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Applied Precalculus MTH 103

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Section 3.3 Vertex Form and Completing the Square

Recall that the standard form of a quadratic function is
\begin{equation*} f(x) = a x^2 + b x + c, \end{equation*}
where \(a\text{,}\) \(b\text{,}\) and \(c\) are real numbers with \(a \neq 0\text{.}\)
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When one first looks upon the function
\begin{equation*} f(x) = (x-1)^2 + 2 \end{equation*}
it may not immediately be apparent that this is in fact a quadratic function. This can be seen by expanding the right-hand side:
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\begin{align*} f(x) \amp\;=\; (x-1)^2 + 2\\ \amp\;=\; (x - 1)(x - 1) + 2\\ \amp\;=\; x^2 - x - x + 1 + 2\\ \amp\;=\; x^2 - 2x + 3 \end{align*}
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Hence,
\begin{equation*} f(x) = (x-1)^2 + 2 \end{equation*}
can be written in the standard form of a quadratic function as
\begin{equation*} f(x) = x^2 - 2x + 3. \end{equation*}
This is the same quadratic function that appeared in part (a) of Example 3.1.3 where we saw that the vertex was \((\textcolor{red}{1}, \textcolor{blue}{2})\text{.}\) Notice that these numbers appear in the alternate formula for \(f(x) = x^2 - 2x + 3\text{:}\)
\begin{equation*} f(x) = (x - \textcolor{red}{1})^2 + \textcolor{blue}{2}. \end{equation*}
This is not a coincidence and brings us to the vertex form of a quadratic function.
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Definition 3.3.1. Vertex Form.

The vertex form of a quadratic function with vertex \((\textcolor{red}{h},\textcolor{blue}{k})\) is given by
\begin{equation*} f(x)=a(x-\textcolor{red}h)^2+\textcolor{blue}k \end{equation*}
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Example 3.3.2.

Identify the vertex of each of the following quadratic functions.
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  1. \(\displaystyle f(t)=-3(t-6)^2-4\)
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  2. \(\displaystyle y=(x+7)^2+3\)
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Solution.
  1. The function can be written as
    \begin{equation*} f(t) = -3(t - \textcolor{red}{6})^2 + \textcolor{blue}{-4}. \end{equation*}
    It can be seen that \(h = \textcolor{red}{6}\) and \(k = \textcolor{blue}{-4}\text{,}\) so its vertex is \((6, -4)\text{.}\)
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  2. Note that for the function
    \begin{equation*} y = (x+7)^2 + 3 \end{equation*}
    addition is being performed inside the parentheses instead of subtraction, as is required to correctly identify \(h\) using the vertex form. Recalling that subtracting a negative number is the same as adding, we can rewrite the above as
    \begin{equation*} y = (x - (\textcolor{red}{-7}))^2 + \textcolor{blue}{3} \end{equation*}
    which allows us to identify that \(h = \textcolor{red}{-7}\) and \(k = \textcolor{blue}{3}\text{.}\) The vertex is \((-7, 3)\text{.}\)
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Subsection Converting from Vertex Form to Standard Form

In the beginning of this section, we converted \(f(x)=(x-1)^2+2\) from vertex form to its standard form \(f(x)=x^2-2x+3\) by expanding the right-hand side. This is what is done in general to convert a quadratic function from vertex form to standard form. To expedite this process, special product formulas may be used. Note that
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\begin{align*} (A+B)^2 \amp\;=\; (A + B)(A + B) \amp (A-B)^2 \amp\;=\; (A - B)(A - B)\\ \amp\;=\; A^2 + AB + AB + B^2 \amp \amp \;=\; A^2 - AB - AB + B^2\\ \amp\;=\; A^2 + 2AB + B^2 \amp\amp \;=\; A^2 - 2AB + B^2 \end{align*}
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Example 3.3.4.

