ORCCA Graphing Quadratic Functions

Section 9.3 Graphing Quadratic Functions

Figure 9.3.1 Alternative Video Lesson

We have learned how to locate the key features of quadratic functions on a graph and find the vertex algebraically. In this section we'll explore how to find the intercepts algebraically and use their coordinates to graph a quadratic function. Then we will see how to interpret the key features in context and distinguish between quadratic and other functions.

Let's start by looking at a quadratic function that models the path of a baseball after it is hit by Ignacio, the batter. The height of the baseball, $H(t)\text{,}$ measured in feet, after $t$ seconds is given by $H(t)=-16t^2+75t+4.7\text{.}$ We know this quadratic function has the shape of a parabola and we want to know the initial height, the maximum height, and the amount of time it takes for the ball to hit the ground if it is not caught. These key features correspond to the vertical intercept, the vertex, and one of the horizontal intercepts.

The graph of this function is shown in Figure 9.3.2. We cannot easily read where the intercepts occur from the graph because they are not integers. We previously covered how to determine the vertex algebraically. In this section, we'll learn how to find the intercepts algebraically. Then we'll come back to this example and find the intercepts for the path of the baseball.

a graph of a parabola that opens downward; the x-intercepts are not integer values

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Figure 9.3.2 Graph of $H(t)=-16t^2+75t+4.7$

Subsection 9.3.1 Finding the Vertical and Horizontal Intercepts Algebraically

In List 9.2.13, we identified that the vertical intercept occurs where the graph of a function intersects the vertical axis. If we're using $x$ and $y$ as our variables, the $x$-value on the vertical axis is $x=0\text{.}$ We will substitute $0$ for $x$ to find the value of $y\text{.}$ In function notation, we find $f(0)\text{.}$

The horizontal intercepts occur where the graph of a function intersects the horizontal axis. If we're using $x$ and $y$ as our variables, the $y$-value on the horizontal axis is $y=0\text{,}$ so we will substitute $0$ for $y$ and find the value(s) of $x\text{.}$ In function notation, we solve the equation $f(x)=0\text{.}$

Here is an example where we find the vertical and horizontal intercepts.

Example 9.3.3

Find the intercepts for the quadratic function $f(x)=x^2-4x-12$ algebraically.

To determine the $y$-intercept, we find $f(0)=\substitute{0}^2-4(\substitute{0})-12=-12\text{.}$ So the $y$-intercept occurs where $y=-12\text{.}$ On a graph, this is the point $(0,-12)\text{.}$

To determine the $x$-intercept(s), we set $f(x)=0$ and solve for $x\text{:}$

\begin{align*} \substitute{0}\amp=x^2-4x-12\\ 0\amp=(x-6)(x+2) \end{align*}

\begin{align*} x-6\amp=0\amp \text{or}\amp\amp x+2\amp=0\\ x\amp=6\amp \text{or}\amp\amp x\amp=-2 \end{align*}

The $x$-intercepts occur where $x=6$ and where $x=-2\text{.}$ On a graph, these are the points $(6,0)$ and $(-2,0)\text{.}$

Notice in Example 9.3.3 that the $y$-intercept was $(0,-12)$ and the value of $c=-12\text{.}$ When we substitute $0$ for $x$ we will always get the value of $c\text{.}$

Fact 9.3.4

The vertical intercept of a quadratic function occurs at the point $(0,c)$ because $f(0)=c\text{.}$

Example 9.3.5

Algebraically determine any horizontal and vertical intercepts of the quadratic function $f(x)=-x^2+5x-7\text{.}$

Explanation

To determine the vertical intercept, we find $f(0)-(0)^2+5(0)-7=-7\text{.}$ Thus the $y$-intercept occurs at the point $(0,-7)\text{.}$

To determine the horizontal intercepts, we'll set $f(x)=0$ and solve for $x\text{:}$

\begin{align*} 0\amp=-x^2+5x-7\\ \end{align*}

This equation cannot be solved using factoring so we'll use the quadratic formula:

\begin{align*} x\amp=\frac{-5\pm \sqrt{5^2-4(-1)(-7)}}{2(-1)}\\ x\amp=\frac{-5\pm \sqrt{-3}}{-2} \end{align*}

The radicand is negative so there are no real solutions to the equation. This means there are no horizontal intercepts.

