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Section 4.2 The Vertex

Subsection 1. 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 4.22.

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 4.23.

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 2. Expand a product of binomials

With practice, you will be able to multiply binomials mentally.

Subsubsection Examples

Example 4.26.

Multiply \(~(x-6)(x+3)\text{.}\)

Solution

We apply the distributive law to multiply each term of the first factor by each term of the second factor. (The "FOIL" method.)

\begin{align*} (x-6)(x+3) \amp = x \cdot x + 3 \cdot x - 6 \cdot x - 6 \cdot 3\\ \amp = x^2+3x-6x-18 \amp \amp \blert{\text{Combine like terms.}}\\ \amp = x^2-3x-18 \end{align*}
Example 4.27.

We multiply \(~(2t^3-5)(3t^3-2)\text{.}\)

Solution

Multiply each term of the first factor by each term of the second factor.

\begin{align*} (2t^3-5)(3t^3-2) \amp = 6t^6-4t^3-15t^3+10 \amp \amp \blert{\text{Combine like terms.}}\\ \amp = 6t^6-19t^3+10 \end{align*}

Subsubsection Exercises

Multiply \(~(4a-7)(3a+8)\text{.}\)

Answer
\(12a^2+11a-56\)

Multiply \(~(1-3b^2)(3-4b^2)\text{.}\)

Answer
\(12b^4-13b^2+3\)

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 4.30.

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 4.31.

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
\(h=30\)