ORCCA Completing the Square

Section 13.3 Completing the Square

Objectives: PCC Course Content and Outcome Guide

MTH 95 CCOG 4.d

permalinkIn this section, we will learn how to “complete the square” with a quadratic expression. This topic is useful for solving quadratic equations and putting quadratic functions in vertex form.

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Figure 13.3.1. Alternative Video Lessons

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Subsection 13.3.1 Solving Quadratic Equations by Completing the Square

When we have an equation like $(x+5)^2=4\text{,}$ we can solve it quickly using the square root property:

\begin{aligned} (x + 5)^{2} & = 4 \end{aligned}

$\begin{align*} (x+5)^2\amp=4 \end{align*}$

\begin{aligned} x + 5 & = - 2 & or & x + 5 & = 2 \\ x & = - 7 & or & x & = - 3 \end{aligned}

$\begin{align*} x+5\amp=-2\amp\text{or}\amp\amp x+5\amp=2\\ x\amp=-7\amp\text{or}\amp\amp x\amp=-3 \end{align*}$

permalinkThe method of completing the square allows us to solve any quadratic equation using the square root property.

Suppose you have a small quadratic expression in the form $x^2+bx\text{.}$ It can be visualized as an “L”-shape as in Figure 13.3.2.

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Figure 13.3.2.

The “missing” square in the upper right corner of Figure 13.3.2 is $\frac{b}{2}$ on each side, so its area is $\left(\frac{b}{2}\right)^2\text{.}$ This means that if we have $x^2+bx$ and add $\left(\frac{b}{2}\right)^2\text{,}$ we are “completing” the larger square.

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Fact 13.3.3. The Term that Completes the Square.

For a polynomial $x^2+bx\text{,}$ the constant term needed to make a perfect square trinomial is $\highlight{\left(\frac{b}{2}\right)^2}\text{.}$

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Process 13.3.4. Completing the Square.

For a quadratic equation simplified to the form $x^2+bx=c\text{,}$ to solve for $x$ by completing the square,

Use Fact 13.3.3 to find the number to add to both sides of the equation to make the left hand side a perfect square. This number is always $\highlight{\left(\frac{b}{2}\right)^2}\text{.}$
Add that number to both sides of $x^2+bx=c$ to get

$\begin{equation*} x^2+bx\addright{\left(\frac{b}{2}\right)^2}=c\addright{\left(\frac{b}{2}\right)^2} \end{equation*}$
The left hand side is now a perfect square that factors as $x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)^2\text{,}$ so the equation becomes

$\begin{equation*} \left(x+\frac{b}{2}\right)^2=c+\left(\frac{b}{2}\right)^2 \end{equation*}$
Solve remaining equation using the Square Root Property.

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Example 13.3.5.

Solve the quadratic equation $x^2+6x=16$ by completing the square.

Explanation

To solve the quadratic equation $x^2+6x=16\text{,}$ on the left side we can complete the square by adding $\left(\frac{b}{2}\right)^2\text{;}$ note that $b=6$ in this case, which makes $\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=3^2=9\text{.}$ We add it to both sides to maintain equality.

\begin{align*} x^2+6x\addright{9}\amp=16\addright{9}\\ x^2+6x+9\amp=25\\ (x+3)^2\amp=25 \end{align*}

Now that we have completed the square, we can solve the equation using the square root property.

\begin{align*} x+3\amp=-5\amp\text{or}\amp\amp x+3\amp=5\\ x\amp=-8\amp\text{or}\amp\amp x\amp=2 \end{align*}

The solution set is $\{-8, 2\}\text{.}$

permalinkNow let's see the process for completing the square when the quadratic equation is given in standard form.

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Example 13.3.6.

Solve $x^2-14x+11=0$ by completing the square.