Subsection 9.3.2 Graphing Quadratic Functions Using Their Key Features

To graph a quadratic function using its key features, we will algebraically determine the following: whether the function opens upward or downward, the vertical intercept, the horizontal intercepts and the vertex. Then we will graph the points and connect them with a smooth curve.

Example 9.3.6

Graph the function $f$ where $f(x)=2x^2+10x+8$ by algebraically determining its key features.

To start, we'll note that this function will open upward, since the leading coefficient is positive.

To find the $y$-intercept, we evaluate $f(0)=2(\substitute{0})^2+10(\substitute{0})+8=8\text{.}$ The $y$-intercept is $(0,8)\text{.}$

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

\begin{align*} 2x^2+10x+8\amp=0\\ 2(x^2+5x+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 $a=2$ and $b=10\text{,}$ we have:

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{10}{2(2)}\\ h\amp=-2.5 \end{align*}

Using this $x$-value to find the $y$-coordinate, we have:

\begin{align*} k\amp=f(-2.5)=2(\substitute{-2.5})^2+10(\substitute{-2.5})+8\\ k\amp=12.5-25+8\\ k\amp=-4.5 \end{align*}

The vertex is the point $(-2.5,-4.5)\text{,}$ and the axis of symmetry is the line $x=-2.5\text{.}$

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 9.3.7. Next, we'll plot these key points as shown in Figure 9.3.8. Finally, we'll note that this parabola opens upward and connect these points with a smooth curve, as shown in Figure 9.3.9.

a Cartesian graph with the x-axis labeled from -5 to 5 and the y-axis labeled from -10 to 10

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the same graph with the key points added; the vertex is at (-2.5,-4.5); the y-intercept is at (0,8); the x-intercepts are at (-4,0) and (-1,0); the axis of symmetry is drawn as a dotted vertical, x=-2.5, line passing through the vertex

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the key points are connected with a solid curve with arrows on both ends

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Figure 9.3.7 Setting up the grid.

Figure 9.3.8 Marking key features.

Figure 9.3.9 Completing the graph.

Example 9.3.10

Graph the function for which $y=-x^2+4x-5$ by algebraically determining its key features.

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

To find the $y$-intercept, we'll substitute $x$ with $0\text{:}$

\begin{align*} y\amp=-(\substitute{0})^2+4(\substitute{0})-5\\ y\amp=-5 \end{align*}

The $y$-intercept is $(0,-5)\text{.}$

Next, we'll find the horizontal intercepts by setting $y=0$ and solving for $x\text{.}$ We cannot use factoring to solve this equation so we'll use the quadratic formula:

\begin{align*} -x^2+4x-5\amp=0\\ x\amp=\frac{-4\pm \sqrt{(4)^2-4(-1)(-5)}}{2(-1)}\\ x\amp=\frac{-4\pm \sqrt{16-20}}{-2}\\ x\amp=\frac{-4\pm \sqrt{-8}}{-2} \end{align*}

The radicand is negative, so there are no real solutions to the equation. This is a parabola that does not have any horizontal intercepts.

To determine the vertex, we'll use $a=-1$ and $b=4\text{:}$

\begin{align*} x\amp=-\frac{4}{2(-1)}\\ x\amp=2 \end{align*}

Using this $x$-value to find the $y$-coordinate, we have:

\begin{align*} y\amp=-(\substitute{2})^2+4(\substitute{2})-5\\ y\amp=-4+8-5\\ y\amp=-1 \end{align*}

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

Plotting this information in an appropriate grid, we have:

a Cartesian graph with both axes labeled from -6 to 6

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the previous graph with the vertex added at (2,-1). The axis of symmetry is drawn as the dotted vertical line x=2; the y-intercept is plotted at (0,-5)

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the previous graph with the symmetric point of the y-intercept plotted at (4,-5)

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Figure 9.3.11 Setting up the grid.

