ORCCA Properties of Quadratic Functions

Section 9.6 Properties of Quadratic Functions

Subsection 9.6.1 Introduction

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.

Definition 9.6.1.

A quadratic function has the form $f(x)=ax^2+bx+c$ where $a, b\text{,}$ and $c$ are real numbers, and $a \neq 0\text{.}$ The graph of a quadratic function has the shape of a parabola.

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

Subsection 9.6.2 Properties of Quadratic Functions

Hannah fired a toy rocket from the ground, which 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:

Table 9.6.2. Function values and points for $f(t)=-16t^2+64t$

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

a coordinate plane with a u-shaped graph that opens downward. The parabola starts at (0,0) and goes up through the point (1,48) until it reaches (2,64) when it starts going down through (3,48) and stops at (4,0) — Figure 9.6.3. Graph of $f(t)=-16t^2+64t$

Now that we have Table 9.6.2 and Figure 9.6.3, 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.

a u-shaped graph opening upward — Figure 9.6.4. The graph of $y=x^2-2x+2$ opens upward

a wide u-shaped graph that opens downward — Figure 9.6.5. The graph of $y=-\frac{1}{4}x^2-\frac{1}{2}x-\frac{1}{4}$ opens downward

a narrow u-shaped graph that opens upward — Figure 9.6.6. The graph of $y=3x^2-18x+23.5$ opens upward

Fact 9.6.7.

We only need to look at the sign of the leading coefficient to determine which way the graph opens. If the leading coefficient is positive, the parabola opens upward. If the leading coefficient is negative, the parabola opens downward.

Checkpoint 9.6.8.

The vertex is the highest or lowest point on the graph. In Figure 9.6.3, the vertex is $(2,64)\text{.}$ This tells us that Hannah's 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 9.6.3, 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 9.6.3, 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 $0\text{,}$ $1\text{,}$ or $2$ horizontal intercepts. The figures below show an example of each.

a u-shaped graph that opens upward and is above the x-axis; there are no horizontal intercepts — Figure 9.6.9. The graph of $y=x^2-2x+2$ has no horizontal intercepts

a wide u-shaped graph that opens downward and has its vertex on the x-axis so there is one x-intercept — Figure 9.6.10. The graph of $y=-\frac{1}{4}x^2-\frac{1}{2}x-\frac{1}{4}$ has one horizontal intercept

a narrow u-shaped graph that has its vertex below the x-axis and opens upward so there are two x-intercepts — Figure 9.6.11. The graph of $y=3x^2-18x+23.5$ has two horizontal intercepts

Here is a summary of the properties of quadratic functions:

List 9.6.12. Summary of Properties of Quadratic Functions

Direction: A parabola opens upward if $a$ is positive and opens downward of $a$ is negative.
Vertex: The vertex of a parabola is the maximum or minimum point on the graph.
Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex.
Vertical Intercept: The vertical intercept is the point where the function intersects the vertical axis. There is exactly one vertical intercept.
Horizontal Intercept(s): The horizontal intercept(s) are the points where a function intersects the horizontal axis. The graph of a parabola can have $0, 1\text{,}$ or $2$ horizontal intercepts.

Example 9.6.13.

Identify the key features of the quadratic function $y=x^2-2x-8$ shown in Figure 9.6.14.

Explanation

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 $(4,0)\text{.}$

the graph of a parabola that opens upward, has a vertex at (-1,9), a y-intercept at (0,-8) and x-intercepts at (-2,0) and (4,0) — Figure 9.6.14. Graph of $y=x^2-2x-8$

Checkpoint 9.6.15.

Subsection 9.6.3 Finding 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.

Fact 9.6.16.

If we denote $(h,k)$ as the coordinates of the vertex of a quadratic function $f(x)=ax^2+bx+c\text{,}$ then $h=-\frac{b}{2a}\text{.}$

To understand why, we can look at the quadratic formula. The vertex is on the axis of symmetry, so it will always occur halfway between the two $x$-intercepts (if there are any). The quadratic formula shows that the $x$-intercepts happen at $-\frac{b}{2a}$ minus some number and at $-\frac{b}{2a}$ plus that same number:

\begin{equation*} x=\frac{\highlight{-b}\lowlight{{}\pm \sqrt{b^2-4ac}}}{{}\highlight{2a}{}} \end{equation*}

Example 9.6.17.

Determine the vertex and axis of symmetry of the quadratic function $f(x)=x^2-4x-12\text{.}$

We will find the $x$-value of the vertex using the formula $h=-\frac{b}{2a}\text{,}$ for $a=1$ and $b=-4\text{.}$

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{(\substitute{-4})}{2(\substitute{1})}\\ h\amp=2 \end{align*}

Now we know the $x$-value of the vertex is $2\text{,}$ so we may evaluate $f(2)$ to determine the $y$-value of the vertex:

\begin{align*} k\amp=f(\substitute{2})=(\substitute{2})^2-4(\substitute{2})-12\\ k=\amp=4-8-12\\ k=\amp=-16 \end{align*}

The vertex is the point $(2,-16)$ and the axis of symmetry is the line $x=2\text{.}$

Example 9.6.18.

