ORCCA Graphing Quadratic Expressions

Section 9.3 Graphing Quadratic Expressions

Objectives: PCC Course Content and Outcome Guide

Figure 9.3.1. Alternative Video Lesson

We have learned how to visually locate the key features of quadratic graphs and how to find the vertex algebraically. In this section we’ll explore how to find the intercepts algebraically and use their coordinates to more precisely graph a quadratic equation. Then we will see how to interpret the key features in context and distinguish between quadratic and other graphs.

Let’s start by looking at a quadratic equation that models the path of a baseball after it is hit by Ignacio, the batter. The height of the baseball, $H\text{,}$ measured in feet, after $t$ seconds is given by $H=-16t^2+75t+4.7\text{.}$ We know the graph will have 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 important ideas correspond to the vertical intercept, the vertex, and one of the horizontal intercepts.

The graph of this equation is shown in Figure 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 — Figure 9.3.2. Graph of $H=-16t^2+75t+4.7$

Subsection 9.3.1 Finding the Vertical and Horizontal Intercepts Algebraically

In List 9.2.12, we identified that the vertical intercept occurs where the graph 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 can substitute $0$ for $x$ to find the value of $y\text{.}$

The horizontal intercepts occur where the graph 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 can substitute $0$ for $y$ and find the value(s) of $x\text{.}$

Example 9.3.3.

Find the intercepts for the quadratic equation $y=x^2-4x-12$ using algebra.

To determine the $y$-intercept, we substitute $x=0$ and find $y=\substitute{0}^2-4(\substitute{0})-12=-12\text{.}$ 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 $y=0$ and solve for $x\text{:}$

\begin{align*} \substitute{0}\amp=x^2-4x-12\\ x\amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-12)}}{2(1)}\\ \amp=\frac{4\pm\sqrt{16+48}}{2}\\ \amp=\frac{4\pm\sqrt{64}}{2}\\ \amp=\frac{4\pm8}{2} \end{align*}

\begin{align*} x\amp=\frac{4-8}{2}\amp \text{or}\amp\amp x\amp=\frac{4+8}{2}\\ x\amp=\frac{-4}{2}\amp \text{or}\amp\amp x\amp=\frac{12}{2}\\ x\amp=-2\amp \text{or}\amp\amp x\amp=6 \end{align*}

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

Notice in Example 3 that the $y$-intercept was $(0,-12)$ and the value of $c$ was $-12\text{.}$ 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 equation occurs at the point $(0,c)$ where $c$ is the constant term, because substituting $x=0$ leaves only the constant term.

Example 9.3.5.

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

Explanation.

To determine the vertical intercept, take the constant term $-7\text{,}$ and recognize that the $y$-intercept is at the point $(0,-7)\text{.}$

To determine the horizontal intercepts, we’ll set $y=0$ and solve for $x\text{:}$

\begin{align*} 0\amp=-x^2+5x-7\\ 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 Equations Using Their Key Features

To graph a quadratic equation using its key features, we can use algebra to determine the following: whether the parabola opens upward or downward, the vertical intercept, the horizontal intercepts and the vertex. Then we can graph the points and connect them with a smooth curve.

Example 9.3.6.

Graph the quadratic equation $y=2x^2+10x+8$ by algebraically determining its key features.

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

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

Next, we’ll find the horizontal intercepts by setting $y=0$ and solving for $x\text{:}$

\begin{equation*} 2x^2+10x+8=0 \end{equation*}

\begin{align*} x\amp=\frac{-10\pm\sqrt{10^2-4(2)(8)}}{2(2)}\\ \amp=\frac{-10\pm\sqrt{100-64}}{4}\\ \amp=\frac{-10\pm\sqrt{36}}{4}\\ \amp=\frac{-10\pm6}{4} \end{align*}

\begin{align*} x\amp=\frac{-10-6}{4}\amp \text{or}\amp\amp x\amp=\frac{-10+6}{4}\\ x\amp=\frac{-16}{4}\amp \text{or}\amp\amp x\amp=\frac{-4}{4}\\ 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}\\ \amp=-\frac{10}{2(2)}\\ \amp=-2.5 \end{align*}

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

\begin{align*} k\amp=2(\substitute{-2.5})^2+10(\substitute{-2.5})+8\\ \amp=12.5-25+8\\ \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 curve. We’ll start by drawing and scaling the axes so all of our key features will be displayed as shown in Figure 7. Next, we’ll plot these key points as shown in Figure 8. Finally, we’ll note that this parabola opens upward and connect these points with a smooth curve, as shown in Figure 9.

a Cartesian graph with the x-axis labeled from -5 to 5 and the y-axis labeled from -10 to 10 — Figure 9.3.7. Setting up the grid.

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 — Figure 9.3.8. Marking key features.

the key points are connected with a solid curve with arrows on both ends — Figure 9.3.9. Completing the graph.

Example 9.3.10.

Graph the quadratic equation $y=-x^2+4x-5$ by algebraically determining its key features.