Figure 9.3.12 Marking key features.

Figure 9.3.13 Using the axis of symmetry to determine one additional point.

Since we don't have any x-intercepts, we would like to have a few more points to graph. We will make a table with a few more values around the vertex, add these, and then draw a smooth curve. This is shown in Table 9.3.14 and Figure 9.3.15.

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

the previous graph with the additional points from the table added; the points are connected with a solid parabola that opens downward with arrows on both ends

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Table 9.3.14 Determine additional function values.

Figure 9.3.15 Completing the graph.

Subsection 9.3.3 Applications of Quadratic Functions

Now we have learned how to find all the key features of a quadratic function algebraically. Here are some applications of quadratic functions so we can learn how to identify and interpret the vertex, intercepts and additional points in context. Let's look at a few examples.

Example 9.3.16

Returning to the path of the baseball in Figure 9.3.2, the function that represents the height of the baseball after Ignacio hit it, is $H(t)=-16t^2+75t+4.7\text{.}$ The height is is feet and the time, $t\text{,}$ is in seconds. Find and interpret the following, in context.

The vertical intercept.
The horizontal intercept(s).
The vertex.
The height of the baseball $1$ second after it was hit.
The time(s) when the baseball is $80$ feet above the ground.

Explanation

To determine the vertical intercept, we'll find $H(0)=-16(\substitute{0})^2+75(\substitute{0})+4.7=4.7\text{.}$ The vertical intercept occurs at $(0,4.7)\text{.}$ This is the height of the baseball at time $t=0\text{,}$ so the initial height of the baseball was $4.7$ feet.
To determine the horizontal intercepts, we'll solve $H(t)=0\text{.}$ Since factoring is not a possibility to solve this equation, we'll use the quadratic formula:

\begin{align*} H(t)\amp=0\\ -16t^2+75t+4.7\amp=0\\ t\amp=\frac{-75\pm \sqrt{75^2-4(-16)(4.7)}}{2(-16)}\\ t\amp=\frac{-75\pm \sqrt{5925.8}}{-32}\\ \end{align*}
Rounding these two values with a calculator, we obtain:
\begin{align*} t\amp\approx -0.06185,\ t\approx 4.749 \end{align*}

The horizontal intercepts occur at approximately $(-0.06185,0)$ and $(4.749,0)\text{.}$ If we assume that the ball was hit when $t=0\text{,}$ a negative time does not make sense. The second horizontal intercept tells us that the ball hit the ground after approximately $4.75$ seconds.
The vertex occurs at $t=h=-\frac{b}{2a}\text{,}$ and for this function $a=-16$ and $b=75\text{.}$ So we have:

\begin{align*} h\amp=-\frac{75}{2(-16)}\\ h\amp=2.34375 \end{align*}

We can now find the output for this input:

\begin{align*} H(\substitute{2.34375})\amp=-16(\substitute{2.34375})^2+75(\substitute{2.34375})+4.7\\ \amp\approx 92.59 \end{align*}

Thus the vertex is $(2.344,92.59)\text{.}$

The vertex tells us that the baseball reached a maximum height of approximately $92.6$ feet about $2.3$ seconds after Ignacio hit it.
To find the height of the baseball after $1$ second, we can compute $H(1)\text{:}$

\begin{align*} H(\substitute{1})\amp=-16(\substitute{1})^2+75(\substitute{1})+4.7\\ \amp=63.7 \end{align*}

The height of the baseball was $63.7$ feet after $1$ second.
If we want to know when the baseball was $80$ feet in the air, then we set $H(t)=80$ and we have:

\begin{align*} H(t)\amp=80\\ -16t^2+75t+4.7\amp=80\\ -16t^2+75t-75.3\amp=0\\ t\amp=\frac{-75\pm \sqrt{75^2-4(-16)(-75.3)}}{2(-16)}\\ t\amp=\frac{-75\pm \sqrt{805.8}}{-32}\\ \end{align*}
Rounding these two values with a calculator, we obtain:
\begin{align*} t\amp\approx 1.457,\ t\approx 3.231 \end{align*}

The baseball was $80$ feet above the ground at two times, at about $1.5$ seconds on the way up and about $3.2$ seconds on the way down.

