Download 5.2 Vertex Form

Section 5.2
Vertex Form
441
5.2 Vertex Form
In the previous section, you learned that it is a simple task to sketch the graph of a
quadratic function if it is presented in vertex form
f (x) = a(x − h)2 + k.
(1)
The goal of the current section is to start with the most general form of the quadratic
function, namely
f (x) = ax2 + bx + c,
(2)
and manipulate the equation into vertex form. Once you have your quadratic function
in vertex form, the technique of the previous section should allow you to construct the
graph of the quadratic function.
However, before we turn our attention to the task of converting the general quadratic
into vertex form, we need to review the necessary algebraic fundamentals. Let’s begin
with a review of an important algebraic shortcut called squaring a binomial.
Squaring a Binomial
A monomial is a single algebraic term, usually constructed as a product of a number
(called a coefficient) and one or more variables raised to nonnegative integral powers,
such as −3x2 or 14y 3 z 5 . The key phrase here is “single term.” A binomial is an algebraic
sum or difference of two monomials (or terms), such as x + 2y or 3ab2 − 2c3 . The key
phrase here is “two terms.”
To “square a binomial,” start with an arbitrary binomial, such as a+b, then multiply
it by itself to produce its square (a + b)(a + b), or, more compactly, (a + b)2 . We can
use the distributive property to expand the square of the binomial a + b.
(a + b)2 = (a + b)(a + b)
= a(a + b) + b(a + b)
= a2 + ab + ba + b2
Because ab = ba, we can add the two middle terms to arrive at the following property.
Property 3. The square of the binomial a + b is expanded as follows.
(a + b)2 = a2 + 2ab + b2
1
(4)
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Chapter 5
I Example 5.
Quadratic Functions
Expand (x + 4)2 .
We could proceed as follows.
(x + 4)2 = (x + 4)(x + 4)
= x(x + 4) + 4(x + 4)
= x2 + 4x + 4x + 16
= x2 + 8x + 16
Although correct, this technique will not help us with our upcoming task. What
we need to do is follow the algorithm suggested by Property 3.
Algorithm for Squaring a Binomial. To square the binomial a + b, proceed
as follows:
1. Square the first term to get a2 .
2. Multiply the first and second terms together, and then multiply the result by
two to get 2ab.
3. Square the second term to get b2 .
Thus, to expand (x + 4)2 , we should proceed as follows.
1. Square the first term to get x2
2. Multiply the first and second terms together and then multiply by two to get 8x.
3. Square the second term to get 16.
Proceeding in this manner allows us to perform the expansion mentally and simply
write down the solution.
(x + 4)2 = x2 + 2(x)(4) + 42 = x2 + 8x + 16
Here are a few more examples. In each, we’ve written an extra step to help clarify
the procedure. In practice, you should simply write down the solution without any
intermediate steps.
(x + 3)2 = x2 + 2(x)(3) + 32 = x2 + 6x + 9
(x − 5)2 = x2 + 2(x)(−5) + (−5)2 = x2 − 10x + 25
2
2
x − 12 = x2 + 2(x) − 21 + − 12 = x2 − x + 41
It is imperative that you master this shortcut before moving on to the rest of the
material in this section.
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Vertex Form
443
Perfect Square Trinomials
Once you’ve mastered squaring a binomial, as in
(a + b)2 = a2 + 2ab + b2 ,
(6)
it’s a simple matter to identify and factor trinomials (three terms) having the form
a2 + 2ab + b2 . You simply “undo” the multiplication. Whenever you spot a trinomial
whose first and third terms are perfect squares, you should suspect that it factors as
follows.
a2 + 2ab + b2 = (a + b)2
(7)
A trinomial that factors according to this rule or pattern is called a perfect square
trinomial.
For example, the first and last terms of the following trinomial are perfect squares.
x2 + 16x + 64
The square roots of the first and last terms are x and 8, respectively. Hence, it makes
sense to try the following.
x2 + 16x + 64 = (x + 8)2
It is important that you check your result using multiplication. So, following the
three-step algorithm for squaring a binomial:
1. Square x to get x2 .
2. Multiply x and 8 to get 8x, then multiply this result by 2 to get 16x.
3. Square 8 to get 64.
Hence, x2 + 16x + 64 is a perfect square trinomial and factors as (x + 8)2 .