To start, we’ll note that this parabola 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\\ \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 $x\text{.}$

\begin{equation*} -x^2+4x-5=0 \end{equation*}

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

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

\begin{align*} k\amp=-(\substitute{2})^2+4(\substitute{2})-5\\ \amp=-4+8-5\\ \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 — Figure 9.3.11. Setting up the grid.

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) — Figure 9.3.12. Marking key features.

the previous graph with the symmetric point of the y-intercept plotted at (4,-5) — 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 make a table with a few more values around the vertex, plot these, and then draw a smooth curve. This is shown in Figure 14 and Figure 15.

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

Figure 9.3.14. Determine additional points to plot.

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 — Figure 9.3.15. Completing the graph.

Subsection 9.3.3 Applications of Quadratic Equations

Now we have learned how to find all the key features of a quadratic equation algebraically. Here are some applications of quadratic equations so we can learn how to identify and interpret the vertex, intercepts and additional points in context.

Example 9.3.16.

Returning to the path of the baseball in Figure 2, the equation that represents the height of the baseball after Ignacio hit it, is $H=-16t^2+75t+4.7\text{.}$ The height is in 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 substitute $t=0$ to find $-16(\substitute{0})^2+75(\substitute{0})+4.7=4.7\text{.}$ 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 $H=0\text{.}$

\begin{align*} H\amp=0\\ -16t^2+75t+4.7\amp=0 \end{align*}

\begin{align*} t\amp=\frac{-75\pm \sqrt{75^2-4(-16)(4.7)}}{2(-16)}\\ \amp=\frac{-75\pm \sqrt{5925.8}}{-32}\\ \end{align*}
Rounding these two values with a calculator, we obtain:
\begin{align*} \amp\approx -0.06185, 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 equation $a=-16$ and $b=75\text{.}$ So we have:

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

And then we can find the vertex’s second coordinate:

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

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

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 $H$ when $t=1\text{:}$

\begin{align*} -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 $H=80$ and we have:

\begin{align*} H\amp=80\\ -16t^2+75t+4.7\amp=80\\ -16t^2+75t-75.3\amp=0 \end{align*}

\begin{align*} t\amp=\frac{-75\pm \sqrt{75^2-4(-16)(-75.3)}}{2(-16)}\\ \amp=\frac{-75\pm \sqrt{805.8}}{-32}\\ \end{align*}
Rounding these two values with a calculator, we obtain:
\begin{align*} \amp\approx 1.457, 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 $1000$ 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 substitute $1000$ for $n$ and we have:

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

If Keenan’s company sells $1000$ 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}\\ \amp=-\frac{520}{2(-0.01)}\\ \amp=26000 \end{align*}

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

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

The maximum profit is $\$6{,}706{,}000\text{,}$ which occurs if $26{,}000$ units are produced.
This question is giving a height of $0$ and asking us to find the time(s). So we will be finding the horizontal intercept(s). We 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)}\\ \amp=\frac{-520\pm\sqrt{268240}}{-0.02}\\ \amp\approx 104, 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 us the profit value. We 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)}\\ \amp=\frac{-520\pm\sqrt{228240}}{-0.02}\\ \amp\approx 2113, 49887 \end{align*}

The company will make $\$1{,}000{,}000$ in profit if they produce about $2113$ 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 after $t$ seconds is given by $H=0.7t^2-23t+200\text{,}$ for $0 \le t \le 30\text{.}$ The height is measured in feet.

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. So we find $H$ when $t=0\text{:}$

\begin{align*} 0.7(\substitute{0})^2-23(\substitute{0})+200\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 telling us to use $t=5$ and find $H\text{:}$

\begin{align*} 0.7(\substitute{5})^2-23(\substitute{5})+200\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 set $H=0$ and solve for $t$ using the quadratic formula:

\begin{align*} H\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 $H=0\text{.}$ That means the plane did not hit the ground.
This question is asking for the lowest point of the airplane so we should find the vertex. Using $a=0.7$ and $b=-23\text{,}$ we have:

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

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

\begin{align*} k\amp=0.7(\substitute{16.43})^2-23(\substitute{16.43})+200\\ \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 corresponding time(s) so we set $H=50$ and solve for $t$ using the quadratic formula:

\begin{align*} H\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)}\\ \amp=\frac{23\pm\sqrt{109}}{1.4}\\ \amp\approx 8.971, 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 Distinguishing Quadratic Equations from Other Equations

So far, we’ve seen that the graphs of quadratic equations are parabolas and have a specific curved with a vertex. We’ve also seen that they have the algebraic form of $y=ax^2+bx+c\text{.}$ Here, we practice recognizing a quadratic equation so that we can call to mind that the equation has these features, which may be useful in some application.

Example 9.3.19.

Determine if each equation is a quadratic equation.

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

Explanation.