Example 9.3.17

The profit that Keenan's manufacturing company makes for producing $n$ refrigerators is given by $P=-0.01n^2+520n-54000\text{,}$ for $0 \le n \le 51{,}896\text{.}$

Determine the profit the company will make when they produce $1{,}000$ refrigerators.
Determine the maximum profit and the number of refrigerators produced that yields this profit.
How many refrigerators need to be produced in order for the company to “break even?” (In other words, for their profit to be $\$0\text{.}$)
How many refrigerators need to be produced in order for the company to make a profit of $\$1{,}000{,}000\text{?}$

Explanation

This question is giving us an input value and asking for the output value. We will substitute $1000$ for $n$ and we have:

\begin{align*} P\amp=-0.01(\substitute{1000})^2+520(\substitute{1000})-54000\\ P\amp=366000 \end{align*}

If Keenan's company sells $1{,}000$ refrigerators it will make a profit of $\$366{,}000\text{.}$
This question is asking for the maximum so we need to find the vertex. This parabola opens downward so the vertex will tell us the maximum profit and the corresponding number of refrigerators that need to be produced. Using $a=-0.01$ and $b=520\text{,}$ we have:

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{520}{2(-0.01)}\\ h\amp=26000 \end{align*}

Now we will find the value of $P$ when $n=26000\text{:}$

\begin{align*} P\amp=-0.01(\substitute{26000})^2+520(\substitute{26000})-54000\\ P\amp=6706000 \end{align*}

The maximum profit is $\$6{,}706{,}000\text{,}$ which occurs if $26{,}000$ units are produced.
This question is giving an output value of $0$ and asking us to find the input(s) so we will be finding the horizontal intercept(s). We will set $P=0$ and solve for $n$ using the quadratic formula:

\begin{align*} 0\amp=-0.01n^2+520n-54000\\ n\amp=\frac{-520\pm \sqrt{520^2-4(-0.01)(-54000)}}{2(-0.01)}\\ n\amp=\frac{-520\pm\sqrt{268240}}{-0.02}\\ n\amp\approx 104, n\approx 51896 \end{align*}

The company will break even if they produce about $104$ refrigerators or $51{,}896$ refrigerators. If the company produces more refrigerators than it can sell its profit will go down.
This question is giving an output value and asking us to find the input. To find the number of refrigerators that need to be produced for the company to make a profit of $\$1{,}000{,}000\text{,}$ we will set $P=1000000$ and solve for $n$ using the quadratic formula:

\begin{align*} 1000000\amp=-0.01n^2+520n-54000\\ 0\amp=-0.01n^2+520n-1054000\\ n\amp=\frac{-520\pm \sqrt{520^2-4(-0.01)(-1054000)}}{2(-0.01)}\\ n\amp\frac{-520\pm\sqrt{228240}}{-0.02}\\ n\amp\approx 2,113, n\approx 49,887 \end{align*}

The company will make $\$1{,}000{,}000$ in profit if they produce about $2{,}113$ refrigerators or $49{,}887$ refrigerators.

Example 9.3.18

Maia has a remote-controlled airplane and she is going to do a stunt dive where the plane dives toward the ground and back up along a parabolic path. The height of the plane is given by the function $H$ where $H(t)=0.7t^2-23t+200\text{,}$ for $0 \le t \le 30\text{.}$ The height is measured in feet and the time, $t\text{,}$ is measured in seconds.

