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## Section9.4Graphs of Quadratic Functions Chapter Review

### Subsection9.4.1Introduction to Functions

In Section 9.1 we covered the definitions of a relation, a function, and domain and range. We then discussed function notation and how to evaluate functions at a particular value as well as how to solve an equation with function notation. Last we introduced the vertical line test

###### Example9.4.1Introduction to Functions

One week in Portland, it was rainy. Shocking, I know. Shown is a diagram for the day of the week and how much rain fell on those days. Convert this diagram to a set of ordered pairs where the first coordinate is the day of the week and the second coordinate is the amount of rain that fell in Portland that day, in inches. Figure 9.4.2 Diagram for the relation “Day of the week to rainfall in PDX”
Explanation

Since Monday has an arrow to $0.2\text{,}$ that must mean that it rained $0.2$ inches of rain on Monday. That gives us the ordered pair $(\text{Monday},0.2)\text{.}$ Likewise, we also have $(\text{Tuesday},0.4)\text{.}$ To write down all of the ordered pairs, we will use a set:

\begin{equation*} \left\{(\text{Monday},0.2), (\text{Tuesday},0.4), (\text{Wednesday},0.0), (\text{Thursday},0.6), (\text{Friday},0.0)\right\} \end{equation*}
###### Example9.4.3Functions as Predictors

Recall that a a function should predict the output perfectly if you know the input. In Example 9.4.1, we saw that the day of a particular week in Portland was related to the rainfall during that week. Was this relation a function?

Explanation

Yes, that relation was a function. If I tell you the day of the week, you can with certainty “predict” how much rain there was on that day by looking at the data. There was a single answer to the question “How much rain fell on Wednesday during that week in Portland?”

At this point, we should note that all historical weather data can be viewed as a function of some sort: On any particular day in the recent past, we can “predict” (by looking it up somewhere) how much rain fell in any city in the world.

However, the opposite question, “If it rained $0.0$ inches, what day of the year was it?” is not going to represent a function because there will be more than one day of the year that it didn't rain at all. Even in Portland.

###### Example9.4.4Algebraic Functions
1. For the equation $y=2x^2+x\text{,}$ will $y$ be a function of $x\text{?}$

2. For the equation $y^2=x\text{,}$ will $y$ be a function of $x\text{?}$

Explanation
1. This question is equivalent to the question “If you input any $x$ value, will there be only one $y$ value?” The answer is “yes.” If you input any number for $x$ you first square that number and multiply it by two, then add the number. For example, if we substitute $\substitute{-3}\text{,}$ we get:

\begin{align*} y\amp=2(\substitute{x})^2+(\substitute{x})\\ y\amp=2(\substitute{-3})^2+(\substitute{-3})\\ \amp=2(9)-3\\ \amp=18-3\\ \amp=15 \end{align*}

There is no way we could substitute a number for $x$ and get more than one value for $y\text{.}$

2. This question is equivalent to the question “If you input any $x$ value, will there be only one $y$ value?” The answer is “no.” If you input any number for $x$ and solve for $y\text{,}$ you might actually get two solutions. For example, if we substitute $\substitute{1}\text{,}$ we get:

\begin{align*} y^2\amp=\substitute{x}\\ y^2\amp=\substitute{1} \end{align*}
\begin{align*} y\amp=\sqrt{1}\amp\text{ or }\amp\amp y\amp=-\sqrt{1}\\ y\amp=1\amp\text{ or }\amp\amp y\amp=-1 \end{align*}

Since there is more than one $y$ value for the $x$ value $1\text{,}$ the relation $y^2=x$ is not a function.

###### Example9.4.5Function Notation

For the function $V=g(x)\text{,}$ what is the name of the function, what is the input variable, and what is the output variable?

Explanation

For the function $V=g(x)\text{,}$ the name of the function is $g\text{,}$ the input variable is $x\text{,}$ and the output variable is $V\text{.}$

###### Example9.4.6Evaluating Functions

Let $a(x)=x^2-2x\text{,}$ let the function $b$ be defined by the graph in Figure 9.4.7, and let the function $c$ be defined by the table in Table 9.4.8. $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Evaluate $a(-3)\text{.}$

2. Evaluate $b(-3)\text{.}$

3. Evaluate $c(-3)\text{.}$

Explanation
1. To evaluate $a(-3)$ we will substitute the value $\substitute{-3}$ in wherever we see $x$ in the equation for $a(x)\text{.}$

\begin{align*} a(\substitute{x})\amp=\substitute{x}^2-2\substitute{x}\\ a(\substitute{-3})\amp=(\substitute{-3})^2-2(\substitute{-3})\\ \amp=9+6\\ \amp=15 \end{align*}
2. To evaluate $b(-3)\text{,}$ we need to look at the graph of $y=b(x)$ and look for the one place on the graph where the $x$-value is $-3\text{.}$ The $y$-value at this point is $1\text{,}$ so we would say that

\begin{equation*} b(-3)=1 \end{equation*}
3. To evaluate $c(-3)$ we examine the table and find the place where the $x$-value is $-3\text{.}$ We conclude that $c(-3)=-2\text{.}$