Determine the vertex and axis of symmetry of the quadratic function $y=-3x^2-3x+7\text{.}$

Explanation

Using the formula $h=-\frac{b}{2a}$ with $a=-3$ and $b=-3\text{,}$ we have :

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{(\substitute{-3})}{2(\substitute{-3})}\\ h\amp=-\frac{1}{2} \end{align*}

Now that we've determined $h=-\frac{1}{2}\text{,}$ we can substitute it for $x$ to find the $y$-value of the vertex:

\begin{align*} y\amp=-3x^2-3x+7\\ y\amp=-3\left(\substitute{-\frac{1}{2}}\right)^2-3\left(\substitute{-\frac{1}{2}}\right)+7\\ y\amp=-3\left(\frac{1}{4}\right)+\frac{3}{2}+7\\ y\amp=-\frac{3}{4}+\frac{3}{2}+7\\ y\amp=-\frac{3}{4}+\frac{6}{4}+\frac{28}{4}\\ y\amp=\frac{31}{4} \end{align*}

The vertex is the point $\left(-\frac{1}{2},\frac{31}{4}\right)$ and the axis of symmetry is the line $x=-\frac{1}{2}\text{.}$

Subsection 9.6.4 Graphing 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.

Example 9.6.19.

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.

Explanation

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

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

To find the $y$-coordinate of the vertex, we substitute $x=-1$ into the equation for our parabola.

\begin{align*} y\amp=-\substitute{x}^2-2\substitute{x}+3\\ y\amp=-(\substitute{-1})^2-2(\substitute{-1})+3\\ \amp=-1+2+3\\ \amp=4 \end{align*}

Now we know that our axis of symmetry is the line $x=-1$ and the vertex is the point $(-1,4)\text{.}$ We will set up our table with two values on each side of $x=-1\text{.}$ We choose $x=-3, -2, -1, 0\text{,}$ and $1$ as shown in Table 9.6.20.

Next, we'll determine the $y$-coordinates by replacing $x$ with each value and we have the complete table as shown in Table 9.6.21. 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.

Table 9.6.20. Setting up the table for $y=-x^2-2x+3$

$x$	$y=-x^2-2x+3$	Point
$-3$
$-2$
$-1$
$0$
$1$

Table 9.6.21. Function values and points for $y=-x^2-2x+3$

$x$	$y=-x^2-2x+3$	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 9.6.22. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.6.23

a coordinate plane with the points from the table plotted; the axis of symmetry is drawn as a vertical dotted line through the vertex — Figure 9.6.22. Plot of the points and axis of symmetry

the points from the previous graph are connected to make a smooth u-shaped parabola opening downward — Figure 9.6.23. Graph of $y=-x^2-2x+3$

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.

Example 9.6.24.

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.

Explanation

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

\begin{align*} h\amp=-\frac{(\substitute{-3})}{2(\substitute{2})}\\ h\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.

Table 9.6.25. Function values and points for $y=2x^2-3x-4$

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

A coordinate plot with the points from the table plotted; the points are (-1,1), (0,-4), (0.75,-5.125), (1,-5), (2,-2); the axis of symmetry is drawn as a dotted line at y=0.75 — Figure 9.6.26. Plot of initial points

The points graphed in Figure 9.6.26 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 9.6.27), which will give it the symmetry we expect. From there, we can complete the sketch of this graph.

The previous plot with symmetric points added; The points are (2.5,1), (1.5,-4), (0.5,-5), (-0.5,-2) — Figure 9.6.27. Plot of symmetric points

The previous plot with all of the points connected by a smooth, u-shaped curve — Figure 9.6.28. Graph of $y=2x^2-3x-4$

Subsection 9.6.5 Applications of Quadratic Functions Involving the Vertex.

We looked at the height of Hannah's 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.

Example 9.6.29.

We looked at the quadratic function $R=(13+0.25x)(1500-50x)$ in Example 6.5.1 of Section 6.5, where $R$ was the revenue (in dollars) for $x$ 25-cent price increases. This function had each jar of Avery's jam priced at 13 dollars, and simplified to

\begin{equation*} R=-12.5x^2-275x+19500\text{.} \end{equation*}

Find the vertex of this quadratic function and explain what it means in the context of this model.