Determine the starting height of the plane as the dive begins.
Determine the height of the plane after $5$ seconds.
Will the plane hit the ground, and if so, at what time?
If the plane does not hit the ground, what is the closest it gets to the ground, and at what time?
At what time(s) will the plane have a height of $50$ feet?

Explanation

This question is asking for the starting height which is the vertical intercept. We will find $H(0)\text{:}$

\begin{align*} H(\substitute{0})\amp=0.7(\substitute{0})^2-23(\substitute{0})+200\\ H(0)\amp=200 \end{align*}

When Maia begins the stunt, the plane has a height of $200$ feet. Recall that we can also look at the value of $c=200$ to determine the vertical intercept.
This question is giving an input of $5$ seconds and asking for the output so we will find $H(5)\text{:}$

\begin{align*} H(\substitute{5})\amp=0.7(\substitute{5})^2-23(\substitute{5})+200\\ H(5)\amp=102.5 \end{align*}

After $5$ seconds, the plane is $102.5$ feet above the ground.
The ground has a height of $0$ feet, so it is asking us to find the horizontal intercept(s) if there are any. We will set $H(t)=0$ and solve for $t$ using the quadratic formula:

\begin{align*} H(t)\amp=0.7t^2-23t+200\\ 0\amp=0.7t^2-23t+200\\ t\amp=\frac{23\pm \sqrt{(-23)^2-4(0.7)(200)}}{2(0.7)}\\ t\amp=\frac{23\pm\sqrt{-31}}{1.4} \end{align*}

The radicand is negative so there are no real solutions to the equation $H(t)=0\text{.}$ That means the plane did not hit the ground.
This question is asking for the lowest point of the plane so we will find the vertex. Using $a=0.7$ and $b=-23\text{,}$ we have:

\begin{align*} h\amp=-\frac{b}{2a}\\ h\amp=-\frac{(-23)}{2(0.7)}\\ h\amp\approx 16.43 \end{align*}

Now we will find the value of $H$ when $t\approx16.43\text{:}$

\begin{align*} H(\substitute{16.43})\amp=0.7(\substitute{16.43})^2-23(\substitute{16.43})+200\\ H(16.43)\amp\approx 11.07 \end{align*}

The minimum height of the plane is about $11$ feet, which occurs after about $16$ seconds.
This question is giving us a height and asking for the time(s) so we will set $H(t)=50$ and solve for $t$ using the quadratic formula:

\begin{align*} H(t)\amp=0.7t^2-23t+200\\ 50\amp=0.7t^2-23t+200\\ 0\amp=0.7t^2-23t+150\\ t\amp=\frac{23\pm \sqrt{(-23)^2-4(0.7)(150)}}{2(0.7)}\\ t\amp=\frac{23\pm\sqrt{109}}{1.4}\\ t\amp\approx 8.971, t\approx 23.89 \end{align*}

Maia's plane will be $50$ feet above the ground about 9 seconds and 24 seconds after the plane begins the stunt.

Subsection 9.3.4 The Domain and Range of Quadratic Applications

Let's identify the domain and range in each of the applications of quadratic functions in this section.

Example 9.3.19

In the baseball example in Figure 9.3.2, Ignacio hit the ball at $0$ seconds, and it lands on the ground at about $4.7$ seconds. The domain is $[0,4.7]\text{.}$

The baseball is at its lowest point when it hits the ground at $0$ feet, and the vertex is its highest point at about $92.6$ feet. The range is $[0,92.6]\text{.}$

Example 9.3.20

Identify the domain and range in Keenan's refrigerator company application in Example 9.3.17. Write them in interval and set-builder notation.

Explanation

The domain is given in the model as $0 \le n \le 51{,}896$ refrigerators. Limits are often stated with a mathematical model because only part of the function fits the real-world situation. The domain is $[0,51896]$ or $\{n\mid 0 \le n \le 51{,}896\}\text{.}$

When $0$ units are produced, the profit is $-\$54{,}000\text{.}$ The profit increases to a maximum value of $\$6{,}706{,}000$ at the vertex, and then goes back down to $\$0$ at $51{,}896$ units produced. So the range is $[-54000,6706000]$ or $\{P\mid -54{,}000 \le P \le 6{,}706{,}000\}\text{.}$

Example 9.3.21

Identify the domain and range of Maia's remote-controlled airplane application in Example 9.3.18. Write them in interval and set-builder notation.