###### Example9.4.9Solving Equations That Have Function Notation

Still let $a(x)=x^2-2x\text{,}$ let the function $b$ be defined by the graph in Figure 9.4.10, and let the function $c$ be defined by the table in Table 9.4.11. $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Solve the equation $a(x)=0\text{.}$

2. Solve the equation $b(x)=2\text{.}$

3. Solve the equation $c(x)=7\text{.}$

Explanation
1. To solve the equation $a(x)=0\text{,}$ we should set the formula for $a(x)$ equal to $0\text{.}$ In this case, factoring will then help.

\begin{align*} x^2-2x\amp=0\\ x(x-2)\amp=0 \end{align*}
\begin{align*} x\amp=0\amp\text{ or }\amp\amp x-2\amp=0\\ x\amp=0\amp\text{ or }\amp\amp x\amp=2 \end{align*}

So, the solution set is $\{0,2\}\text{.}$

2. To solve the equation $b(x)=2\text{,}$ we should examine the graph and find all of the locations where the $y$-value is $2\text{.}$ According to the graph, that happens twice, both when $x=-4$ and when $x=0\text{.}$ So the solution set is $\{-4,0\}.$

3. To solve the equation $c(x)=7\text{,}$ we examine the table and find all the places where the function value is $7\text{.}$ According to the table, that happens twice: once when $x=-5$ and again when $x=9\text{.}$ So the solution set is $\{-5,9\}\text{.}$

###### Example9.4.12Domain and Range

Still let the function $b$ be defined by the graph in Figure 9.4.13, and let the function $c$ be defined by the table in Table 9.4.14. $x$ $c(x)$ $-5$ $7$ $-3$ $-2$ $2$ $2$ $6$ $4$ $9$ $7$
1. Write the domain of $b$ in interval notation.

2. Write the domain of $c$ in interval notation.

3. Write the range of $b$ in interval notation.

4. Write the range of $c$ in interval notation.

Explanation

Recall the definitions of domain and range.

1. To find the the domain of $b\text{,}$ look at the graph of $b$ and see what $x$-values are used for the graph. In other words, what $x$-values could you input and expect to get a $y$-value? It looks like the graph extends forever to the right as well as the left. Therefore, the domain is $(-\infty,\infty)\text{.}$

2. To find the the domain of $c\text{,}$ we list the $x$-values in the table in a set. That would be $\{-5,-3,2,6,9\}\text{.}$

3. To find the the range of $b\text{,}$ look at the graph of $b$ and see what $y$-values are on the graph. The lowest $y$-value is $-3$ and all higher $y$-values are possible outputs. Therefore, the range is $[-3,\infty)\text{.}$

4. To find the the range of $c\text{,}$ we list the $y$-values in the table in a set. That would be $\{7,-2,2,4\}\text{.}$

###### Example9.4.15Determining if a Relation is a Function

Is $y$ a function of $x$ in the following graphs?  Explanation

The graph in Figure 9.4.16 fails the vertical line test and so $y$ is not a function of $x\text{.}$ The graph in Figure 9.4.17 passes the vertical line test and so $y$ is a function of $x\text{.}$

### Subsection9.4.2Properties of Quadratic Functions

In Section 9.2 we covered the definition of a quadratic function, and how to determine the direction, vertex, axis of symmetry, and vertical and horizontal intercepts. We then learned how to algebraically find the vertex, how to graph a parabola using a table, and how to find the domain and range of quadratic functions.

###### Example9.4.18Finding the Vertex and Axis of Symmetry

Algebraically find the vertex of the parabola described by the quadratic function $f(x)=3x^2+8x-7\text{.}$

Explanation

To find the vertex of a parabola algebraically, we use the formula $h=-\frac{b}{2a}$ to find the axis of symmetry first. For our equation, $\substitute{a=3}$ and $\lighthigh{b=8}\text{,}$ so:

\begin{align*} h\amp=-\frac{\lighthigh{8}}{2(\substitute{3})}=-\frac{4}{3} \end{align*}

Next, we evaluate $f\mathopen{}\left(\substitute{-\frac{4}{3}}\right)\mathclose{}$ to find the $y$-coordinate of the vertex.

