\displaystyle{A}{({L})}={80}{L}-{2}{L}^{{2}}[/latex]. Notice that quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area.
In finding the vertex, we take care since the equation is not written in standard polynomial form with decreasing powers. But we know that
a is the coefficient on the squared term, so [latex]\displaystyle{a}={2},{b}={80}[/latex].
Finding the vertex:
[latex-display]\displaystyle{h}=-\frac{{80}}{{{2}{(-{2})}}},{k}={A}{({20})}={80}{({20})}-{2}{({20})}^{{2}}={800}[/latex-display]
The maximum value of the function is an area of 800 square feet, which occurs when
L = 20 feet. When the shorter sides are 20 feet, that leaves 40 feet of fencing for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet, and the longer side parallel to the existing fence has length 40 feet.
Example 3
A local newspaper currently has 84,000 subscribers, at a quarterly charge of $30. Market research has suggested that if they raised the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?
Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the charge per subscription times the number of subscribers. We can introduce variables,
C for charge per subscription and
S for the number subscribers, giving us the equation Revenue =
CS
Since the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently
S = 84,000 and
C = 30, and that if they raise the price to $32 they would lose 5,000 subscribers, giving a second pair of values,
C = 32 and
S = 79,000. From this we can find a linear equation relating the two quantities. Treating
C as the input and
S as the output, the equation will have form
S =
mC +
b. The slope will be
[latex-display]\displaystyle{m}=\frac{{{79},{000}-{84},{000}}}{{{32}-{30}}}=\frac{{-{5},{000}}}{{2}}=-{2},{500}[/latex-display]
This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the vertical intercept
[latex-display]\displaystyle{S}=-{2500}{C}+{b}[/latex-display]
Plug in the point
S = 85,000 and
C = 30
[latex-display]\displaystyle{84},{000}=-{2},{500}{({30})}+{b}[/latex-display]
Solve for
b
[latex-display]\displaystyle{b}={159},{000}[/latex-display]
This gives us the linear equation
S = –2,500
C + 159,000 relating cost and subscribers. We now return to our revenue equation.
Revenue =
CS
Substituting the equation for
S from above, Revenue =
C(–2,500
C +159,000)
Expanding, Revenue = –2,500
C2 + 159,000
C
We now have a quadratic equation for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex:
[latex-display]\displaystyle{h}=-\frac{{{159},{000}}}{{{2}{(-{2},{500})}}}={31.8}[/latex-display]
The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we can evaluate the revenue equation:
Maximum Revenue = –2,500(31.8)
2 + 159,000(31.8) = $2,528,100
Short run Behavior: Intercepts
As with any function, we can find the vertical intercepts of a quadratic by evaluating the function at an input of zero, and we can find the horizontal intercepts by solving for when the output will be zero. Notice that depending upon the location of the graph, we might have zero, one, or two horizontal intercepts.
Example 4
Find the vertical and horizontal intercepts of the quadratic
f(
x) = 3
x2 + 5
x – 2
We can find the vertical intercept by evaluating the function at an input of zero:
f(0) = 3(0)
2 + 5(0) – 2 = 2 Vertical intercept at (0,–2)
For the horizontal intercepts, we solve for when the output will be zero 0 = 3
x2 + 5
x – 2
In this case, the quadratic can be factored easily, providing the simplest method for solution
[latex-display]\displaystyle{0}={({3}{x}-{1})}{({x}+{2})}[/latex-display]
| [latex]\displaystyle{0}={3}{x}-{1}[/latex] |
or |
[latex]\displaystyle{0}={x}+{2}[/latex] |
| [latex]\displaystyle{x}=\frac{{1}}{{3}}[/latex] |
|
[latex]\displaystyle{x}=-{2}[/latex] |
Horizontal intercepts at
[latex-display]\displaystyle{(\frac{{1}}{{3}},{0})}[/latex-display]
Notice that in the standard form of a quadratic, the constant term
c reveals the vertical intercept of the graph.
Example 5
Find the horizontal intercepts of the quadratic
f(
x) = 2
x2 + 4
x – 4
Again we will solve for when the output will be zero
0 = 2
x2 + 4
x – 4
Since the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic into transformation form.