As another example, consider x2 − 10x + 25. The square roots of the first and last
terms are x and 5, respectively. Hence, it makes sense to try
x2 − 10x + 25 = (x − 5)2 .
Again, you should check this result. Note especially that twice the product of x and
−5 equals the middle term on the left, namely, −10x.
Completing the Square
If a quadratic function is given in vertex form, it is a simple matter to sketch the
parabola represented by the equation. For example, consider the quadratic function
f (x) = (x + 2)2 + 3,
which is in vertex form. The graph of this equation is a parabola that opens upward.
It is translated 2 units to the left and 3 units upward. This is the advantage of vertex
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Quadratic Functions
form. The transformations required to draw the graph of the function are easy to spot
when the equation is written in vertex form.
It’s a simple matter to transform the equation f (x) = (x + 2)2 + 3 into the general
form of a quadratic function, f (x) = ax2 + bx + c. We simply use the three-step
algorithm to square the binomial; then we combine like terms.
f (x) = (x + 2)2 + 3
f (x) = x2 + 4x + 4 + 3
f (x) = x2 + 4x + 7
Note, however, that the result of this manipulation, f (x) = x2 + 4x + 7, is not as useful
as vertex form, as it is difficult to identify the transformations required to draw the
parabola represented by the equation f (x) = x2 + 4x + 7.
It’s really the reverse of the manipulation above that is needed. If we are presented
with an equation in the form f (x) = ax2 + bx + c, such as f (x) = x2 + 4x + 7, then an
algebraic method is needed to convert this equation to vertex form f (x) = a(x−h)2 +k;
or in this case, back to its original vertex form f (x) = (x + 2)2 + 3.
The procedure we seek is called completing the square. The name is derived from
the fact that we need to “complete” the trinomial on the right side of y = x2 + 4x + 7
so that it becomes a perfect square trinomial.
Algorithm for Completing the Square The procedure for completing the
square involves three key steps.
1. Take half of the coefficient of x and square the result.
2. Add and subtract the quantity from step one so that the right-hand side of the
equation does not change.
3. Factor the resulting perfect square trinomial and combine constant terms.
Let’s follow this procedure and place f (x) = x2 + 4x + 7 in vertex form.
1. Take half of the coefficient of x. Thus, (1/2)(4) = 2. Square this result. Thus,
22 = 4.
2. Add and subtract 4 on the right side of the equation f (x) = x2 + 4x + 7.
f (x) = x2 + 4x + 4 − 4 + 7
3. Group the first three terms on the right-hand side. These form a perfect square
trinomial.
f (x) = (x2 + 4x + 4) − 4 + 7
Now factor the perfect square trinomial and combine the constants at the end to
get
f (x) = (x + 2)2 + 3.
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That’s it, we’re done! We’ve returned the general quadratic f (x) = x2 + 4x + 7
back to vertex form f (x) = (x + 2)2 + 3.
Let’s try that once more.
I Example 8.
Place the quadratic function f (x) = x2 − 8x − 9 in vertex form.
We follow the three-step algorithm for completing the square.
1. Take half of the coefficient of x and square: i.e.,
[(1/2)(−8)]2 = [−4]2 = 16.
2. Add and subtract this amount to the right-hand side of the function.
f (x) = x2 − 8x + 16 − 16 − 9
3. Group the first three terms on the right-hand side. These form a perfect square
trinomial.
f (x) = (x2 − 8x + 16) − 16 − 9
Factor the perfect square trinomial and combine the coefficients at the end.
f (x) = (x − 4)2 − 25
Now, let’s see how we can use the technique of completing the square to help in
drawing the graphs of general quadratic functions.
Working with f (x) = x2 + bx + c
The examples in this section will have the form f (x) = x2 + bx + c. Note that the
coefficient of x2 is 1. In the next section, we will work with a harder form, f (x) =
ax2 + bx + c, where a 6= 1.
I Example 9. Complete the square to place f (x) = x2 + 6x + 2 in vertex form and
sketch its graph.