\begin{align*} f\mathopen{}\left(\substitute{-\frac{4}{3}}\right)\mathclose{}\amp=3\left(\substitute{-\frac{4}{3}}\right)^2+8\left(\substitute{-\frac{4}{3}}\right)-7\\ \amp=3\left(\frac{16}{9}\right)-\frac{32}{3}-7\\ \amp=\frac{16}{3}-\frac{32}{3}-\frac{21}{3}\\ \amp=\frac{37}{3} \end{align*}

So, the parabola has its axis of symmetry at $x=-\frac{4}{3}$ and has its vertex at the point $\left(-\frac{4}{3},-\frac{37}{3}\right)\text{.}$

###### Example9.4.19Properties of Quadratic Functions

Identify the key features of the quadratic function $y=-2x^2-4x+6$ shown in Figure 9.4.20. Explanation

First, we see that this parabola opens downward because the leading coefficient is negative.

Then we locate the vertex which is the point $(-1,8)\text{.}$ The axis of symmetry is the vertical line $x=-1\text{.}$

The vertical intercept or $y$-intercept is the point $(0,6)\text{.}$

The horizontal intercepts are the points $(-3,0)$ and $(1,0)\text{.}$ ###### Example9.4.22Graphing Quadratic Functions by Making a Table

Determine the vertex and axis of symmetry for the quadratic function $g(x)=2x^2+8x+2\text{.}$ Then make a table of values and sketch the graph of the function.

Explanation

To determine the vertex of $g(x)=2x^2+8x+2\text{,}$ we want to find the $x$-value of the vertex first. We will use $h=-\frac{b}{2a}$ for $\substitute{a=2}$ and $\lighthigh{b=8}\text{:}$

\begin{align*} h\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ h\amp=-\frac{\lighthigh{8}}{2(\substitute{2})}\\ h\amp=-2 \end{align*}

To find the vertex, we evaluate $g(\substitute{-2})\text{.}$

\begin{align*} g(\substitute{-2})\amp=2(\substitute{-2})^2+8(\substitute{-2})+2\\ \amp=2(4)-16+2\\ \amp=8-16+2\\ \amp=-6 \end{align*}

Now we know that our axis of symmetry is the line $x=-2$ and the vertex is the point $(-2,-6)\text{.}$ We will set up our table with two values on each side of $x=-2\text{.}$ We choose $x= -4, -3, \highlight{-2}, -1,$ and $0\text{.}$ Then, we'll determine the $y$-coordinates by replacing $x$ with each value and we have the complete table as shown in Table 9.4.23. 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.

Now that we have our table, we will plot the points and draw in the axis of symmetry as shown in Figure 9.4.24. We complete the graph by drawing a smooth curve through the points and drawing an arrow on each end as shown in Figure 9.4.25  ###### Example9.4.26The Domain and Range of Quadratic Functions

Use the graph of $g(x)=2x^2+8x+2$ in from Example 9.4.22 shown in Figure 9.4.25 to identify the domain and range of $g$ in set-builder and interval notation.

Explanation

For the domain, we look horizontally and see the graph is a continuous curve and one arrow points to the left and the other arrow points to the right. The domain is $\{x\mid x\ \textrm{is a real number}\}$ which is equivalent to $(-\infty,\infty)\text{.}$

For the range we look up and down on the graph, which opens upward. Both arrows point upward and the lowest point on the graph is the vertex at the point $(-2,-6)\text{.}$ The range is $\{y \mid y \ge -6\}$ which is equivalent to $[-6,\infty)\text{.}$

###### Example9.4.27Applications of Quadratic Functions Involving the Vertex

The value of FedEx stock, in dollars, between December 28, 2017 and February 9, 2018 can be very closely approximated by a quadratic function, $F(x)=-0.07x^2+2.7x+248\text{.}$ Find the vertex of this parabola and interpret it in the context of the situation.

Explanation

To find the vertex of $F(x)=-0.07x^2+2.7x+248\text{,}$ we first find the axis of symmetry using the formula $h=-\frac{b}{2a}\text{.}$ For our equation, $\substitute{a=-0.07}$ and $\lighthigh{b=2.7}\text{,}$ so:

\begin{align*} h\amp=-\frac{\lighthigh{2.7}}{2(\substitute{-0.07})}\\ h\amp=-\frac{2.7}{-0.14}\\ h\amp\approx19.3 \end{align*}

Next, we substitute the value of the axis of symmetry, $\substitute{19.3}\text{,}$ into the formula for the parabola.

\begin{align*} F(\substitute{x})\amp=-0.07\substitute{x}^2+2.7\substitute{x}+248\\ F(\substitute{19.3})\amp=-0.07\left(\substitute{19.3}\right)^2+2.7\left(\substitute{19.3}\right)+248\\ \amp\approx274 \end{align*}

So, the parabola has its axis of symmetry at $x=19.3$ and has its vertex at about the point $(19.3,274)\text{.}$ In addition, since $a=-0.07\text{,}$ the parabola will be opening downward. Given this information, the vertex of this parabola is its maximum.

In the reality of the situation, the $19.3$ indicates the number of days after December 28, 2017 which would be sometime on January 16. The $274$ indicates the stock price in dollars. In conclusion, FedEx's stock peaked (between December 28, 2017 and February 9, 2018) on January 16 at a value of $\274\text{.}$

### Subsection9.4.3Graphing Quadratic Functions

In Section 9.3 we covered how to find the vertical and horizontal intercepts of a quadratic function algebraically, how to make a graph of a parabola using key features, applications of quadratic functions and what their real world domains and ranges mean, and how to tell a quadratic function from other types of functions.

###### Example9.4.28Finding the Vertical and Horizontal Intercepts Algebraically
1. Algebraically determine the vertical and horizontal intercepts of the quadratic function $h(x)=6x^2-13x+6\text{.}$

2. Algebraically determine the vertical and horizontal intercepts of the quadratic function $k(x)=2x^2-2x-5\text{.}$

Explanation
1. To find the vertical intercept, we evaluate the function when $x=0\text{.}$

\begin{align*} h(\substitute{0})\amp=6(\substitute{0})^2-13(\substitute{0})+6\\ \amp=6 \end{align*}

So, the vertical intercept of $h$ is $(0,6)\text{.}$

Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Algorithm 8.6.2. For this particular example, we will practice factoring using the AC method. Here, $ac=36$ and factor pairs that add up to $-13$ are $-9$ and $-4\text{,}$ as you will see.

\begin{align*} \substitute{h(x)}\amp=6x^2-13x+6\\ \substitute{0}\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{13x}}+6\\ 0\amp=6x^2\mathbin{\highlight{-}}\mathbin{\highlight{9x}}\mathbin{\highlight{-}}\mathbin{\highlight{4x}}+6\\ 0\amp=3x\highlight{(2x-3)}-2\highlight{(2x-3)}\\ 0\amp=(3x-2)\highlight{(2x-3)} \end{align*}
\begin{align*} 0\amp=3x-2\amp\text{ or }\amp\amp0\amp=2x-3\\ x\amp=\frac{2}{3}\amp\text{ or }\amp\amp x\amp=\frac{3}{2} \end{align*}

So, the horizontal intercepts are $\left(\frac{2}{3},0\right)$ and $\left(\frac{3}{2},0\right)\text{.}$

2. To find the vertical intercept, we have to evaluate the function when $x=0\text{.}$

\begin{align*} k(\substitute{0})\amp=2(\substitute{0})^2-2(\substitute{0})-5\\ \amp=-5 \end{align*}

So, the vertical intercept of $k$ is $(0,-5)\text{.}$

Next, to find the horizontal intercepts, we set the function equal to zero. To solve that equation we will use Algorithm 8.6.2. For this particular example, we will use the quadratic formula because the square root method will not work (because there is a linear term) and factoring fails.

\begin{align*} \substitute{k(x)}\amp=2x^2-2x-5\\ \substitute{0}\amp=2x^2-2x-5\\ \end{align*}

We identify that $\substitute{a=2}\text{,}$ $\lighthigh{b=-2}\text{,}$ and $c=-5$

\begin{align*} x\amp=\frac{-\lighthigh{b}\pm\sqrt{\lighthigh{b}^2-4\substitute{a}c}}{2\substitute{a}}\\ x\amp=\frac{-(\lighthigh{-2})\pm\sqrt{(\lighthigh{-2})^2-4(\substitute{2})(-5)}}{2(\substitute{2})}\\ x\amp=\frac{2\pm\sqrt{4+40}}{4}\\ x\amp=\frac{2\pm\sqrt{44}}{4}\\ x\amp=\frac{2\pm\sqrt{\highlight{4}\cdot11}}{4}\\ x\amp=\frac{2\pm\highlight{2}\sqrt{11}}{4}\\ x\amp=\frac{2}{4}\pm\frac{2\sqrt{11}}{4}\\ x\amp=\frac{1}{2}\pm\frac{\sqrt{11}}{2}\\ x\amp=\frac{1\pm\sqrt{11}}{2} \end{align*}

So, the horizontal intercepts are $\left(\frac{1+\sqrt{11}}{2},0\right)$ and $\left(\frac{1-\sqrt{11}}{2},0\right)\text{.}$ If you wanted to graph these points, you would need to approximate them as $\left(2.16,0\right)$ and $\left(-1.16,0\right)\text{.}$

###### Example9.4.29Graphing Quadratic Functions Using Their Key Features

Graph the function $j(x)=-2x^2+6x+8$ by algebraically determining its key features.

Explanation

To start, we'll note that this function will open downward, as the leading coefficient is negative.

To find the $y$-intercept, we'll evaluate $j(0)\text{:}$

\begin{align*} j(\substitute{0})\amp=-2(\substitute{0})^2+6(\substitute{0})+8\\ \amp=8 \end{align*}

The $y$-intercept is $(0,8)\text{.}$

Next, we'll find the horizontal intercepts by setting $j(x)=0$ and solving for $x\text{:}$

\begin{align*} -2x^2+6x+8\amp=0\\ -2(x^2-3x-4)\amp=0\\ 2(x-4)(x+1)\amp=0 \end{align*}
\begin{align*} x-4\amp=0\amp \text{or}\amp\amp x+1\amp=0\\ x\amp=4\amp \text{or}\amp\amp x\amp=-1 \end{align*}

The $x$-intercepts are $(4,0)$ and $(-1,0)\text{.}$

Lastly, we'll determine the vertex. Noting that $\substitute{a=-2}$ and $\lighthigh{b=6}\text{,}$ we have:

\begin{align*} x\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ x\amp=-\frac{\lighthigh{6}}{2(\substitute{-2})}\\ \amp=1.5 \end{align*}

Using the $x$-value $\substitute{1.5}$ to find the $y$-coordinate, we have:

\begin{align*} j(\substitute{1.5})\amp=-2(\substitute{1.5})^2+6(\substitute{1.5})+8\\ \amp=12.5 \end{align*}

The vertex is the point $(1.5,12.5)\text{,}$ and the axis of symmetry is the line $x=1.5\text{.}$

We're now ready to graph this function. We'll start by drawing and scaling the axes so all of our key features will be displayed as shown in Figure 9.4.30. Next, we'll plot these key points as shown in Figure 9.4.31. Finally, we'll note that this parabola opens downward and connect these points with a smooth curve, as shown in Figure 9.4.32.   ###### Example9.4.33Applications of Quadratic Functions

The Five-hundred-meter Aperture Spherical Telescope 1 en.wikipedia.org/wiki/Five_hundred_meter_Aperture_Spherical_Telescope (FAST) located in SW China is the worlds largest radio telescope. The name is actually incorrect because it is not spherical: it is parabolic! Parabolic telescopic dishes are very common because parabolas have the unique feature that they focus light coming in at a single point where a collecting instrument is placed.

If a scientist somehow managed to climb to the rim of the dish and roll a ball from the top down toward the center, her ball would travel along the parabola $h(x)=0.0018x^2-0.9x\text{,}$ where $x$ is the horizontal distance from the rim of the dish where the ball was released, in meters, and $h(x)$ is the height of the ball above above the rim, in meters (so negative height would mean below the rim). Note that this is an approximation based on real data.

1. Find and interpret the vertical intercept of the graph.

2. Find and interpret the horizontal intercepts of the graph.

3. Find and interpret the vertex of the graph.

Explanation
1. The vertical intercept is found when $\substitute{x=0}\text{.}$

\begin{align*} h(\substitute{x})\amp=0.0018\substitute{x}^2-0.9\substitute{x}\\ h(\substitute{0})\amp=0.0018(\substitute{0})^2-0.9(\substitute{0})\\ \amp=0 \end{align*}

So the vertical intercept is $(0,0)$ which means that the ball was released $0$ meters from the rim (on the rim) at a height of $0$ meters above the rim (again, on the rim).

2. The horizontal intercepts are found when $h(x)=0\text{.}$ This will be a quadratic equation that we can solve using factoring.

\begin{align*} h(x)\amp=0\\ 0.0018x^2-0.9x\amp=0\\ x\left(0.0018x-0.9\right)\amp=0 \end{align*}
\begin{align*} x\amp=0\amp\text{ or }\amp\amp 0.0018x-0.9\amp=0\\ x\amp=0\amp\text{ or }\amp\amp 0.0018x\amp=0.9\\ x\amp=0\amp\text{ or }\amp\amp x\amp=500 \end{align*}

The horizontal intercepts are $(0,0)$ and $(500,0)\text{.}$ The interpretation of $(0,0)$ is the same as before. The interpretation of $(500,0)$ is that the other side of the rim of the dish is $500$ meters away. If the ball were to somehow roll all the way down, and then back up the other side, it would pop up at the opposite rim exactly $500$ meters horizontally away.

3. The vertex is found when the $x$-value is at the axis of symmetry. To find that, we use the formula $x=-\frac{\lighthigh{b}}{2\substitute{a}}$ where $\substitute{a=0.0018}$ and $\lighthigh{b=-0.9}$

\begin{align*} x\amp=-\frac{\lighthigh{b}}{2\substitute{a}}\\ x\amp=-\frac{\lighthigh{-0.9}}{2(\substitute{0.0018})}\\ \amp=250 \end{align*}

Then, to find the $y$-value, we substitute that $x$-value, $\substitute{250}$ into the original function.

\begin{align*} h(\substitute{x})\amp=0.0018\substitute{x}^2-0.9\substitute{x}\\ h(\substitute{250})\amp=0.0018(\substitute{250})^2-0.9(\substitute{250})\\ \amp=-112.5 \end{align*}

The vertex is the point $(250,-112.5)\text{.}$ Since the vale of $a$ is positive, we know that this parabola opens upward. This means that the vertex is the lowest point on the graph. So the lowest point of the telescope dish is $250$ meters from the rim (horizontally) and $112.5$ meters below the height of the rim. That is one big dish!

###### Example9.4.34The Domain and Range of Quadratic Applications

In the FAST telescope example, find the real world domain and range of the function $h\text{.}$

Explanation

Real world domain and range problems are sometimes subjective. The domain in this case would represent all possible $x$-values that make sense in the reality of the situation. Recall that $x$ is the horizontal distance from the rim of the dish where the ball was released, in meters. Since we found out that the dish is $500$ meters across we could say that the domain is $[0,500]\text{.}$

The range will represent all possible $y$ values, that make sense in reality. Recall that $h(x)$ is the height above above the rim, in meters. Since the ball won't roll any higher than the rim of the dish, $0$ will be the largest $y$-value. We found out that the lowest point on the graph is $-112.5$ meters below the rim. So the range is [-112.5,0]. Figure 9.4.35 A diagram of the parabolic function representing the telescope's dish
###### Example9.4.36Distinguishing Quadratic Functions from Other Functions and Relations

Decide if the equations represent quadratic functions or something else.

1. $y-1=3(x-2)-5$

2. $y=3(x-2)^2-5$

3. $y=\sqrt{x-2}-5$

4. $y^2=x^2-5$

Explanation

Recall that Definition 9.2.2 says that a quadratic function has the form $f(x)=ax^2+bx+c$ where $a\text{,}$ $b\text{,}$ and $c$ are real numbers, and $a \neq 0\text{.}$

1. The equation $y-1=3(x-2)-5$ is not quadratic. It is in fact a linear equation.

2. The equation $y=3(x-2)^2-5$ is quadratic. We could simplify the right hand side and would have something of the form $y=ax^2+bx+c\text{.}$

3. The equation $y=\sqrt{x-2}-5$ is not quadratic. The $x$ is inside a radical, not squared, so it cannot be converted into the form $y=ax^2+bx+c\text{.}$

4. The equation $y^2=x^2-5$ is not quadratic. It cannot be re-written in the form $y=ax^2+bx+c$ (due to the $y^2$ term).

### Subsection9.4.4Exercises

###### 1

Which of the following graphs show $y$ as a function of $x\text{?}$      ###### 2

Which of the following graphs show $y$ as a function of $x\text{?}$      ###### 3

Evaluate the function at the given values.

$F(x)={-5x+4}$

1. $F(1)=$

2. $F(-3)=$

3. $F(0)=$

###### 4

Evaluate the function at the given values.

$G(x)={-3x+8}$

1. $G(5)=$

2. $G(-3)=$

3. $G(0)=$

###### 5

Evaluate the function at the given values.

$\displaystyle{H(x)=\frac{{7x}}{{-6x+3}}}$

1. $H(4)=$ .

2. $H(-6)=$ .

###### 6

Evaluate the function at the given values.

$\displaystyle{K(x)=\frac{{2x}}{{3x-3}}}$

1. $K(4)=$ .

2. $K(-1)=$ .

###### 7

Evaluate the function at the given values.

$K(x)={-2x^{2}-5x-6}$

1. $K(2)=$

2. $K(-4)=$

###### 8

Evaluate the function at the given values.

$f(x)={-3x^{2}+3x-1}$

1. $f(4)=$

2. $f(-5)=$

###### 9

Solve for $x\text{,}$ where $g(x)={x^{2}-2x-82}\text{.}$

If $g(x)=-2\text{,}$ then $x=$.

###### 10

Solve for $x\text{,}$ where $h(x)={x^{2}-x-17}\text{.}$

If $h(x)=-5\text{,}$ then $x=$.

###### 11

A function is graphed. This function has domain and range .

###### 12

A function is graphed. This function has domain and range .

###### 13

Function $f$ is graphed. 1. $\displaystyle{f(2)={}}$

2. Solve $\displaystyle{f(x)=1}\text{.}$

###### 14

Function $f$ is graphed. 1. $\displaystyle{f(0)={}}$

2. Solve $\displaystyle{f(x)=1}\text{.}$

###### 15

Heather started saving in a piggy bank on her birthday. The function $f(x)={5x+3}$ models the amount of money, in dollars, in Heather’s piggy bank. The independent variable represents the number of days passed since her birthday.

Interpret the meaning of $f(2)=13\text{.}$

• A. Two days after Heather started her piggy bank, there were $\13$ in it.

• B. The piggy bank started with $\2$ in it, and Heather saves $\13$ each day.

• C. Thirteen days after Heather started her piggy bank, there were $\2$ in it.

• D. The piggy bank started with $\13$ in it, and Heather saves $\2$ each day.

###### 16

Haley started saving in a piggy bank on her birthday. The function $f(x)={4x+3}$ models the amount of money, in dollars, in Haley’s piggy bank. The independent variable represents the number of days passed since her birthday.

Interpret the meaning of $f(4)=19\text{.}$

• A. Nineteen days after Haley started her piggy bank, there were $\4$ in it.

• B. The piggy bank started with $\4$ in it, and Haley saves $\19$ each day.

• C. Four days after Haley started her piggy bank, there were $\19$ in it.

• D. The piggy bank started with $\19$ in it, and Haley saves $\4$ each day.

###### 17

The following figure has the graph $y=d(t)\text{,}$ which models a particle’s distance from the starting line in feet, where $t$ stands for time in seconds since timing started. 1. $d(2)=$

2. Interpret the meaning of $d(2)\text{:}$

• A. In the first $8$ seconds, the particle moved a total of $2$ feet.

• B. In the first $2$ seconds, the particle moved a total of $8$ feet.

• C. The particle was $8$ feet away from the starting line $2$ seconds since timing started.

• D. The particle was $2$ feet away from the starting line $8$ seconds since timing started.

3. Solve $d(t)={4}$ for $t\text{.}$ $t=$

4. Interpret the meaning of part c’s solution(s):

• A. The article was $4$ feet from the starting line $1$ seconds since timing started, or $8$ seconds since timing started.

• B. The article was $4$ feet from the starting line $8$ seconds since timing started.

• C. The article was $4$ feet from the starting line $1$ seconds since timing started.

• D. The article was $4$ feet from the starting line $1$ seconds since timing started, and again $8$ seconds since timing started.

###### 18

The following figure has the graph $y=d(t)\text{,}$ which models a particle’s distance from the starting line in feet, where $t$ stands for time in seconds since timing started. 1. $d(5)=$

2. Interpret the meaning of $d(5)\text{:}$

• A. The particle was $5$ feet away from the starting line $7.5$ seconds since timing started.

• B. The particle was $7.5$ feet away from the starting line $5$ seconds since timing started.

• C. In the first $7.5$ seconds, the particle moved a total of $5$ feet.

• D. In the first $5$ seconds, the particle moved a total of $7.5$ feet.

3. Solve $d(t)={3}$ for $t\text{.}$ $t=$

4. Interpret the meaning of part c’s solution(s):

• A. The article was $3$ feet from the starting line $8$ seconds since timing started.

• B. The article was $3$ feet from the starting line $1$ seconds since timing started, and again $8$ seconds since timing started.

• C. The article was $3$ feet from the starting line $1$ seconds since timing started, or $8$ seconds since timing started.

• D. The article was $3$ feet from the starting line $1$ seconds since timing started.

###### 19

The function $C$ models the the number of customers in a store $t$ hours since the store opened.

 $t$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $C(t)$ $0$ $41$ $75$ $97$ $97$ $73$ $48$ $0$
1. $C(1)=$

2. Interpret the meaning of $C(1)\text{:}$

• A. There were $41$ customers in the store $1$ hour after the store opened.

• B. In $1$ hour since the store opened, the store had an average of $41$ customers per hour.

• C. In $1$ hour since the store opened, there were a total of $41$ customers.

• D. There was $1$ customer in the store $41$ hours after the store opened.

3. Solve $C(t)=97$ for $t\text{.}$ $t=$

4. Interpret the meaning of Part c’s solution(s):

• A. There were $97$ customers in the store $3$ hours after the store opened.

• B. There were $97$ customers in the store $4$ hours after the store opened.

• C. There were $97$ customers in the store either $4$ hours after the store opened, or $3$ hours after the store opened.

• D. There were $97$ customers in the store $4$ hours after the store opened, and again $3$ hours after the store opened.

###### 20

The function $C$ models the the number of customers in a store $t$ hours since the store opened.