[latex-display]\displaystyle{h}=-\frac{{b}}{{{2}{a}}}=-\frac{{4}}{{{2}{({2})}}}=-{1},{k}={f{{(-{1})}}}={2}{(-{1})}^{{2}}+{4}{(-{1})}-{4}=-{6}[/latex-display]
[latex-display]\displaystyle{f{{({x})}}}={2}{({x}+{1})}^{{2}}-{6}[/latex-display]
Now we can solve for when the output will be zero
[latex-display]\displaystyle{0}={2}{({x}+{1})}^{{2}}-{6}[/latex-display]
[latex-display]\displaystyle{6}={2}{({x}+{1})}^{{2}}[/latex-display]
[latex-display]\displaystyle{3}={({x}+{1})}^{{2}}[/latex-display]
[latex-display]\displaystyle{x}+{1}=\pm\sqrt{{3}}[/latex-display]
[latex-display]\displaystyle{x}=-{1}\pm\sqrt{{3}}[/latex-display]
The graph has horizontal intercepts at
[latex-display]\displaystyle{(-{1}-sqrt{{3}},{0})}[/latex-display]
Try it Now 2
In Try it Now 1, we found the standard and transformation form for the function
g(x) = 13 +
x2 – 6
x. Now find the Vertical & Horizontal intercepts (if any).
This process is done commonly enough that sometimes people find it easier to solve the problem once in general and remember the formula for the result, rather than repeating the process each time. Based on our previous work we showed that any quadratic in standard form can be written into transformation form as:
[latex-display]\displaystyle{f{{({x})}}}={a}{({x}+\frac{{b}}{{{2}{a}}})}^{{2}}+{c}-\frac{{{b}^{{2}}}}{{{4}{a}}}[/latex-display]
Solving for the horizontal intercepts using this general equation gives:
| [latex]\displaystyle{0}={a}{({x}+\frac{{b}}{{{2}{a}}})}^{{2}}+{c}-\frac{{{b}^{{2}}}}{{{4}{a}}}[/latex] |
start to solve for x by moving the constants to the other side. |
| [latex]\displaystyle\frac{{{b}^{{2}}}}{{{4}{a}}}-{c}={a}{({x}+\frac{{{b}}}{{{2}{a}}})}^{{2}}[/latex] |
divide both sides by a |
| [latex]\displaystyle\frac{{{b}^{{2}}}}{{{4}{a}^{{2}}}}-\frac{{c}}{{a}}={({x}+\frac{{{b}}}{{{2}{a}}})}^{{2}}[/latex] |
find a common denominator to combine fractions |
| [latex]\displaystyle\frac{{{b}^{{2}}}}{{{4}{a}^{{2}}}}-\frac{{{4}{a}{c}}}{{{4}{a}^{{2}}}}={({x}+\frac{{{b}}}{{{2}{a}}})}^{{2}}[/latex] |
combine the fractions on the left side of the equation |
| [latex]\displaystyle\frac{{{b}^{{2}}-{4}{a}{c}}}{{{4}{a}^{{2}}}}={({x}+\frac{{{b}}}{{{2}{a}}})}^{{2}}[/latex] |
take the square root of both sides |
| [latex]\displaystyle\pm\sqrt{{\frac{{{b}^{{2}}-{4}{a}{c}}}{{{4}{a}^{{2}}}}}}={x}+\frac{{b}}{{{2}{a}}}[/latex] |
subtract [latex]\displaystyle\frac{{b}}{{{2}{a}}}[/latex] from both sides |
| [latex]\displaystyle-\frac{{b}}{{{2}{a}}}\pm\frac{{\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}={x}[/latex] |
combine the fractions |
| [latex]\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{({b}^{{2}}-{4}{a}{c})}}}}}{{{2}{a}}}[/latex] |
Notice that this can yield two different answers for x |
Quadratic Formula
For a quadratic function given in standard form
[latex]\displaystyle{f{{({x})}}}={a}{x}^{{2}}+{b}{x}+{c}[/latex], the
quadratic formula gives the horizontal intercepts of the graph of this function.
[latex-display]\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}[/latex-display]
Try it Now 3
For these two equations determine if the vertex will be a maximum value or a minimum value.
- [latex]\displaystyle{g{{({x})}}}=-{8}{x}+{x}^{{2}}+{7}[/latex]
- [latex]\displaystyle{g{{({x})}}}=-{3}{({3}-{x})}^{{2}}+{2}[/latex]
Important Topics of this Section
- Quadratic functions
- Transformation form/Vertex form
- Vertex as a maximum / Vertex as a minimum
- Quadratic formula
Try it Now Answers
- [latex]\displaystyle{g{{({x})}}}={x}^{{2}}-{6}{x}+{13}[/latex]in Transformation form
- Vertical intercept at (0, 13), NO horizontal intercepts.
- 1. Vertex is a minimum value2. Vertex is a maximum value
David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, "
Chapter 3: Polynomial and Rational Functions," licensed under a
CC BY-SA 3.0 license.
