Intermediate Algebra: Functions and Graphs

Subsection Introduction

The graph of the quadratic equation \(y = ax^2 + bx + c\) is called a parabola. Some parabolas are shown below.

All these parabolas share certain features.

• The graph has either a highest point (if the parabola opens downward, as in figure (a) or a lowest point (if the parabola opens upward, as in figure (b). This high or low point is called the vertex of the graph.
• The parabola is symmetric about a vertical line, called the axis of symmetry, that runs through the vertex.
• A parabola has a \(y\)-intercept, and it may have zero, one, or two \(x\)-intercepts.
• If there are two \(x\)-intercepts, they are equidistant from the axis of symmetry.

The values of the constants \(a\text{,}\) \(b\text{,}\) and \(c\) determine the location and orientation of the parabola. We'll consider each of these constants separately.

Subsection The Graph of \(y = ax^2\)

Use your calculator to graph the following three equations in the standard window, as shown below:

You can see that the graph of \(y=3x^2\) is narrower than the basic parabola, and the graph of \(y=0.1x^2\) is wider. As \(x\) increases, the graph of \(y=3x^2\) increases faster than the basic parabola, and the graph of \(y=0.1x^2\) increases more slowly. Compare the corresponding \(x\)-values for the three graphs shown in the table.

For each \(x\)-value, the points on the graph of \(y=3x^2\) are higher than the points on the basic parabola, while the points on the graph of \(y=0.1x^2\) are lower. Multiplying by a positive constant greater than 1 stretches the graph vertically, and multiplying by a positive constant less than 1 squashes the graph vertically.

What about negative values for \(a\text{?}\) Consider the graphs of

\begin{align*} y \amp = x^2\\ y \amp = -2x^2\\ y \amp = -0.5x^2 \end{align*}

We see that multiplying \(x^2\) by a negative constant reflects the graph about the \(x\)-axis. These parabolas open downward.

Example 3.41.

Sketch by hand a graph of each quadratic equation.

1. \(\displaystyle y = 2x^2\)
2. \(\displaystyle y = -\dfrac{1}{2}x^2\)

Solution

Both equations have the form \(y = ax^2\text{.}\) The graph of \(y = 2x^2\) opens upward because \(a = 2 \gt 0\text{,}\) and the graph of \(y = -\dfrac{1}{2}x^2\) opens downward because \(a = -\dfrac{1}{2}\lt 0\text{.}\)

To make a reasonable sketch by hand, it is enough to plot a few guidepoints; the points with \(x\)-coordinates \(1\) and \(-1\) are easy to compute.

\(x\)	\(y=2x^2\)	\(y=-\frac{1}{2}x^2\)
\(-1\)	\(2\)	\(-\dfrac{1}{2}\)
\(0\)	\(0\)	\(0\)
\(1\)	\(2\)	\(-\dfrac{1}{2}\)

We sketch parabolas through each set of guidepoints, as shown at left.

Subsection The Graph of \(y= x^2 + c\)

Next, we consider the effect of the constant term, \(c\text{,}\) on the graph. Compare the graphs of

\begin{align*} y \amp = x^2\\ y \amp = x^2+4\\ y \amp = x^2-4 \end{align*}

The graph of \(y=x^2+4\) is shifted upward four units compared to the basic parabola, and the graph of \(y=x^2-4\) is shifted downward four units. Look at the table, which shows the \(y\)-values for the three graphs.

Each point on the graph of \(y=x^2+4\) is four units higher than the corresponding point on the basic parabola, and each point on the graph of \(y=x^2-4\) is four units lower. In particular, the vertex of the graph of \(y=x^2+4\) is the point \((0,4)\text{,}\) and the vertex of the graph of \(y=x^2-4\) is the point \((0,-4)\text{.}\)

Subsection The Graph of \(y = ax^2 + bx\)

How does the linear term, \(bx\text{,}\) affect the graph?

Example 3.47.

Describe the graph of the equation

\begin{equation*} y = 2x^2 + 8x \end{equation*}

Solution

The graph in the standard window is shown below. We see that the axis of symmetry for this parabola is not the \(y\)-axis: the graph is shifted to the left, compared to the basic parabola. We find the \(y\)-intercepts of the graph by setting \(y\) equal to zero:

\begin{align*} 0 \amp= 2x^2 + 8x\\ \amp= 2x(x + 4) \end{align*}

The solutions of this equation are \(0\) and \(-4\text{,}\) so the \(x\)-intercepts are the points \((0, 0)\) and \((-4, 0)\text{.}\)

We can find the vertex of the graph by using the symmetry of the parabola. The \(x\)-coordinate of the vertex lies exactly half-way between the \(x\)-intercepts, so we average their \(x\)-coordinates to find:

To find the \(y\)-coordinate of the vertex, substitute \(x = -2\) into the equation for the parabola:

Thus, the vertex is the point \((-2,-8)\text{.}\)

Subsection A Formula for the Vertex

We can find a formula for the vertex of any parabola of the form

\begin{equation*} y = ax^2 + bx \end{equation*}

First, find the \(x\)-intercepts of the graph by setting \(y\) equal to zero and solving for \(x\text{.}\)

The \(x\)-intercepts are the points \((0, 0)\) and \((\dfrac{-b}{a}, 0)\text{.}\)

Next, we find the \(x\)-coordinate of the vertex by taking the average of the two \(x\)-intercepts.

Now we have a formula for the \(x\)-coordinate of the vertex.

We find the \(y\)-coordinate of the vertex by substituting its \(x\)-coordinate into the equation for the parabola.

Example 3.50.

1. Find the vertex of the graph of \(f(x) = -1.8x^2 - 16.2x\text{.}\)
2. Find the \(x\)-intercepts of the graph.

Solution

1. The \(x\)-coordinate of the vertex is
   \begin{equation*} x_v=\frac{-b}{2a}=\frac{-(-16.2)}{2(-1.8)}=-4.5 \end{equation*}
   To find the \(y\)-coordinate of the vertex, evaluate \(f(x)\) at \(x = \alert{-4.5}\text{.}\)
   \begin{equation*} y_v=-1.8(\alert{-4.5})^2 - 16.2(\alert{-4.5}) = 36.45 \end{equation*}
   The vertex is \((-4.5, 36.45)\text{.}\)
2. To find the \(x\)-intercepts of the graph, set \(y = 0\) and solve.
   \begin{align*} -1.8x^2 - 16.2x \amp= 0\amp\amp \blert{\text{Factor.}}\\ -x(1.8x + 16.2) \amp= 0\amp\amp \blert{\text{Set each factor equal to zero.}}\\ -x =0 \hphantom{blank} 1.8x + 16.2 \amp = 0\amp\amp \blert{\text{Solve each equation.}}\\ x = 0 \hphantom{blankblankblan} x \amp= -9 \end{align*}
   The \(x\)-intercepts of the graph are \((0, 0)\) and \((-9, 0)\text{.}\) The graph is shown below.