First, take half of the coefficient of x and square; i.e., [(1/2)(6)]2 = 9. On the right
side of the equation, add and subtract this amount so as to not change the equation.
f (x) = x2 + 6x + 9 − 9 + 2
Group the first three terms on the right-hand side.
f (x) = (x2 + 6x + 9) − 9 + 2
The first three terms on the right-hand side form a perfect square trinomial that is
easily factored. Also, combine the constants at the end.
f (x) = (x + 3)2 − 7
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This is a parabola that opens upward. We need to shift the parabola 3 units to the left
and then 7 units downward, placing the vertex at (−3, −7) as shown in Figure 1(a).
The axis of symmetry is the vertical line x = −3. The table in Figure 1(b) calculates
two points to the right of the axis of symmetry, and mirror points on the left of the
axis of symmetry make for an accurate plot of the parabola.
y
10
10
x
x
−2
−1
y = (x + 3)2 − 7
−6
−3
(−3, −7)
x = −3
(a)
(b)
Figure 1. Plotting the graph of the
quadratic function f (x) = (x + 3)2 − 7.
Let’s look at another example.
I Example 10. Complete the square to place f (x) = x2 − 8x + 21 in vertex form
and sketch its graph.
First, take half of the coefficient of x and square; i.e., [(1/2)(−8)]2 = 16. On
the right side of the equation, add and subtract this amount so as to not change the
equation.
f (x) = x2 − 8x + 16 − 16 + 21
Group the first three terms on the right-hand side of the equation.
f (x) = (x2 − 8x + 16) − 16 + 21
The first three terms form a perfect square trinomial that is easily factored. Also,
combine constants at the end.
f (x) = (x − 4)2 + 5
This is a parabola that opens upward. We need to shift the parabola 4 units to the
right and then 5 units upward, placing the vertex at (4, 5), as shown in Figure 2(a).
The table in Figure 2(b) calculates two points to the right of the axis of symmetry,
and mirror points on the left of the axis of symmetry make for an accurate plot of the
parabola.
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y
10
(4, 5)
10
x
x
5
6
y = (x − 4)2 + 5
6
9
x=4
(a)
(b)
Figure 2. Plotting the graph of the
quadratic function f (x) = (x − 4)2 + 5.
Working with f (x) = ax2 + bx + c
In the last two examples, the coefficient of x2 was 1. In this section, we will learn how
to complete the square when the coefficient of x2 is some number other than 1.
I Example 11. Complete the square to place f (x) = 2x2 + 4x − 4 in vertex form
and sketch its graph.
In the last two examples, we gained some measure of success when the coefficient
of x2 was a 1. We were just getting comfortable with that situation and we’d like
to continue to be comfortable, so let’s start by factoring a 2 from each term on the
right-hand side of the equation.
f (x) = 2 x2 + 2x − 2
If we ignore the factor of 2 out front, the coefficient of x2 in the trinomial expression
inside the parentheses is a 1. Ah, familiar ground! We will proceed as we did before,
but we will carry the factor of 2 outside the parentheses in each step. Start by taking
half of the coefficient of x and squaring the result; i.e., [(1/2)(2)]2 = 1.
Add and subtract this amount inside the parentheses so as to not change the equation.
f (x) = 2 x2 + 2x + 1 − 1 − 2
Group the first three terms inside the parentheses and combine constants.
f (x) = 2 (x2 + 2x + 1) − 3
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Quadratic Functions
The grouped terms inside the parentheses form a perfect square trinomial that is easily
factored.
f (x) = 2 (x + 1)2 − 3
Finally, redistribute the 2.
f (x) = 2(x + 1)2 − 6
This is a parabola that opens upward. In addition, it is stretched by a factor of
2, so it will be somewhat narrower than our previous examples. The parabola is also
shifted 1 unit to the left, then 6 units downward, placing the vertex at (−1, −6), as
shown in Figure 3(a). The table in Figure 3(b) calculates two points to the right of
the axis of symmetry, and mirror points on the left of the axis of symmetry make for
an accurate plot of the parabola.
y
10
10
x
x
0
1
y = 2(x + 1)2 − 6
−4
2
(−1, −6)
x = −1
(a)
(a)
Figure 3. Plotting the graph of the
quadratic function f (x) = 2x2 + 4x − 4.