Explanation

We see that the polynomial on the left side is not a perfect square trinomial, so we need to complete the square. We subtract $11$ from both sides so we can add the missing term on the left.

\begin{align*} x^2-14x+11\amp=0\\ x^2-14x\amp=-11 \end{align*}

Next comes the completing-the-square step. We need to add the correct number to both sides of the equation to make the left side a perfect square. Remember that Fact 13.3.3 states that we need to use $\left(\frac{b}{2}\right)^2$ for this. In our case, $b=-14\text{,}$ so $\left(\frac{b}{2}\right)^2=\left(\frac{-14}{2}\right)^2=49$

\begin{align*} x^2-14x\addright{49}\amp=-11\addright{49}\\ (x-7)^2\amp=38 \end{align*}

\begin{align*} x-7\amp=-\sqrt{38}\amp\text{or}\amp\amp x-7\amp=\sqrt{38}\\ x\amp=7-\sqrt{38}\amp\text{or}\amp\amp x\amp=7+\sqrt{38} \end{align*}

The solution set is $\{7-\sqrt{38}, 7+\sqrt{38}\}\text{.}$

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Checkpoint 13.3.7.

So far, the value of $b$ has been even each time, which makes $\frac{b}{2}$ a whole number. When $b$ is odd, we end up adding a fraction to both sides. Here is an example.

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Example 13.3.8.

Complete the square to solve for $z$ in $z^2-3z-10=0\text{.}$

Explanation

First move the constant term to the right side of the equation:

\begin{align*} z^2-3z-10\amp=0\\ z^2-3z\amp=10 \end{align*}

Next, to complete the square, we need to find the right number to add to both sides. According to Fact 13.3.3, we need to divide the value of $b$ by $2$ and then square the result to find the right number. First, divide by $2\text{:}$

\begin{equation} \frac{b}{2}=\frac{-3}{2}=-\frac{3}{2}\label{mRk}\tag{13.3.1} \end{equation}

and then we square that result:

\begin{equation} \left(-\frac{3}{2}\right)^2=\frac{9}{4}\label{SYt}\tag{13.3.2} \end{equation}

Now we can add the $\frac{9}{4}$ from Equation (13.3.2) to both sides of the equation to complete the square.

\begin{align*} z^2-3z\addright{\frac{9}{4}}\amp=10\addright{\frac{9}{4}} \end{align*}

Now, to factor the seemingly complicated expression on the left, just know that it should always factor using the number from the first step in the completing the square process, Equation (13.3.1).

\begin{align*} \left(z\mathbin{\highlight{-}}\highlight{\frac{3}{2}}\right)^2\amp=\frac{49}{4} \end{align*}

\begin{align*} z-\frac{3}{2}\amp=-\frac{7}{2}\amp\text{or}\amp\amp z-\frac{3}{2}\amp=\frac{7}{2}\\ z\amp=\frac{3}{2}-\frac{7}{2}\amp\text{or}\amp\amp z\amp=\frac{3}{2}+\frac{7}{2}\\ z\amp=-\frac{4}{2}\amp\text{or}\amp\amp z\amp=\frac{10}{2}\\ z\amp=-2\amp\text{or}\amp\amp z\amp=5 \end{align*}

The solution set is $\{-2,5\}\text{.}$

In each of the previous examples, the value of $a$ was equal to $1\text{.}$ This is necessary for our missing term formula to work. When $a$ is not equal to $1$ we will divide both sides by $a\text{.}$ Let's look at an example of that.

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Example 13.3.9.

Solve for $r$ in $2r^2+2r=3$ by completing the square.

Explanation

Because there is a leading coefficient of $2\text{,}$ we divide both sides by $2\text{.}$

\begin{align*} 2r^2+2r\amp=3\\ \divideunder{2r^2}{2}+\divideunder{2r}{2}\amp=\divideunder{3}{2}\\ r^2+r\amp=\frac{3}{2} \end{align*}

Next, we complete the square. Since $b=1\text{,}$ first,

\begin{equation} \frac{b}{2}=\frac{1}{2}\label{XIm}\tag{13.3.3} \end{equation}

and next, squaring that, we have

\begin{equation} \left(\frac{1}{2}\right)^2=\frac{1}{4}\text{.}\label{DPv}\tag{13.3.4} \end{equation}

