ORCCA Key Features of Quadratic Graphs

Section 9.2 Key Features of Quadratic Graphs

Figure 9.2.1. Alternative Video Lesson

permalinkIn this section we will learn about quadratic graphs and their key features, including vertex, axis of symmetry and intercepts.

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Subsection 9.2.1 Properties of Quadratic Graphs

Hannah fired a toy rocket from the ground, which launched into the air with an initial speed of $64$ feet per second. The height of the rocket can be modeled by the equation $y=-16t^2+64t\text{,}$ where $t$ is how many seconds had passed since the launch. To see the shape of the graph made by this equation, we make a table of values and plot the points.

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

Figure 9.2.2. Points for

y = - 16 t^{2} + 64 t

$y=-16t^2+64t$

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) — Graph of $y=-16t^2+64t$

permalinkA curve with the shape that we see in Figure 9.2.3 is called a parabola. Notice the symmetry in Figure 9.2.2, how the $y$ -values in rows above the middle row match those below the middle row. Also notice the symmetry in the shape of the graph, how its left side is a mirror image of its right side.

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. That means that for large values of $t\text{,}$ the $at^2$ term will be large and negative, and the resulting $y$ -value will be low on the $y$ -axis. So the negative leading coefficient causes the arms of the parabola to point downward.

permalinkHere are some more quadratic graphs so we can see which way they open.

a U-shaped graph opening upward — The graph of $y=x^2-2x+2$ opens upward. Its leading coefficient is positive.

a wide U-shaped graph that opens downward — The graph of $y=-\frac{1}{4}x^2-\frac{1}{2}x-\frac{1}{4}$ opens downward. Its leading coefficient is negative.

a narrow U-shaped graph that opens upward — The graph of $y=3x^2-18x+23.5$ opens upward. Its leading coefficient is positive.

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Fact 9.2.7.

The graph of a quadratic equation $y=ax^2+bx+c$ opens upward or downward according to the sign of the leading coefficient $a\text{.}$ If the leading coefficient is positive, the parabola opens upward. If the leading coefficient is negative, the parabola opens downward.

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Checkpoint 9.2.8.

permalinkThe vertex of a quadratic graph is the highest or lowest point on the graph, depending on whether the graph opens downward or upward. In Figure 9.2.3, 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, cutting the quadratic graph into two symmetric halves. We write the axis of symmetry as an equation of a vertical line so it always starts with “ $x=\text{.}$ ” In Figure 9.2.3, 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 vertical axis is labeled $y\text{.}$ In Figure 9.2.3, the point $(\firsthighlight{0},\secondhighlight{0})$ is the starting point of the rocket, and it is where the graph crosses the $y$ -axis, so it is the vertical intercept. The $y$ -value of $\secondhighlight{0}$ means the rocket was on the ground when the $t$ -value was $\firsthighlight{0}\text{,}$ which was when the rocket launched.

The horizontal intercept(s) are the points where the parabola crosses the horizontal axis. They are the $x$ -intercepts if the horizontal axis is labeled $x\text{.}$ 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 came back to hit the ground after $4$ seconds.

permalinkIt is possible for a quadratic graph to have zero, one, or two 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 — 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 — 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 — The graph of $y=3x^2-18x+23.5$ has two horizontal intercepts

permalinkHere is a summary of the key features of quadratic graphs.

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List 9.2.12. Summary of Key Features of Quadratic Graphs

Consider a quadratic equation in the form $y=ax^2+bx+c$ and the parabola that it makes when graphed.

Direction: The parabola opens upward if $a$ is positive and opens downward of $a$ is negative.
Vertex: The vertex of the 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 graph intersects the vertical axis. There is exactly one vertical intercept.
Horizontal Intercept(s): The horizontal intercept(s) are the point(s) where a graph intersects the horizontal axis. The graph of a parabola can have zero, one, or two horizontal intercepts.

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Example 9.2.13.

Identify the key features of the quadratic graph of $y=x^2-2x-8$ shown in Figure 9.2.14.

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 9.2.14. Graph of $y=x^2-2x-8$

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Checkpoint 9.2.15.

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Subsection 9.2.2 Finding the Vertex and Axis of Symmetry Algebraically

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

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Fact 9.2.16.

If we denote $(h,k)$ as the coordinates of the vertex of a quadratic graph defined by $y=ax^2+bx+c\text{,}$ then $h=-\frac{b}{2a}\text{.}$ Then we can find $k$ by substituting $h$ in for $x\text{.}$

To understand why, we can look at the quadratic formula 7.2.2. 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. So $-\frac{b}{2a}$ is right in the middle, and it must be the horizontal coordinate of the vertex, $h\text{.}$ If we have already memorized the quadratic formula, this new formula for $h$ is not hard to remember:

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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

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Example 9.2.17.

Determine the vertex and axis of symmetry of the parabola $y=x^2-4x-12\text{.}$

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

\begin{aligned} h & = - \frac{b}{2 a} \\ = - \frac{(- 4)}{2 (1)} \\ = 2 \end{aligned}

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

Now we know the first coordinate of the vertex is $2\text{,}$ so we may substitute $x=2$ to determine the second coordinate of the vertex:

\begin{aligned} k & = (2)^{2} - 4 (2) - 12 \\ = 4 - 8 - 12 \\ = - 16 \end{aligned}

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

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

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Example 9.2.18.