Convert each quadratic function from vertex form to standard form and identify \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)
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  1. \(\displaystyle y=\left(x-\frac{1}{2}\right)^2+5\)
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  2. \(\displaystyle g(t)=-3(t-5)^2-10\)
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Solution.
  1. Note that
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    \begin{align*} y \amp\;=\; \left( x - \frac{1}{2} \right)^2 + 5\\ \amp\;=\; \left( (x)^2 -x + \left( \frac{1}{2} \right)^2 \right) + 5\\ \amp\;=\; x^2 - x + \frac{1}{4} + 5\\ \amp\;=\; x^2 - x + \frac{1}{4} + \frac{5}{1} \cdot \textcolor{blue}{\frac{4}{4}}\\ \amp\;=\; x^2 - x + \frac{21}{4} \end{align*}
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    so the standard form is \(y = x^2 - x + \frac{21}{4}\text{.}\) Here \(a = 1\text{,}\) \(b = -1\text{,}\) and \(c = \frac{21}{4}\text{.}\)
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  2. Note that
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    \begin{align*} g(t) \amp\;=\; -3 ( t - 5 )^2 - 10\\ \amp\;=\; -3 \left( (t)^2 - 2 (t) (5) + (5)^2 \right) - 10\\ \amp\;=\; -3 ( t^2 - 10 t + 25 ) - 10\\ \amp\;=\; -3 ( t^2 b - 10 c t + 25 d) - 10\\ \amp\;=\; -3 t^2 + 30 t - 75 - 10\\ \amp\;=\; -3 t^2 + 30 t - 85 \end{align*}
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    so the standard form is \(g(t)=-3t^2+30t-85\text{.}\) Here \(a=-3\text{,}\) \(b=30\text{,}\) and \(c=-85\text{.}\)
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Subsection Converting from Standard Form to Vertex Form

Converting from standard form to vertex form requires completing the square. This process will be explained below, but first an observation. Note that
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\begin{align*} x^2 + Bx + \left( \frac{B}{2} \right)^2 \amp= x^2 + 2(x) \left( \frac{B}{2} \right) + \left( \frac{B}{2} \right)^2 \\ \amp= \left( x + \frac{B}{2} \right)^2 \end{align*}
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by the first special product formula (see Theorem 3.3.3).
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Example 3.3.6.

Convert each of the following quadratic functions from standard form to vertex form. Then identify the vertex.
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  1. \(\displaystyle f(x)=x^2+6x+10\)
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  2. \(\displaystyle y = t^2 - 8t + 1\)
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  3. \(\displaystyle g(x) = x^2 + 5x - 3\)
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Solution.
  1. The quadratic function \(f(x)=x^2+6x+10\) is of the form \(x^2+Bx+C\) with \(B=6\text{.}\) Now
    \begin{equation*} \textcolor{blue}{\left( \frac{B}{2} \right)^2 } = \left( \frac{6}{2} \right)^2 = (3)^2 = 9. \end{equation*}
    Adding and subtracting this quantity from the quadratic yields:
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    \begin{align*} x^2 + 6x + 10 \amp\;=\; \textcolor{red}{\underbrace{ {\color{black}x^2 + 6x }\textcolor{blue}{+9}}_{\textcolor{red}{\text{factors into } (x+3)^2}} } + 10 \textcolor{blue}{-} \textcolor{blue}{9} \\ \amp\;=\; (x+3)^2 + 1. \end{align*}
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    Hence, the vertex form of \(f(x)=x^2+6x+10\) is
    \begin{equation*} f(x) = (x+3)^2 + 1 \end{equation*}
    and the vertex is \((-3, 1)\text{.}\)
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  2. The quadratic function \(y = t^2 - 8t + 1\) is of the form \(t^2 + Bt + C\) with \(B = -8\text{.}\) Now
    \begin{equation*} \textcolor{blue}{\left( \frac{B}{2} \right)^2 } = \left( \frac{-8}{2} \right)^2 = (-4)^2 = 16. \end{equation*}
    Adding and subtracting this quantity from the quadratic yields:
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    \begin{align*} t^2 - 8t + 1 \amp\;=\; \textcolor{red}{\underbrace{ {\color{black}t^2 - 8t} \textcolor{blue}{+16} }_{\textcolor{red}{\text{factors into } (t-4)^2}} } + 1 \textcolor{blue}{-} \textcolor{blue}{16} \\ \amp\;=\; (t-4)^2 - 15. \end{align*}
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    Hence, the vertex form of \(y = t^2 - 8t + 1\) is
    \begin{equation*} y = (t-4)^2 - 15 \end{equation*}
    and the vertex is \((4, -15)\text{.}\)
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  3. The quadratic function \(g(x) = x^2 + 5x - 3\) is of the form \(x^2 + Bx + C\) with \(B = 5\text{.}\) Now
    \begin{equation*} \textcolor{blue}{\left( \frac{B}{2} \right)^2 } = \left( \frac{5}{2} \right)^2 = \frac{25}{4} \end{equation*}
    Adding and subtracting this quantity from the quadratic yields:
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    By the way...
    \begin{align*} x^2 + 5x - 3 \amp\;=\; \textcolor{red}{\underbrace{ {\color{black}x^2 + 5x} \textcolor{blue}{+\frac{25}{4}} }_{\textcolor{red}{\text{factors into } \left( x + \frac{B}{2} \right)^2=\left(x+\frac{5}{2}\right)^2}} } \hspace{-20pt}- 3 \textcolor{blue}{-} \textcolor{blue}{\frac{25}{4}}\\ \amp\;=\; \left(x + \frac{5}{2}\right)^2 - \frac{3}{1} \textcolor{orange}{\cdot \frac{4}{4}} - \frac{25}{4} \\ \amp\;=\; \left(x + \frac{5}{2}\right)^2 - \frac{12}{4} - \frac{25}{4}\\ \amp\;=\; \left(x + \frac{5}{2}\right)^2 - \frac{37}{4}. \end{align*}
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    Hence, the vertex form of \(g(x) = x^2 + 5x - 3\) is
    \begin{equation*} g(x) = \left(x + \frac{5}{2}\right)^2 - \frac{37}{4} \end{equation*}
    and the vertex is \(\left( -\frac{5}{2}, -\frac{37}{4} \right)\text{.}\)
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By the way...