So we add $\frac{1}{4}$ from Equation (13.3.4) to both sides of the equation:

\begin{align*} r^2+r\addright{\frac{1}{4}}\amp=\frac{3}{2}\addright{\frac{1}{4}}\\ r^2+r+\frac{1}{4}\amp=\frac{6}{4}+\frac{1}{4} \end{align*}

Here, remember that we always factor with the number found in the first step of completing the square, Equation (13.3.3).

\begin{align*} \left(r+\highlight{\frac{1}{2}}\right)^2\amp=\frac{7}{4} \end{align*}

\begin{align*} r+\frac{1}{2}\amp=-\frac{\sqrt{7}}{2}\amp\text{or}\amp\amp r+\frac{1}{2}\amp=\frac{\sqrt{7}}{2}\\ r\amp=-\frac{1}{2}-\frac{\sqrt{7}}{2}\amp\text{or}\amp\amp r\amp=-\frac{1}{2}+\frac{\sqrt{7}}{2}\\ r\amp=\frac{-1-\sqrt{7}}{2}\amp\text{or}\amp\amp r\amp=\frac{-1+\sqrt{7}}{2} \end{align*}

The solution set is $\left\{\frac{-1-\sqrt{7}}{2}, \frac{-1+\sqrt{7}}{2}\right\}\text{.}$

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Subsection 13.3.2 Deriving the Quadratic Formula by Completing the Square

In Section 7.2, we learned the Quadratic Formula. You may have wondered where the formula comes from, and now that we know how to complete the square, we can derive it. We will solve the standard form equation $ax^2+bx+c=0$ for $x\text{.}$

First, we subtract $c$ from both sides and divide both sides by $a\text{.}$

\begin{aligned} a x^{2} + b x + c & = 0 \\ a x^{2} + b x & = - c \\ \frac{a x^{2}}{a} + \frac{b x}{a} & = - \frac{c}{a} \\ x^{2} + \frac{b}{a} x & = - \frac{c}{a} \end{aligned}

$\begin{align*} ax^2+bx+c\amp=0\\ ax^2+bx\amp=\subtractright{c}\\ \divideunder{ax^2}{a}+\divideunder{bx}{a}\amp=-\divideunder{c}{a}\\ x^2+\frac{b}{a}x\amp=-\frac{c}{a} \end{align*}$

permalinkNext, we complete the square by taking half of the middle coefficient and squaring it. First,

\begin{matrix} (13.3.5) & \frac{\frac{b}{a}}{2} = \frac{b}{2 a} \end{matrix}

$\begin{equation} \frac{\frac{b}{a}}{2}=\frac{b}{2a}\label{Izo}\tag{13.3.5} \end{equation}$

permalinkand then squaring that we have

\begin{matrix} (13.3.6) & {(\frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} \end{matrix}

$\begin{equation} \left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}\label{oGx}\tag{13.3.6} \end{equation}$

We add the $\frac{b^2}{4a^2}$ from Equation (13.3.6) to both sides of the equation:

\begin{aligned} x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} & = + \frac{b^{2}}{4 a^{2}} - \frac{c}{a} \end{aligned}

$\begin{align*} x^2+\frac{b}{a}x\addright{\frac{b^2}{4a^2}}\amp=\addright{\frac{b^2}{4a^2}}-\frac{c}{a} \end{align*}$

permalinkRemember that the left side always factors with the value we found in Equation (13.3.5). So we have:

\begin{aligned} {(x + \frac{b}{2 a})}^{2} & = \frac{b^{2}}{4 a^{2}} - \frac{c}{a} \end{aligned}

$\begin{align*} \highlight{\left(\unhighlight{x+\frac{b}{2a}}\right)^2}\amp=\frac{b^2}{4a^2}-\frac{c}{a} \end{align*}$

To find a common denominator on the right, we multiply by $4a$ in the numerator and denominator on the second term.