Explanation

Note that if we tried to use $R=(13+0.25x)(1500-50x)\text{,}$ we would not be able to immediately identify the values of $a$ and $b$ needed to determine the vertex. Using the expanded form of $R=-12.5x^2-275x+19500\text{,}$ we see that $a=-12.5$ and $b=-275\text{,}$ so the vertex occurs at:

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{\substitute{-275}}{2(\substitute{-12.5})}\\ h\amp=-11 \end{align*}

We will now find the value of $R$ for $x=-11\text{:}$

\begin{align*} R\amp=-12.5(\substitute{-11})^2-275(\substitute{-11})+19500\\ R\amp=21012.5 \end{align*}

Thus, the vertex occurs at $(-11,21012.5)\text{.}$

Literally interpreting this, we can state that $-11$ of the $25$-cent price increases result in a maximum revenue of $\$21{,}012.50\text{.}$

We can calculate “$-11$ of the 25-cent price increases” to be a decrease of $\$2.75\text{.}$ The price was set at $\$13$ per jar, so the maximum revenue of $\$21{,}012.50$ would occur when Avery sets the price at $\$10.25$ per jar.

Example 9.6.30.

Imagine that Jae got a new air rifle to shoot targets. The first thing they 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 ¹, the bulls-eye is a 0.5 mm diameter dot, about the size of the head of a pin, so accuracy is key.

en.wikipedia.org/wiki/ISSF_10_meter_air_rifle

Jae would like to set up the air rifle scope to be accurate at a level distance of $35$ yards (from the muzzle, which is the tip of the barrel), but they also need 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 the scope reticule (cross-hairs) at $35$ yards so that they can consistently hit the bulls-eye with the reticule directly over it, they set up targets at various distances to test the gun. Jae then shoots at the targets with the cross-hairs directly on the bulls-eye and measures the distance that the pellet hit above or below the bulls-eye when shot at those distances.

Table 9.6.31. Shooting Distance vs Pellet Rise/Fall

Distance to Target in Yards	$5$	$10$	$20$	$30$	$35$	$40$	$50$
Above/Below Bulls-eye	$\downarrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\odot$	$\downarrow$	$\downarrow$
Distance Above/Below in Inches	$0.1$	$0.6$	$1.1$	$0.6$	$0$	$0.8$	$3.2$

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

Explanation

(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.)

Since the input values seem to be going up by $5$s or $10$s, we will scale the $x$-axis by $10$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 Jae hit was $1.1$ inches when the target was $20$ yards away.

a Cartesian grid with points (5,-0.1), (10,0.6),(20,1.1),(30,0.6),(35,0),(40,-0.8),(50,-3.2); the points are connected with a smooth u-shaped curve that opens downward — Figure 9.6.32. Graph of Target Data

Example 9.6.33.

Kali has $500$ feet of fencing and she needs to build a rectangular pen for her goats. What are the dimensions of the rectangle that would give her goats the largest area?

Explanation

We will use $\ell$ for the length of the pen and $w$ for the width, in feet. We know that the perimeter must be $500$ feet, so that gives us

\begin{equation*} 2\ell+2w=500 \end{equation*}

First we will solve for the length:

\begin{align*} 2\ell+2w\amp=500\\ 2\ell\amp=500-2w\\ \ell\amp=250-w \end{align*}

Now we can build a function for the rectangle's area, using the formula for area:

\begin{align*} A(w)\amp=\ell\cdot w\\ A(w)\amp=(250-w)\cdot w\\ A(w)\amp=250w-w^2\\ A(w)\amp=-w^2+250w \end{align*}

The area is a quadratic function so we can identify $a=-1$ and $b=250$ and find the vertex:

\begin{align*} w\amp=-\frac{(\substitute{250})}{2(\substitute{-1})}\\ w\amp=\frac{250}{2}\\ w\amp=125 \end{align*}

Since the width of the rectangle is 125 feet, we can find the length using our expression:

\begin{align*} \ell\amp=250-w\\ \ell\amp=250-\substitute{125}\\ \ell\amp=125 \end{align*}

To find the maximum area we can either substitute the width into the area function or multiply the length by the width:

\begin{align*} A\amp=\ell\cdot w\\ A\amp=125\cdot 125\\ A\amp=15{,}625 \end{align*}

The maximum area that Kali can get is $15{,}625$ square feet if she builds her pen to be a square with a length and width of $125$ feet.

Exercises 9.6.6 Exercises

$x$	$y=\frac{5}{8} x +9$	Points
$-24$	$-6$	$\left(-24,-6\right)$
$-16$
$-8$
$0$
$8$
$16$

5.

Evaluate the expression $\displaystyle \frac{1}{5} \big( x + 3 \big)^2 - 4$ when $x = -8\text{.}$

6.

Evaluate the expression $\displaystyle \frac{1}{3} \big( x + 3 \big)^2 - 2$ when $x = -6\text{.}$

7.

Evaluate the expression $-16t^{2}+64t+128$ when $t=4\text{.}$

8.

Axis of symmetry:

Vertex: