Skip to main content

Section 6.4 Problem Solving

Subsection 1. Complete the square

To write a quadratic equation in vertex form, we need to complete the square.

Subsubsection Examples

Example 6.52.

Write the equation \(~y=x^2+8x+10~\) in vertex form.

Solution

Complete the square on the variable terms.

\begin{align*} y \amp =(x^2+8x+ \underline{\hspace{1.818181818181818em}})+10 \amp \amp ~\blert{2p=8,~ \text{so}~p^2=4^2=16.}\\ y \amp =(x^2+8x+ \alert{16})+10 - \alert{16} \amp \amp ~\blert{\text{Add and subtract 16.}}\\ y \amp =(x+4)^2-6 \amp \amp \begin{array}{l} \blert{\text{Write } x^2+8x+16 \text{ as the}}\\ \blert{\text{square of a binomial.}} \end{array} \end{align*}
Example 6.53.

Write the equation \(~y=2x^2-12x+10~\) in vertex form.

Solution

First, factor 2 from the variable terms.

\begin{equation*} y=2(x^2-6x)+10 \end{equation*}

Next, complete the square inside parentheses.

\begin{align*} y \amp =2(x^2-6x+ \underline{\hspace{1.818181818181818em}})+10 \amp \amp \blert{2p=-6,~ \text{so}~p^2=(-3)^2=9.}\\ y \amp =2(x^2-6x+ \alert{9})+10 - 2(\alert{9}) \amp \amp \blert{\text{Add and subtract}~2(9).}\\ y \amp =2(x-3)^2-8 \amp \amp \begin{array}{l} \blert{\text{Write } x^2-6x+9 \text{ as the}}\\ \blert{\text{square of a binomial.}} \end{array} \end{align*}

Subsubsection Exercises

Write the equation \(~y=x^2-12x+24~\) in vertex form.

Answer
\(y=(x-6)^2-12\)

Write the equation \(~y=3x^2+12x+4~\) in vertex form.

Answer
\(y=3(x+2)^2-8\)

Subsection 2. Graph parabolas in vertex form

We can sketch the graph of a parabola with the vertex, the \(y\)-intercept, and its symmetric point.

Subsubsection Examples

Example 6.56.

Graph the equation \(~y=\dfrac{-1}{2}(x+3)^2-2\)

Solution

The vertex is the point \((-3,2)\text{.}\) We can find the \(y\)-intercept by setting \(x=0\text{.}\)

\begin{equation*} y = \dfrac{-1}{2}(\alert{0}+3)^2-2 = \dfrac{-9}{2}-2 = -6\dfrac{1}{2} \end{equation*}

The \(y\)-intercept is the point \(\left(0, -6\dfrac{1}{2}\right)\text{.}\) The axis of symmetry is the vertical line \(x=-3\text{,}\) and there is a symmetric point equidistant from the axis, namely \(\left(-6, -6\dfrac{1}{2}\right)\text{.}\) We plot these three points and sketch the parabola through them.

parabola
Example 6.57.

Graph the equation \(~y=2(x-15)^2-72\)

Solution

The vertex is \((15,-72)\text{.}\) We find the \(y\)-intercept by setting \(x=0\text{:}\)

\begin{equation*} y=2(\alert{0}-15)^2-72=378 \end{equation*}

The \(y\)-intercept is \((0, 378)\text{.}\) We find the \(x\)-intercepts by setting \(y=0\text{:}\)

\begin{align*} 2(x-15)^2-72 \amp = \alert{0}\\ (x-15)^2 \amp = 36\\ x \amp = \pm 6+15 \end{align*}

The \(x\)-intercepts are \((9,0)\) and \((21,0)\text{.}\) We plot these three points and sketch the parabola through them.

parabola

Subsubsection Exercises

Find the vertex, the \(y\)-intercept, and the \(x\)-intercepts of \(~y=(x-2)^2-5\text{,}\) and sketch its graph.

grid
Answer
parabola

Find the vertex, the \(y\)-intercept, and the \(x\)-intercepts of \(~y=-2(x+1)^2+3\text{,}\) and sketch its graph.

grid
Answer
parabola

Subsection 3. Solve an equation for a parameter

We can find the equation for a parabola if we know, for example, the vertex and one other point.

Subsubsection Examples

Example 6.60.

The point \((6,2)\) lies on the graph of \(~y=a(x-4)^2+1.~\) Solve for \(a\text{.}\)

Solution

Substitute 6 for \(x\) and 2 for \(y\text{,}\) then solve for \(a\text{.}\)

\begin{align*} \alert{2} \amp = a(\alert{6}-4)^2+1\\ 2 \amp = a(4) + 1\\ 1 \amp = 4a \end{align*}

The solution is \(a = \dfrac{1}{4}\text{.}\)

Example 6.61.

The point \((-2,11)\) lies on the graph of \(~y=x^2+bx-3.~\) Solve for \(b\text{.}\)

Solution

Substitute -2 for \(x\) and 11 for \(y\text{,}\) then solve for \(b\text{.}\)

\begin{align*} (\alert{-2})^2 +b(\alert{-2}) -3 \amp = 11\\ 4-2b-3 \amp = 11\\ -2b \amp = 10 \end{align*}

The solution is \(b = -5\text{.}\)

Subsubsection Exercises

The point \((-6,10)\) lies on the graph of \(~y=a(x+3)^2-2.~\) Solve for \(a\text{.}\)

Answer
\(a=\dfrac{4}{3}\)

The point \((-3,8)\) lies on the graph of \(~y=-x^2+bx+5.~\) Solve for \(b\text{.}\)

Answer
\(b=-4\)

The point \((8,-12)\) lies on the graph of \(~y=ax^2-4x+36.~\) Solve for \(a\text{.}\)

Answer
\(a=\dfrac{-1}{3}\)

The point \((60,-480)\) lies on the graph of \(~y=\dfrac{-2}{3}(x-h)^2+120.~\) Solve for \(h\text{.}\)

Answer
\(h=30\)