Explanation

The domain is given in the model as $[0,30]$ seconds, because this parabola opens upward and the plane cannot keep flying up forever. In set-builder notation the domain is $\{t\mid 0 \le t \le 30\}\text{.}$

When $t=0$ seconds, the plane is $200$ feet above the ground. It dives down to a height of about $11$ feet and then flies up again. We need to know how high the plane is at 30 seconds to determine the range, so we find $H(30)\text{:}$

\begin{align*} H(\substitute{30})\amp=0.7(\substitute{30})^2-23(\substitute{30})+200\\ H(30)\amp=140 \end{align*}

The plane has returned to a height of $140$ feet after $30$ seconds. The starting point of $200$ feet is still the highest point, so the range is $[11,200]$ or $\{H(t)\mid 11 \le H(t) \le 200\}\text{.}$

Subsection 9.3.5 Distinguishing Quadratic Functions from Other Functions and Relations

So far, we've seen that the graphs of quadratic functions are parabolas and have a specific, curved shape. We've also seen that they have the algebraic form of $y=ax^2+bx+c\text{.}$ Here, we will learn to tell the difference between quadratic functions and other relations and functions.

Example 9.3.22

Determine if each relation represented algebraically is a quadratic function.

$y+5x^2-4=0$
$x^2+y^2=9$
$y=-5x+1$
$y=(x-6)^2+3$
$y=\sqrt{x+1}+5$

Explanation

As $y+5x^2-4=0$ can be re-written as $y=-5x^2+4\text{,}$ this equation represents a quadratic function.
The equation $x^2+y^2=9$ cannot be re-written in the form $y=ax^2+bx+c$ (due to the $y^2$ term), so this equation does not represent a quadratic function.
The equation $y=-5x+1$ represents a linear function, not a quadratic function.
The equation $y=(x-6)^2+3$ can be re-written as $y=x^2-12x+39\text{,}$ so this does represent a quadratic function.
The equation $y=\sqrt{x+1}+5$ does not represent a quadratic function as $x$ is inside a radical, not squared.

Example 9.3.23

Determine if each function represented graphically could represent a quadratic function.

a graph of a function that oscillates up and down between the y-values of -1 and 1;the pattern continues from -infinity to infinity

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the graph of a curve that opens upward and has a minimum at (2,0)

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the graph of a line with a slope of -1/3

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Figure 9.3.24 Function a.

Figure 9.3.25 Function b.

Figure 9.3.26 Function c.

Explanation

Since this graph has multiple maximum points and minimum points, it is not a parabola and it is not possible that it represents a quadratic function.
This graph looks like a parabola, and it's possible that it represents a quadratic function.
This graph does not appear to be a parabola, but looks like a straight line. It's not likely that it represents a quadratic function.

Subsection 9.3.6 Exercises

33

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

34

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

35

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

36

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

37

$y=x^2-8x+16$

38

$y=x^2+6x+9$

39

$y=x^2-4$

40

$y=x^2-9$

41

$y=x^2+6x$

42

$y=x^2-8x$

43

$y=-x^2+5x$

44

$y=-x^2+16$

45

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

46

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

47

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

48

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

49

$y=-x^2+4x-1$

50

$y=-x^2-x+3$

51

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

52

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

Applications of Quadratic Functions

53

An object was shot up into the air at an initial vertical speed of $320$ feet per second. Its height as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-16t^{2}+320t}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height in feet.

After , this object reached its maximum height of .
This object flew for before it landed on the ground.
This object was in the air ${12\ {\rm s}}$ after its release.
This object was ${1584\ {\rm ft}}$ high at two times: once after its release, and again later after its release.

54

An object was shot up into the air at an initial vertical speed of $384$ feet per second. Its height as time passes can be modeled by the quadratic function $f\text{,}$ where $f(t)={-16t^{2}+384t}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $f(t)$ represents the object’s height in feet.

After , this object reached its maximum height of .
This object flew for before it landed on the ground.
This object was in the air ${7\ {\rm s}}$ after its release.
This object was ${1520\ {\rm ft}}$ high at two times: once after its release, and again later after its release.

55

From a clifftop over the ocean ${200\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${156.8\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ 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 $f(t)={-4.9t^{2}+156.8t+200}\text{.}$ 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.

After , this object reached its maximum height of .
This object flew for before it landed in the ocean.
This object was above sea level ${26\ {\rm s}}$ after its release.
This object was ${748.8\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.

56

From a clifftop over the ocean ${160\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${176.4\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ 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 $f(t)={-4.9t^{2}+176.4t+160}\text{.}$ 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.

After , this object reached its maximum height of .
This object flew for before it landed in the ocean.
This object was above sea level ${8\ {\rm s}}$ after its release.
This object was ${1742.7\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.

57

A remote control aircraft will perform a stunt by flying toward the ground and then up. Its height can be modeled by the function $h(t)={1.2t^{2}-16.8t+55.8}\text{.}$ The plane

will
will not

hit the ground during this stunt.

58

A remote control aircraft will perform a stunt by flying toward the ground and then up. Its height can be modeled by the function $h(t)={0.1t^{2}-1.6t+10.4}\text{.}$ The plane

will
will not

hit the ground during this stunt.

59

A submarine is traveling in the sea. Its depth can be modeled by $d(t)={-0.9t^{2}+16.2t-72.9}\text{,}$ where $t$ stands for time in seconds. The submarine

will
will not

hit the sea surface along this route.

60

A submarine is traveling in the sea. Its depth can be modeled by $d(t)={-1.6t^{2}+28.8t-133.6}\text{,}$ where $t$ stands for time in seconds. The submarine

will
will not

hit the sea surface along this route.

61

An object is launched upward at the height of $400$ meters. It’s height can be modeled by

\begin{equation*} h=-4.9t^2+90t+400\text{,} \end{equation*}

where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $420$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $420$ meters. Round your answers to two decimal places if needed.

The object’s height would be $420$ meters the first time at seconds, and then the second time at seconds.

62

An object is launched upward at the height of $210$ meters. It’s height can be modeled by

\begin{equation*} h=-4.9t^2+70t+210\text{,} \end{equation*}

where $h$ stands for the object’s height in meters, and $t$ stands for time passed in seconds since its launch. The object’s height will be $250$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $250$ meters. Round your answers to two decimal places if needed.

The object’s height would be $250$ meters the first time at seconds, and then the second time at seconds.

63

Currently, an artist can sell $220$ paintings every year at the price of ${\$60.00}$ per painting. Each time he raises the price per painting by ${\$5.00}\text{,}$ he sells $5$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $220-5x$ paintings every year at the price of $60+5x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(60+5x)(220-5x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$18{,}375.00}$ per year from selling paintings, what new price should he set?

To earn ${\$18{,}375.00}$ per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.

64

Currently, an artist can sell $250$ paintings every year at the price of ${\$90.00}$ per painting. Each time he raises the price per painting by ${\$15.00}\text{,}$ he sells $5$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $250-5x$ paintings every year at the price of $90+15x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(90+15x)(250-5x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$37{,}125.00}$ per year from selling paintings, what new price should he set?

To earn ${\$37{,}125.00}$ per year, the artist could sell his paintings at two different prices. The lower price is per painting, and the higher price is per painting.

Challenge

65

Consider the function $f(x) = x^{2}+nx+p\text{.}$ Let $n$ and $p$ be real numbers. Give your answers as points.

Suppose the function has two real $x$-intercepts. What are they?
What is its $y$-intercept?
What is its vertex?

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