Let’s look at an example where the coefficient of x2 is negative.
I Example 12. Complete the square to place f (x) = −x2 + 6x − 2 in vertex form
and sketch its graph.
In the last example, we factored out the coefficient of x2 . This left us with a
trinomial having leading coefficient 2 1, which enabled us to proceed much as we did
before: halve the middle coefficient and square, add and subtract this amount, factor
the resulting perfect square trinomial. Since we were successful with this technique in
the last example, let’s begin again by factoring out the leading coefficient, in this case
a −1.
2
The leading coefficient of a quadratic function is the coefficient of x2 . That is, if f (x) = ax2 + bx + c,
then the leading coefficient is “a.” We’ll have more to say about the leading coefficient in Chapter 6.
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Vertex Form
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f (x) = −1 x2 − 6x + 2
If we ignore the factor of −1 out front, the coefficient of x2 in the trinomial expression
inside the parentheses is a 1. Again, familiar ground! We will proceed as we did before,
but we will carry the factor of −1 outside the parentheses in each step. Start by taking
half of the coefficient of x and squaring the result; i.e., [(1/2)(−6)]2 = 9.
Add and subtract this amount inside the parentheses so as to not change the equation.
f (x) = −1 x2 − 6x + 9 − 9 + 2
Group the first three terms inside the parentheses and combine constants.
f (x) = −1 (x2 − 6x + 9) − 7
The grouped terms inside the parentheses form a perfect square trinomial that is easily
factored.
f (x) = −1 (x − 3)2 − 7
Finally, redistribute the −1.
f (x) = −(x − 3)2 + 7
This is a parabola that opens downward. The parabola is also shifted 3 units to
the right, then 7 units upward, placing the vertex at (3, 7), as shown in Figure 4(a).
The table in Figure 4(b) calculates two points to the right of the axis of symmetry,
and mirror points on the left of the axis of symmetry make for an accurate plot of the
parabola.
y
10
(3, 7)
10
x
x
4
5
y = −(x − 3)2 + 7
6
3
x=3
(a)
(b)
Figure 4. Plotting the graph of the
quadratic function f (x) = −(x − 3)2 + 7.
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Quadratic Functions
Let’s try one more example.
I Example 13. Complete the square to place f (x) = 3x2 + 4x − 8 in vertex form
and sketch its graph.
Let’s begin again by factoring out the leading coefficient, in this case a 3.
f (x) = 3 x2 + 34 x − 38
Fractions add a degree of difficulty, but, if you follow the same routine as in the previous
examples, you should be able to get the needed result. Take half of the coefficient of x
and square the result; i.e., [(1/2)(4/3)]2 = [2/3]2 = 4/9.
Add and subtract this amount inside the parentheses so as to not change the equation.
f (x) = 3 x2 + 43 x + 94 − 49 − 38
Group the first three terms inside the parentheses. You’ll need a common denominator
to combine constants.
f (x) = 3 x2 + 43 x + 49 − 49 − 24
9
The grouped terms inside the parentheses form a perfect square trinomial that is easily
factored.
i
h
2
f (x) = 3 x + 23 − 28
9
Finally, redistribute the 3.
f (x) = 3 x +
2 2
3
−
28
3
This is a parabola that opens upward. It is also stretched by a factor of 3, so it
will be narrower than all of our previous examples. The parabola is also shifted 2/3
units to the left, then 28/3 units downward, placing the vertex at (−2/3, −28/3), as
shown in Figure 5(a). The table in Figure 5(b) calculates two points to the right of
the axis of symmetry, and mirror points on the left of the axis of symmetry make for
an accurate plot of the parabola.
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Section 5.2
Vertex Form
451
y
5
10
x
x
0
1
y = 3(x + 2/3)2 − 28/3
−8
−1
(−2/3, −28/3)
x = −2/3
(a)
(b)
Figure 5. Plotting the graph of the quadratic
function f (x) = 3(x + 2/3)2 − 28/3.
Version: Fall 2007