\begin{aligned} {(x + \frac{b}{2 a})}^{2} & = \frac{b^{2}}{4 a^{2}} - \frac{c}{a} \cdot \frac{4 a}{4 a} \\ {(x + \frac{b}{2 a})}^{2} & = \frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}} \\ {(x + \frac{b}{2 a})}^{2} & = \frac{b^{2} - 4 a c}{4 a^{2}} \end{aligned}

$\begin{align*} \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2}{4a^2}-\frac{c}{a}\multiplyright{\frac{4a}{4a}}\\ \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\ \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2-4ac}{4a^2} \end{align*}$

Now that we have completed the square, we can see that the $x$ -value of the vertex is $-\frac{b}{2a}\text{.}$ That is the vertex formula. Next, we solve the equation using the square root property to find the Quadratic Formula.

\begin{aligned} x + \frac{b}{2 a} & = \pm \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \\ x + \frac{b}{2 a} & = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \\ x & = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \\ x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}

$\begin{align*} x+\frac{b}{2a}\amp=\highlight{\pm\sqrt{\unhighlight{\frac{b^2-4ac}{4a^2}}}}\\ x+\frac{b}{2a}\amp=\pm\divideunder{\sqrt{b^2-4ac}}{2a}\\ x\amp=\subtractright{\frac{b}{2a}}\pm\frac{\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{align*}$

This shows us that the solutions to the equation $ax^2+bx+c=0$ are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}\text{.}$

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Subsection 13.3.3 Putting Quadratic Functions in Vertex Form

permalinkIn Section 13.2, we learned about the vertex form of a parabola, which allows us to quickly read the coordinates of the vertex. We can now use the method of completing the square to put a quadratic function in vertex form. Completing the square with a function is a little different than with an equation so we will start with an example.

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Example 13.3.10.

Write a formula in vertex form for the function $q$ defined by $q(x)=x^2+8x$

Explanation

The formula is in the form $x^2+bx\text{,}$ so we need to add $\left(\frac{b}{2}\right)^2$ to complete the square by Fact 13.3.3. When we had an equation, we could add the same quantity to both sides. But now we do not wish to change the left side, since we are trying to end up with a formula that still says $q(x)=\ldots\text{.}$ Instead, we add and subtract the term from the right side in order to maintain equality. In this case,

\begin{align*} \left(\frac{b}{2}\right)^2\amp=\left(\frac{8}{2}\right)^2\\ \amp=4^2\\ \amp=16 \end{align*}

To maintain equality, we both add and subtract $16$ on the same side of the equation. It is functionally the same as adding $0$ on the right, but the $16$ makes it possible to factor the expression in a particular way:

\begin{align*} q(x)\amp=x^2+8x\addright{16}\subtractright{16}\\ \amp=\highlight{\left(\unhighlight{x^2+8x+16}\right)}-16\\ \amp=(x+4)^2-16 \end{align*}

Now that we have completed the square, our function is in vertex form. The vertex is $(-4,-16)\text{.}$ One way to verify that our work is correct is to graph the original version of the function and check that the vertex is where it should be.

a plot of y=x^2+8x with the vertex labeled at (-4,-16) — Figure 13.3.11. Graph of $y=x^2+8x$

permalinkLet's look at a function that has a constant term and see how to complete the square.

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Example 13.3.12.

Write a formula in vertex form for the function $f$ defined by $f(x)=x^2-12x+3$

Explanation

To complete the square, we need to add and subtract $\left(-\frac{12}{2}\right)^2=(-6)^2=36$ on the right side.

\begin{align*} f(x)\amp=x^2-12x\addright{36}\subtractright{36}+3\\ \amp=\highlight{\left(\unhighlight{x^2-12x+36}\right)}-36+3\\ \amp=(x-6)^2-33 \end{align*}

The vertex is $(6,-33)\text{.}$

In the first two examples, $a$ was equal to $1\text{.}$ When $a$ is not equal to one, we have an additional step. Since we are working with an expression where we intend to preserve the left side as $f(x)=\ldots\text{,}$ we cannot divide both sides by $a\text{.}$ Instead we factor $a$ out of the first two terms. Let's look at an example of that.