As $y+5x^2-4=0$ can be re-written as $y=-5x^2+4\text{,}$ this equation is a quadratic equation.
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 is not a quadratic equation.
The equation $y=-5x+1$ is a linear equation, not a quadratic equation.
The equation $y=(x-6)^2+3$ can be re-written as $y=x^2-12x+39\text{,}$ so this is a quadratic equation.
The equation $y=\sqrt{x+1}+5$ is not a quadratic equation as $x$ is inside a radical, not squared.

Example 9.3.20.

Determine if each graph could be the graph of a quadratic equation.

a graph of a curve that oscillates up and down between the y-values of -1 and 1;the pattern continues from -infinity to infinity — Figure 9.3.21.

the graph of a curve that opens upward and has a minimum at (2,0) — Figure 9.3.22.

the graph of a line with a slope of -1/3 — Figure 9.3.23.

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 equation.
This graph looks like a parabola, and it’s possible that it represents a quadratic equation.
This graph does not appear to be a parabola, but looks like a straight line. It’s not likely that it represents a quadratic equation.

Reading Questions 9.3.5 Reading Questions

1.

Explain how to find a parabola’s $y$-intercept when you have the equation for the parabola.

2.

Why does a parabola sometimes have zero $x$-intercepts, sometimes have one, and sometimes have two?

3.

When you have the equation for a quadratic graph, what can you always try to use to find any horizontal intercepts?

Exercises 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 Equations.

53.

An object was shot up into the air with an initial vertical speed of $480$ feet per second. Its height as time passes can be modeled by the quadratic equation $y={-16t^{2}+480t}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $y$ 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 ${22\ {\rm s}}$ after its release.
This object was ${2576\ {\rm ft}}$ high at two times: once after its release, and again later after its release.

54.

An object was shot up into the air with an initial vertical speed of $512$ feet per second. Its height as time passes can be modeled by the quadratic equation $y={-16t^{2}+512t}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $y$ 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 ${13\ {\rm s}}$ after its release.
This object was ${1392\ {\rm ft}}$ high at two times: once after its release, and again later after its release.

55.

From an oceanside clifftop ${100\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${245\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ It fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic equation $y={-4.9t^{2}+245t+100}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $y$ 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 ${14\ {\rm s}}$ after its release.
This object was ${3118.4\ {\rm m}}$ above sea level twice: once after its release, and again later after its release.

56.

From an oceanside clifftop ${170\ {\rm m}}$ above sea level, an object was shot into the air with an initial vertical speed of ${264.6\ {\textstyle\frac{\rm\mathstrut m}{\rm\mathstrut s}}}\text{.}$ It fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic equation $y={-4.9t^{2}+264.6t+170}\text{.}$ Here $t$ represents the number of seconds since the object’s release, and $y$ 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 ${37\ {\rm s}}$ after its release.
This object was ${3252.1\ {\rm m}}$ 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, in feet, can be modeled by the equation $h={0.2t^{2}-4t+18}\text{,}$ where $t$ is in seconds. 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, in feet, can be modeled by the equation $h={t^{2}-6t+4}\text{,}$ where $t$ is in seconds. The plane

?
will
will not

hit the ground during this stunt.

59.

A submarine is traveling in the sea. Its depth, in meters, can be modeled by $d={-1.7t^{2}+13.6t-25.2}\text{,}$ 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, in meters, can be modeled by $d={-0.7t^{2}+7t-19.5}\text{,}$ 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 $280$ meters. Its height can be modeled by

\begin{equation*} h=-4.9t^2+100t+280\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 $320$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $320$ meters. Round your answers to two decimal places if needed.

The object’s height would be $320$ meters the first time at seconds, and then the second time at seconds.

62.

An object is launched upward at the height of $300$ meters. Its height can be modeled by

\begin{equation*} h=-4.9t^2+80t+300\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 $310$ meters twice before it hits the ground. Find how many seconds since the launch would the object’s height be $310$ meters. Round your answers to two decimal places if needed.

The object’s height would be $310$ meters the first time at seconds, and then the second time at seconds.

63.

Currently, an artist can sell $260$ paintings every year at the price of ${\$50.00}$ per painting. Each time he raises the price per painting by ${\$10.00}\text{,}$ he sells $10$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $260-10x$ paintings every year at the price of $50+10x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(50+10x)(260-10x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$23{,}400.00}$ per year from selling paintings, what new price should he set?

To earn ${\$23{,}400.00}$ 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 $220$ paintings every year at the price of ${\$50.00}$ per painting. Each time he raises the price per painting by ${\$5.00}\text{,}$ he sells $10$ fewer paintings every year.

Assume he will raise the price per painting $x$ times, then he will sell $220-10x$ paintings every year at the price of $50+5x$ dollars. His yearly income can be modeled by the equation:

\begin{equation*} i=(50+5x)(220-10x) \end{equation*}

where $i$ stands for his yearly income in dollars. If the artist wants to earn ${\$12{,}350.00}$ per year from selling paintings, what new price should he set?

To earn ${\$12{,}350.00}$ 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 equation $y = x^{2}+nx+p\text{.}$ Let $n$ and $p$ be real numbers. Give your answers as points.

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

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