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Intermediate Algebra: Functions and Graphs

Katherine Yoshiwara

Section 3.3 Graphing Parabolas

Investigation 3.3.1. Perimeter and Area.

Do all rectangles with the same perimeter, say 36 inches, have the same area? Two different rectangles with perimeter 36 inches are shown. The first rectangle has base 10 inches and height 8 inches, and its area is 80 square inches. The second rectangle has base 12 inches and height 6 inches. Its area is 72 square inches.
graph
  1. The table shows the bases of various rectangles, in inches. Each rectangle has a perimeter of 36 inches. Fill in the height and the area of each rectangle. (To find the height of the rectangle, reason as follows: The base plus the height makes up half of the rectangle’s perimeter.)
    Base Height Area
    \(10\) \(8\) \(80\)
    \(12\) \(6\) \(72\)
    \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(14\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(17\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(19\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(2\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(11\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(16\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(15\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(1\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(6\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(8\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(13\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(7\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    grid
  2. What happens to the area of the rectangle when we change its base? On the grid above, plot the points with coordinates (Base, Area). (For this graph we will not use the heights of the rectangles.) The first two points, \((10, 80)\) and \((12, 72)\text{,}\) are shown. Connect your data points with a smooth curve.
  3. What are the coordinates of the highest point on your graph?
  4. Each point on your graph represents a particular rectangle with perimeter 36 inches. The first coordinate of the point gives the base of the rectangle, and the second coordinate gives the area of the rectangle. What is the largest area you found among rectangles with perimeter 36 inches? What is the base for that rectangle? What is its height?
  5. Describe the rectangle that corresponds to the point \((13, 65)\text{.}\)
  6. Find two points on your graph with vertical coordinate 80.
  7. If the rectangle has area 80 square inches, what is its base? Why are there two different answers here? Describe the rectangle corresponding to each answer.
  8. Now we’ll write an algebraic expression for the area of the rectangle in terms of its base. Let \(x\) represent the base of the rectangle. First, express the height of the rectangle in terms of \(x\text{.}\) (Hint: If the perimeter of the rectangle is 36 inches, what is the sum of the base and the height?) Now write an expression for the area of the rectangle in terms of \(x\text{.}\)
  9. Use your formula from part (8) to compute the area of the rectangle when the base is 5 inches. Does your answer agree with the values in your table and the point on your graph?
  10. Use your formula to compute the area of the rectangle when \(x=0\) and when \(x=18\text{.}\) Describe the “rectangles” that correspond to these data points.
  11. Continue your graph to include the points corresponding to \(x=0\) and to \(x=18\text{.}\)

Subsection 3.3.1 Introduction

The graph of the quadratic equation \(y = ax^2 + bx + c\) is called a parabola. Some parabolas are shown below.
two parabolas with labeled features
All these parabolas share certain features.
  • The graph has either a highest point (if the parabola opens downward, as in figure (a) or a lowest point (if the parabola opens upward, as in figure (b). This high or low point is called the vertex of the graph.
  • The parabola is symmetric about a vertical line, called the axis of symmetry, that runs through the vertex.
  • A parabola has a \(y\)-intercept, and it may have zero, one, or two \(x\)-intercepts.
  • If there are two \(x\)-intercepts, they are equidistant from the axis of symmetry.
The values of the constants \(a\text{,}\) \(b\text{,}\) and \(c\) determine the location and orientation of the parabola. We’ll consider each of these constants separately.

Checkpoint 3.3.1. QuickCheck 1.

Which point on a parabola always lies on the axis of symmetry?
  • The \(x\)-intercept
  • The \(y\)-intercept
  • The vertex
  • The origin
Answer.
\(\text{Choice 3}\)
Solution.
The vertex

Subsection 3.3.2 The Graph of \(y = ax^2\)

Use your calculator to graph the following three equations in the standard window, as shown below:
\begin{align*} y \amp = x^2\\ y \amp = 3x^2\\ y \amp = 0.1x^2 \end{align*}
3 parabolas
You can see that the graph of \(y=3x^2\) is narrower than the basic parabola, and the graph of \(y=0.1x^2\) is wider. As \(x\) increases, the graph of \(y=3x^2\) increases faster than the basic parabola, and the graph of \(y=0.1x^2\) increases more slowly. Compare the corresponding \(x\)-values for the three graphs shown in the table.
\(x\) \(y=x^2\) \(y=3x^2\) \(y=0.1x^2\)
\(-2\) \(4\) \(12\) \(0.4\)
\(1\) \(1\) \(3\) \(0.1\)
\(3\) \(9\) \(27\) \(0.9\)
For each \(x\)-value, the points on the graph of \(y=3x^2\) are higher than the points on the basic parabola, while the points on the graph of \(y=0.1x^2\) are lower. Multiplying by a positive constant greater than 1 stretches the graph vertically, and multiplying by a positive constant less than 1 squashes the graph vertically.
What about negative values for \(a\text{?}\) Consider the graphs of
\begin{align*} y \amp = x^2\\ y \amp = -2x^2\\ y \amp = -0.5x^2 \end{align*}
parabolas
We see that multiplying \(x^2\) by a negative constant reflects the graph about the \(x\)-axis. These parabolas open downward.