Example 3.3.8.

Convert each of the following quadratic functions from standard form to vertex form. Then identify the vertex.
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  1. \(\displaystyle y = 3x^2 - 6x + 1\)
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  2. \(\displaystyle f(t) = -2t^2 + t - 4\)
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Solution.
  1. We begin by factoring the \(3\) out of the first two terms to obtain
    \begin{equation*} 3 \left( x^2 - 2x \right) + 1. \end{equation*}
    We then complete the square inside the parentheses; in other words, we will complete the square on \(x^2 - 2x\text{.}\) Here \(B = -2\) so
    \begin{equation*} {\textcolor{blue}{\left( \frac{B}{2} \right)^2}} = \left( \frac{-2}{2} \right)^2 = (-1)^2 = 1 \end{equation*}
    and
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    \begin{align*} \textcolor{orange}{x^2 - 2x} \amp\;=\; \textcolor{red}{\underbrace{ \textcolor{black}{x^2 - 2x} \textcolor{blue}{+1} }_{\textcolor{red}{\text{factors into } (x-1)^2}} } \textcolor{blue}{-} \textcolor{blue}{1} \\ \amp\;=\; \textcolor{orange}{(x-1)^2 - 1}. \end{align*}
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    Now
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    \begin{align*} 3 \left( \textcolor{orange}{x^2 - 2x} \right) + 1 \amp\;=\; 3 \left( \textcolor{orange}{(x-1)^2 - 1} \right) + 1\\ \amp\;=\; 3 (x-1)^2 - 3 + 1 \\ \amp\;=\; 3 (x-1)^2 - 2. \end{align*}
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    Hence, the vertex form of \(y = 3x^2 - 6x + 1\) is
    \begin{equation*} y = 3(x-1)^2 - 2 \end{equation*}
    and the vertex is \(\left( 1, -2 \right)\text{.}\)
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  2. We begin by factoring the \(-2\) out of the first two terms to obtain
    \begin{equation*} -2 \left( t^2 - \frac{1}{2} t \right) - 4. \end{equation*}
    We then complete the square inside the parentheses; in other words, we will complete the square on \(t^2 - \frac{1}{2} t\text{.}\) Here \(B = -\frac{1}{2}\) so
    \begin{equation*} {\textcolor{blue}{\left( \frac{B}{2} \right)^2}} = \left( -\frac{1/2}{2} \right)^2 = \left( -\frac{1}{4} \right)^2 = \frac{1}{16} \end{equation*}
    and
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    \begin{align*} \textcolor{orange}{t^2 - \frac{1}{2} t} \amp\;=\; \textcolor{red}{\underbrace{ \textcolor{black}{t^2 - \frac{1}{2} t} \textcolor{blue}{+\frac{1}{16}} }_{\textcolor{red}{\text{factors into } \left(t+\frac{B}{2}\right)^2=\left( t - \frac{1}{4} \right)^2}} } \hspace{-20pt} \textcolor{blue}{-} \textcolor{blue}{\frac{1}{16}}\\ \amp\;=\; \textcolor{orange}{\left( t - \frac{1}{4} \right)^2 - \frac{1}{16}}. \end{align*}
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    Now
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    \begin{align*} -2 \left( \textcolor{orange}{t^2 - \frac{1}{2} t} \right) - 4 \amp\;=\; -2 \left( \textcolor{orange}{\left( t - \frac{1}{4} \right)^2 - \frac{1}{16}} \right) - 4\\ \amp\;=\; -2 \left( t - \frac{1}{4} \right)^2 + \frac{2}{16} - 4 \\ \amp\;=\; -2 \left( t - \frac{1}{4} \right)^2 + \frac{1}{8} - \frac{4}{1} \textcolor{red}{\cdot \frac{8}{8}} \\ \amp\;=\; -2 \left( t - \frac{1}{4} \right)^2 + \frac{1}{8} - \frac{32}{8} \\ \amp\;=\; -2 \left( t - \frac{1}{4} \right)^2 - \frac{31}{8}. \end{align*}
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    Hence, the vertex form of \(f(t) = -2t^2 + t - 4\) is
    \begin{equation*} f(t) = -2 \left( t - \frac{1}{4} \right)^2 - \frac{31}{8} \end{equation*}
    and the vertex is \(\left( \frac{1}{4}, -\frac{31}{8} \right)\text{.}\)
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Subsection Using Vertex Form to Find Horizontal Intercepts