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Example 13.3.13.

Write a formula in vertex form for the function $g$ defined by $g(x)=5x^2+20x+25$

Explanation

Before we can complete the square, we factor the $\highlight{5}$ out of the first two terms.

\begin{align*} g(x)\amp=\highlight{5}\left(x^2+4x\right)+25 \end{align*}

Now we complete the square inside the parentheses by adding and subtracting $\left(\frac{4}{2}\right)^2=2^2=4\text{.}$

\begin{align*} g(x)\amp=5\left(x^2+4x\addright{4}\subtractright{4}\right)+25 \end{align*}

Notice that the constant that we subtracted is inside the parentheses, but it will not be part of our perfect square trinomial. In order to bring it outside, we need to multiply it by $5\text{.}$ We are distributing the $5$ to that term so we can combine it with the outside term.

\begin{align*} g(x)\amp=5\left(\highlight{\left(\unhighlight{x^2+4x+4}\right)}-\highlight{4}\right)+25\\ \amp=5\highlight{\left(\unhighlight{x^2+4x+4}\right)}-\multiplyleft{5}4+25\\ \amp=5\left(x+2\right)^2-20+25\\ \amp=5\left(x+2\right)^2+5 \end{align*}

The vertex is $(-2,5)\text{.}$

permalinkHere is an example that includes fractions.

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Example 13.3.14.

Write a formula in vertex form for the function $h$ defined by $h(x)=-3x^2-4x-\frac{7}{4}$

Explanation

First, we factor the leading coefficient out of the first two terms.

\begin{align*} h(x)\amp=-3x^2-4x-\frac{7}{4}\\ \amp=-3\left(x^2+\frac{4}{3}x\right)-\frac{7}{4} \end{align*}

Next, we complete the square for $x^2+\frac{4}{3}x$ inside the grouping symbols by adding and subtracting the right number. To find that number, we divide the value of $b$ by two and square the result. That looks like:

\begin{equation} \frac{b}{2}=\frac{\frac{4}{3}}{2}=\frac{4}{3}\cdot\frac{1}{2}=\frac{2}{3}\label{KoB}\tag{13.3.7} \end{equation}

and then,

\begin{equation} \left(\frac{2}{3}\right)^2=\frac{2^2}{3^2}=\frac{4}{9}\label{qvK}\tag{13.3.8} \end{equation}

Adding and subtracting the value from Equation (13.3.8), we have:

\begin{align*} h(x)\amp=-3\left(x^2+\frac{4}{3}x\addright{\frac{4}{9}}\subtractright{\frac{4}{9}}\right)-\frac{7}{4}\\ \amp=-3\left(\highlight{\left(\unhighlight{x^2+\frac{4}{3}x+\frac{4}{9}}\right)}-\frac{4}{9}\right)-\frac{7}{4}\\ \amp=-3\highlight{\left(\unhighlight{x^2+\frac{4}{3}x+\frac{4}{9}}\right)}-\left(3\cdot-\frac{4}{9}\right)-\frac{7}{4}\\ \end{align*}

Remember that when completing the square, the expression should always factor with the number found in the first step of the completing-the-square process, Equation (13.3.7).

\begin{align*} \amp=-3\highlight{\left(\unhighlight{x+\frac{2}{3}}\right)^2}+\frac{4}{3}-\frac{7}{4}\\ \amp=-3\left(x+\frac{2}{3}\right)^2+\frac{16}{12}-\frac{21}{12}\\ \amp=-3\left(x+\frac{2}{3}\right)^2-\frac{5}{12} \end{align*}

The vertex is $\left(-\frac{2}{3},-\frac{5}{12}\right)\text{.}$

permalinkCompleting the square can also be used to find a minimum or maximum in an application.

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Example 13.3.15.

In Example 5.4.16, we learned that artist Tyrone's annual income from paintings can be modeled by $I(x)=-100x^2+1000x+20000\text{,}$ where $x$ is the number of times he will raise the price per painting by $20.00. To maximize his income, how should Tyrone set his price per painting? Find the maximum by completing the square.

