ORCCAOpen Resources for Community College Algebra

Subsection9.2.1Introduction

In this section we will learn about quadratic functions and how to identify their key features on a graph. We will identify their direction, vertex, axis of symmetry and intercepts. We will also see how to graph a parabola by finding the vertex and making a table of function values. We will look at applications that involve the vertex of a quadratic function.

Notice that a quadratic function has a squared term that linear functions do not have. If \(a=0,\) the function is linear. To understand the shape and features of a quadratic function, let's look at an example.

Subsection9.2.2Properties of Quadratic Functions

A toy rocket was fired from the ground and then flew into the air at a speed of 64 feet per second. The path of the rocket can be modeled by the function \(f\) where \(f(t)=-16t^2+64t\text{.}\) To see the shape of the function we will make a table of values and plot the points. For the table we we will choose some values for \(t\) and then evaluate the function at each \(t\)-value:

\(t\)	\(f(t)=-16t^2+64t\)	Point
\(0\)	\(f(0)=-16(0)^2+64(0)=0\)	\((0,0)\)
\(1\)	\(f(1)=-16(1)^2+64(1)=48\)	\((1,48)\)
\(2\)	\(f(2)=-16(2)^2+64(2)=64\)	\((2,64)\)
\(3\)	\(f(3)=-16(3)^2+64(3)=48\)	\((3,48)\)
\(4\)	\(f(4)=-16(4)^2+64(4)=0\)	\((4,0)\)

Now that we have Table 9.2.3 and Figure 9.2.4, we can see the features of this parabola. Notice the symmetry in the shape of the graph and the \(y\)-values in the table. Consecutive \(y\)-values do not increase by a constant amount in the way that linear functions do.

The first feature that we will talk about is the direction that a parabola opens. All parabolas open either upward or downward. This parabola in the rocket example opens downward because \(a\) is negative. Here are some more quadratic functions graphed so we can see which way they open.

The vertex is the highest or lowest point on the graph. In Figure 9.2.4, the vertex is \((2,64)\text{.}\) This tells us that the rocket reached its maximum height of \(64\) feet after \(2\) seconds. If the parabola opens downward, as in the rocket example, then the \(y\)-value of the vertex is the maximum \(y\)-value. If the parabola opens upward then the \(y\)-value of the vertex is the minimum \(y\)-value.

The axis of symmetry is a vertical line that passes through the vertex, dividing it in half. The vertex is the only point that does not have a symmetric point. We write the axis of symmetry as an equation of a vertical line so it always starts with "\(x=\text{.}\)" In Figure 9.2.4, the equation for the axis of symmetry is \(x=2\text{.}\)

The vertical intercept is the point where the parabola crosses the vertical axis. The vertical intercept is the \(y\)-intercept if the axes are labeled \(x\) and \(y\text{.}\) In Figure 9.2.4, the point \((0,0)\) is the starting point of the rocket. The \(y\)-value of \(0\) means the rocket started on the ground.

The horizontal intercept(s) are the points where the parabola crosses the horizontal axis. They are the \(x\)-intercepts if the axes are labeled \(x\) and \(y\text{.}\) The point \((0,0)\) on the path of the rocket is also a horizontal intercept. The \(t\)-value of \(0\) indicates the time when the rocket was launched from the ground. There is another horizontal intercept at the point \((4,0)\text{,}\) which means the rocket hit the ground after \(4\) seconds.

It is possible for a quadratic function to have \(0\text{,}\) \(1\) or \(2\) horizontal intercepts. The figures below show an example of each.

Here is a summary of the properties of quadratic functions:

Example9.2.14

Identify the key features of the quadratic function \(y=x^2-2x-8\) shown in Figure 9.2.15.

Solution

First, we see that this parabola opens upward because the leading coefficient is positive.

Then we locate the vertex which is the point \((1,-9)\text{.}\) The axis of symmetry is the vertical line \(x=1\text{.}\)

The vertical intercept or \(y\)-intercept is the point \((0,-8)\text{.}\)

The horizontal intercepts are the points \((-2,0)\) and \((4,0)\text{.}\)

Subsection9.2.3Finding the Vertex and Axis of Symmetry Algebraically

The coordinates of the vertex are not easy to identify on a graph if they are not integers. Another way to find it is by using a formula.