If a quadratic function is given in vertex form, then its horizontal intercepts can be found by following the process illustrated in the example below.
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Example 3.3.9.

Find the horizontal intercepts of each of the following functions.
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  1. \(\displaystyle f(x)=x^2-9\)
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  2. \(\displaystyle g(x)=(x+1)^2-10\)
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  3. \(\displaystyle y(t)=4(t-1)^2-64\)
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  4. \(\displaystyle h(x)=(x-5)^2+8\)
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Solution.
  1. To find the horizontal or \(x\)-intercept(s) of this quadratic function, we set \(f(x)\) equal to \(0\) and solve for \(x\text{.}\) We could treat the left-hand side as a difference of squares and proceed by factoring, but we also could take an alternate approach:
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    \begin{align*} x^2 - 9 \amp \; = \; 0\\ x^2 - 9 \textcolor{blue}{+9} \amp\;=\; 0 \textcolor{blue}{+ 9}\\ x^2 \amp\;=\; 9. \end{align*}
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    This equation can be solved by taking the square root of both sides, remembering that there will be both a positive solution and a negative solution since both \((\sqrt{9})^2 = 9\) and \((-\sqrt{9})^2 = 9\text{:}\)
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    \begin{align*} x^2 \amp\;=\; 9\\ x \amp\;=\; \pm \sqrt{9}\\ x \amp\;=\; \pm 3. \end{align*}
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    Hence, the horizontal intercepts of the function \(f(x) = x^2 - 9\) and the solutions of the equation \(x^2 - 9 = 0\) are \(x = -3\) and \(x = 3\text{.}\)
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  2. To find the horizontal intercepts, we set \(g(x)\) equal to \(0\) and solve for \(x\text{:}\)
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    \begin{align*} (x+1)^2 - 10 \amp\;=\; 0\\ (x+1)^2 - 10 \textcolor{blue}{+ 10} \amp\;=\; 0 \textcolor{blue}{+ 10}\\ (x+1)^2 \amp\;=\; 10\\ x+1 \amp\;=\; \pm \sqrt{10}\\ x+1 \textcolor{blue}{- 1} \amp\;=\; \pm \sqrt{10} \textcolor{blue}{- 1}\\ x \amp\;=\; -1 \pm \sqrt{10}. \end{align*}
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    We obtain two solutions to our equation and two horizontal intercepts of \(g(x)\text{:}\) \(x = -1 - \sqrt{10}\) and \(x = -1 + \sqrt{10}\text{.}\)
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  3. To find the horizontal intercepts, we set \(y(t)\) to \(0\) and solve for \(t\text{:}\)
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    \begin{align*} 4(t-1)^2 - 64 \amp\;=\; 0\\ 4(t-1)^2 - 64 \textcolor{blue}{+64} \amp\;=\; 0 \textcolor{blue}{+64}\\ 4(t-1)^2 \amp\;=\; 64\\ \dfrac{\cancel{4}(t-1)^2}{\cancel{4}} \amp\;=\; \dfrac{64}{4}\\ (t-1)^2 \amp\;=\; 16\\ t-1 \amp\;=\; \pm \sqrt{16}\\ t-1 \amp\;=\; \pm 4\\ t-1 \textcolor{blue}{+ 1} \amp\;=\; \pm 4 \textcolor{blue}{+ 1}\\ t \amp\;=\; 1 \pm 4. \end{align*}
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    Hence, the two solutions and therefore the two horizontal intercepts are \(t = 1 + 4 = 5\) and \(t = 1 - 4 = -3\text{.}\)
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  4. Once more, we set the function \(h(x)\) equal to \(0\) and solve for \(x\text{:}\)
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    \begin{align*} (x-5)^2 + 8 \amp\;=\; 0\\ (x-5)^2 + 8 \textcolor{blue}{- 8} \amp\;=\; 0 \textcolor{blue}{- 8}\\ (x-5)^2 \amp\;=\; -8. \end{align*}
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    In this case there is no real solution, since \(\sqrt{-8}\) is not a real number. This means that the equation has no real solutions and the function \(h(x) = (x-5)^2 + 8\) has no horizontal intercepts. If we take a moment to visualize the graph of the function, this makes sense. Its graph is an upward facing parabola with vertex \((5, 8)\text{.}\) Since the lowest point (the vertex) of this parabola is above the \(x\)-axis, there cannot be any horizontal intercepts.
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Exercises Exercises