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

Explanation

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

\begin{align*} h\amp=-\frac{b}{2a}\\ \amp=-\frac{(\substitute{-3})}{2(\substitute{-3})}\\ \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*} k\amp=-3x^2-3x+7\\ \amp=-3\left(\substitute{-\frac{1}{2}}\right)^2-3\left(\substitute{-\frac{1}{2}}\right)+7\\ \amp=-3\left(\frac{1}{4}\right)+\frac{3}{2}+7\\ \amp=-\frac{3}{4}+\frac{3}{2}+7\\ \amp=-\frac{3}{4}+\frac{6}{4}+\frac{28}{4}\\ \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 $x=-\frac{1}{2}\text{.}$

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Subsection 9.2.3 Graphing Quadratic Equations by Making a Table

When we learned how to graph lines, we could choose any $x$ -values to build a table of values. For quadratic equations, we want to make sure the vertex is present in the table, since it is such a special point. So we find the vertex first and then choose our $x$ -values surrounding it. We can use the property of symmetry to speed things up.

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Example 9.2.19.

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

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})}\\ \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*} k\amp=-\substitute{x}^2-2\substitute{x}+3\\ \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 set up our table with two values on each side of $x=-1\text{.}$ We choose $x=-3\text{,}$ $-2\text{,}$ $-1\text{,}$ $0\text{,}$ and $1$ as shown in Figure 9.2.20.

Next, we determine the $y$-coordinates by replacing $x$ with each value and we have the complete table as shown in Figure 9.2.21. 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.

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

Figure 9.2.20. Setting up the table

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

Figure 9.2.21. Values and points for $y=-x^2-2x+3$

Now that we have our table, we plot the points and draw in the axis of symmetry as shown in Figure 9.2.22. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.2.23.

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 9.2.22. 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 9.2.23. 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.

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Example 9.2.24.

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

Explanation

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

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

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

\begin{align*} k\amp=2\left(\substitute{\frac{3}{4}}\right)^2-3\left(\substitute{\frac{3}{4}}\right)-4\\ \amp=2\left(\frac{9}{16}\right)-\frac{9}{4}-4\\ \amp=\frac{9}{8}-\frac{18}{8}-\frac{32}{8}\\ \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 create a table. We choose $x$-values on both sides of $x=0.75\text{,}$ but we choose integers because it will be easier to find the $y$-values.

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

Figure 9.2.25. Values and points for $y=2x^2-3x-4$

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 9.2.26. Plot of initial points

The points graphed in Figure 9.2.26 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 9.2.27), 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 9.2.27. Plot of symmetric points

The previous plot with all of the points connected by a smooth, U-shaped curve — Figure 9.2.28. Graph of $y=2x^2-3x-4$

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Subsection 9.2.4 Applications of Quadratic Graphs 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 $y$ -value on a quadratic graph.

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Example 9.2.29.

Jae got a new air rifle for target practice. The first thing they did with it was some testing to find out how accurate the targeting cross-hairs were. In Olympic $10$ -meter air rifle shooting ¹en.wikipedia.org/wiki/ISSF_10_meter_air_rifle, the bulls-eye is a 0.5 mm diameter dot, about the size of the head of a pin, so accuracy is key. To test the accuracy, Jae stood at certain specific distances from a bullseye target, aimed the cross-hairs on the bullseye, and fired. Jae recorded how far above or how far below the pellet hit relative to the bullseye.

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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	$0.1$	$0.6$	$1.1$	$0.6$	$0$	$0.8$	$3.2$

Figure 9.2.30. Shooting Distance vs Pellet Rise/Fall

Make a graph of the height of the pellet relative to the bulls-eye at the shooting distances Jae used in Figure 9.2.30 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 increasing by $5$s or $10$s, we 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 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 9.2.31. Graph of Target Data

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Example 9.2.32.

We looked at the quadratic equation $R=(13+0.25x)(1500-50x)$ in Example 5.4.2 of Section 5.4, where $R$ was the revenue (in dollars) for $x$ 25-cent price increases from an initial price of $13. The expression simplified to

R = - 12.5 x^{2} - 275 x + 19500 .

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

Find the vertex of this quadratic expression 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}\\ \amp=-\frac{\substitute{-275}}{2(\substitute{-12.5})}\\ \amp=-11 \end{align*}

And the second coordinate for the vertex is at:

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

So 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. 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.

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Example 9.2.33.

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 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 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 write a formula for the rectangle's area:

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

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

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

Since the width of the rectangle that will maximize area is 125 ft, we can find the length using our expression:

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

To find the maximum area we can either substitute the width into the area formula 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.

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Reading Questions 9.2.5 Reading Questions

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1.

There are four key features of a quadratic graph discussed in this section. What are they?

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2.

Explain how the formula for the first coordinate of a parabola's vertex is similar to the quadratic formula.

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3.

If a parabola's vertex is at $(4,6)\text{,}$ and you know the coordinates of some points on the parabola where $x=1,2,3\text{,}$ at what other $x$ -values do you know coordinates on the parabola?

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Exercises 9.2.6 Exercises

6.

Evaluate the expression $\displaystyle \frac{1}{3} \big( x + 3 \big)^2 - 8$ when $x = -6\text{.}$

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7.

Evaluate the expression $-16t^{2}+64t+128$ when $t=3\text{.}$

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8.

$y=(x+3)^2$

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36.

$y=(x+2)^2$

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

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

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