To understand this formula we can look at the quadratic formula. The vertex is on the axis of symmetry, so it will always occur between the the two horizontal intercepts. Looking at the quadratic formula we can see that this value is in the middle of the two values obtained from the quadratic formula:

Subsection9.2.4Graphing Quadratic Functions by Making a Table

When we learned how to graph lines, we could choose any \(x\)-values. For quadratic functions, though, we want to find the vertex and choose our \(x\)-values around it. Then we can use the property of symmetry to help us. Let's look at an example.

Example9.2.20

Determine the vertex and axis of symmetry for the quadratic function \(y=-x^2-2x+3\text{.}\) Then make a table of values and sketch the graph of the function.

To determine the vertex of \(y=-x^2-2x+3\text{,}\) we want to find the \(x\)-value of the vertex first. We will use \(x=-\frac{b}{2a}\) for \(a=-1\) and \(b=-2\text{:}\)

\begin{align*} x\amp=-\frac{(\substitute{-2})}{2(\substitute{-1})}\\ x\amp=\frac{2}{-2}\\ x\amp=-1 \end{align*}

Now we know that our axis of symmetry is the line \(x=-1\) and the axis of symmetry is the line \(x=-1\text{.}\) We will set up our table with two values on each side of \(x=-1\text{.}\) We choose \(x=-3, -2, -1, 0\) and \(1\) as shown in Table 9.2.21.

Next, we'll determine the \(y\)-coordinates by replacing \(x\) with each value and we have the complete table as shown in Table 9.2.22. Notice that each pair of \(y\)-values on either side of the vertex match. This helps us to check that our vertex and \(y\)-values are correct.

\(x\)	\(y=-x^2-2x+3\)	Point
\(-3\)
\(-2\)
\(-1\)
\(0\)
\(1\)

\(x\)	\(y=-x^2-2x+3\)	Point
\(-3\)	\(y=-(\substitute{-3})^2-2(\substitute{-3})+3=0\)	\((-3,0)\)
\(-2\)	\(y=-(\substitute{-2})^2-2(\substitute{-2})+3=3\)	\((-2,3)\)
\(-1\)	\(y=-(\substitute{-1})^2-2(\substitute{-1})+3=4\)	\((-1,4)\)
\(0\)	\(y=-(\substitute{0})^2-2(\substitute{0})+3=3\)	\((0,3)\)
\(1\)	\(y=-(\substitute{1})^2-2(\substitute{1})+3=0\)	\((1,0)\)

Now that we have our table, we will plot the points and draw in the axis of symmetry as shown in Figure 9.2.23. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.2.24

The method we used works best when the \(x\)-value of the vertex is an integer. We can still make a graph if that is not the case as we will demonstrate in the next example.

Example9.2.25

Determine the vertex and axis of symmetry for the quadratic function \(y=2x^2-3x-4\text{.}\) Use this to create a table of values and sketch the graph of this function.

Solution

To determine the vertex of \(y=2x^2-3x-4\text{,}\) we'll find \(x=-\frac{b}{2a}\) for \(a=2\) and \(b=-3\text{:}\)

\begin{align*} x\amp=-\frac{(\substitute{-3})}{2(\substitute{2})}\\ x\amp=\frac{3}{4} \end{align*}

Next, we'll determine the \(y\)-coordinate by replacing \(x\) with \(\frac{3}{4}\) in \(y=2x^2-3x-4\text{:}\)

\begin{align*} y\amp=2\left(\substitute{\frac{3}{4}}\right)^2-3\left(\substitute{\frac{3}{4}}\right)-4\\ y\amp=2\left(\frac{9}{16}\right)-\frac{9}{4}-4\\ y\amp=\frac{9}{8}-\frac{18}{8}-\frac{32}{8}\\ y\amp=-\frac{41}{8} \end{align*}

Thus the vertex occurs at \(\left(\frac{3}{4},-\frac{41}{8}\right)\text{,}\) or at \((0.75,-5.125)\text{.}\) The axis of symmetry is then the line \(x=\frac{3}{4}\text{,}\) or \(x=0.75\text{.}\)

Now that we know the \(x\)-value of the vertex, we will create a table. We will choose \(x\)-values on both sides of \(x=0.75\text{,}\) but we will choose integers because it will be easier to find the function values.

\(x\)	\(y=2x^2-3x-4\)	Point
\(-1\)	\(1\)	\((-1,1)\)
\(0\)	\(-4\)	\((0,-4)\)
\(0.75\)	\(-5.125\)	\((0.75,-5.125)\)
\(1\)	\(-5\)	\((1,-5)\)
\(2\)	\(-2\)	\((2,-2)\)

The points graphed in Figure 9.2.27 don't have the symmetry we'd expect from a parabola. This is because the vertex occurs at an \(x\)-value that is not an integer, and all of the chosen values in the table are integers. We can use the axis of symmetry to determine more points on the graph (as shown in Figure 9.2.28), which will give it the symmetry we expect. From there, we can complete the sketch of this graph.