Explanation

To find the maximum is essentially the same as finding the vertex, which we can find by completing the square. To complete the square for $I(x)=-100x^2+1000x+20000\text{,}$ we start by factoring out the $-100$ from the first two terms:

\begin{align*} I(x)\amp=-100x^2+1000x+20000\\ \amp=-100\left(x^2-10x\right)+20000\\ \end{align*}

Next, we complete the square for $x^2-10x$ by adding and subtracting $\left(-\frac{10}{2}\right)^2=(-5)^2=\highlight{25}\text{.}$

\begin{align*} I(x)\amp=-100\left(x^2-10x\addright{25}\subtractright{25}\right)+20000\\ \amp=-100\left(\highlight{\left(\unhighlight{x^2-10x+25}\right)}-25\right)+20000\\ \amp=-100\highlight{\left(\unhighlight{x^2-10x+25}\right)}-\left(100\cdot-25\right)+20000\\ \amp=-100(x-5)^2+2500+20000\\ \amp=-100(x-5)^2+22500 \end{align*}

The vertex is the point $(5,22500)\text{.}$ This implies Tyrone should raise the price per painting $\substitute{5}$ times, which is $\substitute{5}\cdot20=100$ dollars. He would sell $100-5(\substitute{5})=75$ paintings. This would make the price per painting $200+100=300$ dollars, and his annual income from paintings would become $22,500 by this model.

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Subsection 13.3.4 Graphing Quadratic Functions by Hand

permalinkNow that we know how to put a quadratic function in vertex form, let's review how to graph a parabola by hand.

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Example 13.3.16.

Graph the function $h$ defined by $h(x)=2x^2+4x-6$ by determining its key features algebraically.

Explanation

To start, we'll note that this function opens upward because the leading coefficient, $2\text{,}$ is positive.

Now we may complete the square to find the vertex. We factor the $2$ out of the first two terms, and then add and subtract $\left(\frac{2}{2}\right)^2=1^2=\highlight{1}$ on the right side.

\begin{align*} h(x)\amp=2\left(x^2+2x\right)-6\\ \amp=2\left[x^2+2x\addright{1}\subtractright{1}\right]-6\\ \amp=2\left[\highlight{\left(\unhighlight{x^2+2x+1}\right)}-1\right]-6\\ \amp=2\highlight{\left(\unhighlight{x^2+2x+1}\right)}-\left(2\cdot1\right)-6\\ \amp=2\left(x+1\right)^2-2-6\\ \amp=2\left(x+1\right)^2-8 \end{align*}

The vertex is $(-1,-8)$ so the axis of symmetry is the line $x=-1\text{.}$

To find the $y$-intercept, we'll replace $x$ with $0$ or read the value of $c$ from the function in standard form:

\begin{align*} h(\substitute{0})\amp=2(\substitute{0})^2+2(\substitute{0})-6\\ \amp=-6 \end{align*}

The $y$-intercept is $(0,-6)$ and we can find its symmetric point on the graph, which is $(-2,-6)\text{.}$

Next, we'll find the horizontal intercepts. We see this function factors so we write the factored form to get the horizontal intercepts.

\begin{align*} h(x)\amp=2x^2+4x-6\\ \amp=2\left(x^2+2x-3\right)\\ \amp=2(x-1)(x+3) \end{align*}

The $x$-intercepts are $(1,0)$ and $(-3,0)\text{.}$

Now we plot all of the key points and draw the parabola.

a graph of the parabola y=2x^2+4x-6 opens upward and the key features are labeled;the vertex is (-1,-8);the axis of symmetry is drawn at x=-1;the y-intercept is (0,-6) and its symmetric point is (-2,-6);the x-intercepts are (1,0) and (-3,0) — Figure 13.3.17. The graph of $y=2x^2+4x-6\text{.}$

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Example 13.3.18.

Write a formula in vertex form for the function $p$ defined by $p(x)=-x^2-4x-1\text{,}$ and find the graph's key features algebraically. Then sketch the graph.