 $t$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $C(t)$ $0$ $43$ $78$ $97$ $95$ $81$ $43$ $0$
1. $C(3)=$

2. Interpret the meaning of $C(3)\text{:}$

• A. There were $97$ customers in the store $3$ hours after the store opened.

• B. There were $3$ customers in the store $97$ hours after the store opened.

• C. In $3$ hours since the store opened, there were a total of $97$ customers.

• D. In $3$ hours since the store opened, the store had an average of $97$ customers per hour.

3. Solve $C(t)=43$ for $t\text{.}$ $t=$

4. Interpret the meaning of Part c’s solution(s):

• A. There were $43$ customers in the store $1$ hours after the store opened, and again $6$ hours after the store opened.

• B. There were $43$ customers in the store either $1$ hours after the store opened, or $6$ hours after the store opened.

• C. There were $43$ customers in the store $1$ hours after the store opened.

• D. There were $43$ customers in the store $6$ hours after the store opened.

###### 21

Find the axis of symmetry and vertex of the quadratic function.

${y}={4x^{2}+32x-3}$

Axis of symmetry:

Vertex:

###### 22

Find the axis of symmetry and vertex of the quadratic function.

${y}={x^{2}-4x+5}$

Axis of symmetry:

Vertex:

###### 23

Find the axis of symmetry and vertex of the quadratic function.

${y}={3x^{2}-15x+2}$

Axis of symmetry:

Vertex:

###### 24

Find the axis of symmetry and vertex of the quadratic function.

${y}={4x^{2}+12x-4}$

Axis of symmetry:

Vertex:

###### 25

For $y=4x^2-8x+5\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 26

For $y=2x^2+4x+7\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 27

For $y=-x^2+4x+2\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 28

For $y=-x^2+2x-5\text{,}$ determine the vertex, create a table of ordered pairs, and then graph the function.

###### 29

You will build a rectangular sheep enclosure next to a river. There is no need to build a fence along the river, so you only need to build on three sides. You have a total of $500$ feet of fence to use. Find the dimensions of the pen such that you can enclose the maximum possible area. One approach is to let $x$ represent the length of fencing that runs perpendicular to the river, and write a formula for a function of $x$ that outputs the area of the enclosure. Then find its vertex and interpret it.

The length of the pen (parallel to the river) should be , the width (perpendicular to the river) should be , and the maximum possible area is .

###### 30

You will build a rectangular sheep enclosure next to a river. There is no need to build a fence along the river, so you only need to build on three sides. You have a total of $400$ feet of fence to use. Find the dimensions of the pen such that you can enclose the maximum possible area. One approach is to let $x$ represent the length of fencing that runs perpendicular to the river, and write a formula for a function of $x$ that outputs the area of the enclosure. Then find its vertex and interpret it.

The length of the pen (parallel to the river) should be , the width (perpendicular to the river) should be , and the maximum possible area is .

###### 31

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+4x+3}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 32

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={-x^{2}-3x-2}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 33

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+x+5}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 34

Find the $y$-intercept and any $x$-intercept(s) of the quadratic function ${y}={x^{2}+8x+4}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 35

Find the $y$-intercept and any $x$-intercept(s) of the parabola with equation ${y}={16x^{2}-8x+1}\text{.}$

$y$-intercept:

$x$-intercept(s):

###### 36

Find the $y$-intercept and any $x$-intercept(s) of the parabola with equation ${y}={25x^{2}-1}\text{.}$

$y$-intercept:

$x$-intercept(s):

Graph each curve by algebraically determining its key features.

###### 37

$y=x^2-7x+12$

###### 38

$y=x^2+5x-14$

###### 39

$y=-x^2-x+20$

###### 40

$y=-x^2+4x+21$

###### 41

$y=x^2+6x$

###### 42

$y=x^2-8x$

###### 43

$y=x^2+4x+7$

###### 44

$y=x^2-2x+6$

###### 45

$y=2x^2-4x-30$

###### 46

$y=3x^2+21x+36$

###### 47

From a clifftop over the ocean ${170\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${264.6\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-4.9t^{2}+264.6t+170}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height (above sea level) in meters.

1. After , this object reached its maximum height of .

2. This object flew for before it landed in the ocean.

3. This object was above sea level ${47\ {\rm s}}$ after its release.

4. This object was ${3663.7\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.

###### 48

From a clifftop over the ocean ${130\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${294\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ On its way down it fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-4.9t^{2}+294t+130}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height (above sea level) in meters.

1. After , this object reached its maximum height of .

2. This object flew for before it landed in the ocean.

3. This object was above sea level ${18\ {\rm s}}$ after its release.

4. This object was ${967.9\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.