Subsection9.2.5The Domain and Range of Quadratic Functions

In Example 9.1.31, we found the domain and range of different types of functions using their graphs. Now that we have graphed some quadratic functions, let's practice identifying the domain and range.

Since all parabolas have the same shape, they all have the same domain of \(\{x\mid x\ \textrm{is a real number}\}\) or \((-\infty,\infty)\text{.}\) The range depends on which way the parabola opens and the \(y\)-coordinate of the vertex. When we look at application problems, however, the domain and range will depend on the values that make sense in the given context. For example, times and lengths do not usually have negative values. We will revisit this after looking at some applications.

Subsection9.2.6Applications of Quadratic Functions Involving the Vertex.

We looked at the height of a toy rocket with respect to time at the beginning of this section and saw that it reached a maximum height of \(64\) feet after \(2\) seconds. Let's look at some more applications that involve finding the minimum or maximum value of a quadratic function.

Example9.2.32

Imagine that Yang got a new air rifle to shoot targets. The first thing he did with it was to sight the scope at a certain distance so the pellets consistently hit where the cross hairs are pointed. In Olympic \(10\)-meter air rifle shooting, the bulls-eye is a \(0.5\)mm dot, about the size of the head of a pin, so accuracy is key.¹Visit en.wikipedia.org/wiki/ISSF_10_meter_air_rifle for more.)

Yang would like to set up his air rifle scope to be accurate at a level distance of \(35\) yards (from the muzzle, which is the tip of the barrel), but he also needs to know how much to correct for gravity at different distances. Since the projectile will be affected by gravity, knowing the distance that the target will be set up is essential to be accurate. After zeroing his scope reticule (cross-hairs) at \(35\) yards so that he can consistently hit the bulls-eye with the reticule directly over it, he sets up targets at various distances to test his gun. He then shoots at the targets with the cross-hairs directly on the bulls-eye and measured the distance that the pellet hit above or below the bulls-eye when shot at those distances.

Distance to Target in Yards	Above/Below Bulls-eye	Distance Above/Below in Inches
\(5\) yds	Below	\(0.1\) in
\(10\) yds	Above	\(0.6\) in
\(20\) yds	Above	\(1.1\) in
\(30\) yds	Above	\(0.6\) in
\(35\) yds	On Bulls-eye	\(0\) in
\(40\) yds	Below	\(0.8\) in
\(50\) yds	Below	\(3.2\) in

Make a graph of the height above the bulls-eye that the air rifle shoots at the distances collected in the table and find the vertex. What does the vertex mean in this context?

(Note that values measured below the bulls-eye should be graphed as negative \(y\)-values. Keep in mind that the units on the axes are different: along the \(x\)-axis, the units are yards, whereas on the \(y\)-axis, the units are inches.)

Solution

Since the input values seem to be going up by \(5\)s or \(10\)s, we will scale the \(x\)-axis by \(5\)s. The \(y\)-axis needs to be scaled by \(1\)s.

From the graph we can see that the point \((20,1.1)\) is our best guess for the real life vertex. This means the highest above the cross-hairs he hit was \(1.1\) inches when the target was \(20\) yards away.

Returning to the domain and range, we will look at the path of the toy rocket in Graph 9.2.4. Looking horizontally, the \(t\)-values make sense from \(0\) seconds, when the rocket is fired, until \(4\) seconds, when it comes back to the ground. This give us a domain of \(\{t\mid 0 \le t \le 4\}\) or \([0,4]\text{.}\) For the range, the height of the rocket goes from \(0\) feet on the ground and reaches a maximum height of \(64\) feet. The range is \(\{f(t)\mid 0 \le f(t) \le 64\}\) or \([0,64]\text{.}\)

In the air-rifle application in Example 9.2.32, the \(x\)-values are connected from \(5\) to \(50\) yards. If we assume that Yang will never be competing in target shoots beyond \(50\) yards, the domain will be \([5,50]\text{.}\) The \(y\)-values go from \(-3.2\) to \(1.1\) inches so the range is \([-3.2,1.1]\text{.}\)

In order to find the domain and range for many applications we need to know how to find the vertical and horizontal intercepts. We will look at that in the next section.