Explanation

In this function, the leading coefficient is negative so it will open downward. To complete the square we first factor $-1$ out of the first two terms.

\begin{align*} p(x)\amp=-x^2-4x-1\\ \amp=-\left(x^2+4x\right)-1 \end{align*}

Now, we add and subtract the correct number on the right side of the function: $\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2=2^2=\highlight{4}\text{.}$

\begin{align*} p(x)\amp=-\left(x^2+4x\addright{4}\subtractright{4}\right)-1\\ \amp=-\left(\highlight{\left(\unhighlight{x^2+4x+4}\right)}-4\right)-1\\ \amp=-\highlight{\left(\unhighlight{x^2+4x+4}\right)}-(-4)-1\\ \amp=-\left(x+2\right)^2+4-1\\ \amp=-\left(x+2\right)^2+3 \end{align*}

The vertex is $(-2,3)$ so the axis of symmetry is the line $x=-2\text{.}$

We find the $y$-intercept by looking at the value of $c\text{,}$ which is $-1\text{.}$ So, the $y$-intercept is $(0,-1)$ and we can find its symmetric point on the graph, $(-4,-1)\text{.}$

The original expression, $-x^2-4x-1\text{,}$ does not factor so to find the $x$-intercepts we need to set $p(x)=0$ and complete the square or use the quadratic formula. Since we just went through the process of completing the square above, we can use that result to save several repetitive steps.

\begin{align*} p(x)\amp=0\\ -\left(x+2\right)^2+3\amp=0\\ -(x+2)^2\amp=-3\\ (x+2)^2\amp=3 \end{align*}

\begin{align*} x+2\amp=-\sqrt{3}\amp\text{or}\amp\amp x+2\amp=\sqrt{3}\\ x\amp=-2-\sqrt{3}\amp\text{or}\amp\amp x\amp=-2+\sqrt{3}\\ x\amp\approx-3.73\amp\text{or}\amp\amp x\amp\approx-0.268 \end{align*}

The $x$-intercepts are approximately $(-3.7,0)$ and $(-0.3,0)\text{.}$ Now we can plot all of the points and draw the parabola.

a graph of the parabola y=-x^2-4x-1 opens downward and the key features are labeled;the vertex is (-2,3);the axis of symmetry is drawn at x=-2;the y-intercept is (0,-1) and its symmetric point is (-4,-1);the x-intercepts are (-3.7,0) and (-0.3,0) — Figure 13.3.19. The graph of $y=-x^2-4x-1\text{.}$

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Reading Questions 13.3.5 Reading Questions

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1.

For the expression $y=x^2+10x-9\text{,}$ explain in words what is the next step to complete the square.

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2.

Why is completing the square called completing the square?

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3.

How can you check that they completed the square correctly?

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Exercises 13.3.6 Exercises

Review and Warmup

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1.

Use a square root to solve ${\left(t+6\right)^{2}}={16}\text{.}$

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2.

Use a square root to solve ${\left(x-1\right)^{2}}={49}\text{.}$

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3.

Use a square root to solve ${\left(3x-7\right)^{2}}={49}\text{.}$

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4.

Use a square root to solve ${\left(8y+1\right)^{2}}={25}\text{.}$

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5.

Use a square root to solve ${\left(y-1\right)^{2}}={19}\text{.}$

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6.

Use a square root to solve ${\left(r-8\right)^{2}}={7}\text{.}$

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7.

Use a square root to solve ${r^{2}+8r+16}={64}\text{.}$

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8.

Use a square root to solve ${r^{2}-4r+4}={9}\text{.}$

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9.

Use a square root to solve ${4t^{2}+8t+4}={4}\text{.}$

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10.

Use a square root to solve ${49t^{2}-126t+81}={49}\text{.}$

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11.

Use a square root to solve ${16x^{2}-8x+1}={10}\text{.}$

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12.

Use a square root to solve ${81x^{2}+126x+49}={5}\text{.}$