Vertex.

For each of the following, identify the vertex of the given quadratic function.
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4.
\(h(x)=-(x+\frac{1}{2})^2+\frac{5}{3}\)
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Solution.
\(\left(-\frac{1}{2},\frac{5}{3}\right)\)
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Vertex Form to Standard Form.

Convert each quadratic function from vertex form to standard form and identify \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)
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5.
\(f(x)=(x+1)^2+2\)
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Solution.
\(f(x)=x^2+2x+3\text{;}\) \(a=1\text{,}\) \(b=2\text{,}\) \(c=3\)
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6.
\(g(x)=5(x-2)^2+6\)
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Solution.
\(g(x)=5x^2-20x+26\text{;}\) \(a=5\text{,}\) \(b=-20\text{,}\) \(c=26\)
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7.
\(y=-2(t-4)^2+5\)
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Solution.
\(y=-2t^2+16t-27\text{;}\) \(a=-2\text{,}\) \(b=16\text{,}\) \(c=-27\)
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8.
\(h(x)=-(x+\frac{1}{2})^2+\frac{5}{3}\)
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Solution.
\(h(x)=-x^2-x+\frac{17}{12}\text{;}\) \(a=-1\text{,}\) \(b=-1\text{,}\) \(c=\frac{17}{12}\)
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Standard Form to Vertex Form.

Convert each quadratic function from standard form to vertex form by completing the square. Then identify the vertex.
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9.
\(f(x)=x^2+2 x-5\)
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Solution.
\(f(x)=(x+1)^2-6\text{;}\) vertex: \((-1,-6)\)
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10.
\(g(x)=x^2-6 x+10\)
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Solution.
\(g(x)=\left(x-3\right)^2+1\text{;}\) vertex \((3,1)\)
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11.
\(h(x)=x^2+3x+\frac{5}{4}\)
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Solution.
\(h(x)=\left(x+\frac{3}{2}\right)^2-1\text{;}\) vertex \(\left(-\frac{3}{2},-1\right)\)
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12.
\(F(x)=-x^2+10 x+15\)
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Solution.
\(F(x)=-\left(x-5\right)^2+40\text{;}\) vertex \((5,40)\)
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13.
\(G(x)=-3 x^2+6 x-4\)
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Solution.
\(G(x)=-3(x-1)^2-1\text{;}\) vertex \((1,-1)\)
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14.
\(H(x)=-2 x^2-16 x-17\)
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Solution.
\(H(x)=-2\left(x+4\right)^2+15\text{;}\) vertex \((-4,15)\)
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Horizontal Intercepts.

Find the horizontal intercepts of each quadratic function. Give exact and approximate values rounded off to three decimal places.
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16.
\(g(x)=(x-1)^2-2\)
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Solution.
\(1-\sqrt{2}\approx-0.414\text{,}\) \(1+\sqrt{2}\approx2.414\)
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17.
\(h(x)=(x-\frac{1}{2})^2-1\)
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Solution.
\(-0.5\text{,}\) \(1.5\)
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18.
\(F(x)=x^2+4x-6\)
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Solution.
\(-2-\sqrt{10}\approx-5.162\text{,}\) \(-2+\sqrt{10}\approx1.162\)
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20.
\(H(x)=-2 x^2-16 x-27\)
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Solution.
\(-4-\sqrt{\frac{5}{2}}\approx-5.581\text{,}\) \(-4+\sqrt{\frac{5}{2}}\approx-2.419\)
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