The Graph of \(y=ax^2\).

  • The parabola opens upward if \(a \gt 0\)
  • The parabola opens downward if \(a \lt 0\)
  • The magnitude of \(a\) determines how wide or narrow the parabola is.
  • The vertex, the \(x\)-intercepts, and the \(y\)-intercept all coincide at the origin.

Example 3.3.2.

Sketch by hand a graph of each quadratic equation.
  1. \(\displaystyle y = 2x^2\)
  2. \(\displaystyle y = -\dfrac{1}{2}x^2\)
Solution.
Both equations have the form \(y = ax^2\text{.}\) The graph of \(y = 2x^2\) opens upward because \(a = 2 \gt 0\text{,}\) and the graph of \(y = -\dfrac{1}{2}x^2\) opens downward because \(a = -\dfrac{1}{2}\lt 0\text{.}\)
To make a reasonable sketch by hand, it is enough to plot a few guidepoints; the points with \(x\)-coordinates \(1\) and \(-1\) are easy to compute.
two vertically scaled parabolas
\(x\) \(y=2x^2\) \(y=-\frac{1}{2}x^2\)
\(-1\) \(2\) \(-\dfrac{1}{2}\)
\(0\) \(0\) \(0\)
\(1\) \(2\) \(-\dfrac{1}{2}\)
We sketch parabolas through each set of guidepoints, as shown at left.

Checkpoint 3.3.3. QuickCheck 2.

True or false.
  1. If \(a\gt 1\text{,}\) the graph of \(y=ax^2\) is wider than the basic parabola.
    • True
    • False
  2. The vertex of a parabola lies on its axis of symmetry.
    • True
    • False
  3. If the \(y\)-intercept is negative, the parabola opens downward.
    • True
    • False
  4. A parabola may have one, two, or three \(x\)-intercepts.
    • True
    • False
Answer 1.
\(\text{False}\)
Answer 2.
\(\text{True}\)
Answer 3.
\(\text{False}\)
Answer 4.
\(\text{False}\)
Solution.
  1. False
  2. True
  3. False
  4. False

Checkpoint 3.3.4. Practice 1.

basic parabola and four vertically scaled parabolas
Match each parabola in the figure above with its equation. The basic parabola is shown in black.
  1. \(y = -\dfrac{3}{4}x^2\)
    • I
    • II
    • III
    • IV
  2. \(y = \dfrac{1}{4}x^2\)
    • I
    • II
    • III
    • IV
  3. \(y = \dfrac{5}{2}x^2\)
    • I
    • II
    • III
    • IV
  4. \(y = -\dfrac{5}{4}x^2\)
    • I
    • II
    • III
    • IV
Answer 1.
\(\text{III}\)
Answer 2.
\(\text{II}\)
Answer 3.
\(\text{I}\)
Answer 4.
\(\text{IV}\)
Solution.
  1. III
  2. II
  3. I
  4. IV

Subsection 3.3.3 The Graph of \(y= x^2 + c\)

Next, we consider the effect of the constant term, \(c\text{,}\) on the graph. Compare the graphs of
\begin{align*} y \amp = x^2\\ y \amp = x^2+4\\ y \amp = x^2-4 \end{align*}
vertically shifted parabolas
The graph of \(y=x^2+4\) is shifted upward four units compared to the basic parabola, and the graph of \(y=x^2-4\) is shifted downward four units. Look at the table, which shows the \(y\)-values for the three graphs.
\(x\) \(y=x^2\) \(y=x^2+4\) \(y=x^2-4\)
\(-1\) \(1\) \(5\) \(-3\)
\(0\) \(0\) \(4\) \(-4\)
\(2\) \(4\) \(8\) \(0\)
Each point on the graph of \(y=x^2+4\) is four units higher than the corresponding point on the basic parabola, and each point on the graph of \(y=x^2-4\) is four units lower. In particular, the vertex of the graph of \(y=x^2+4\) is the point \((0,4)\text{,}\) and the vertex of the graph of \(y=x^2-4\) is the point \((0,-4)\text{.}\)

The Graph of \(y=x^2+c\).

Compared to the graph of \(y=x^2\text{,}\) the graph of \(y=x^2+c\)
  • is shifted upward by \(c\) units if \(c \gt 0\text{.}\)
  • is shifted downward by \(c\) units if \(c \lt 0\text{.}\)

Example 3.3.5.

Sketch graphs for the following quadratic equations.
  1. \(\displaystyle y = x^2 - 2\)
  2. \(\displaystyle y = -x^2 + 4\)
Solution.
  1. The graph of \(y = x^2 - 2\) is shifted downward by two units, compared to the basic parabola. The vertex is the point \((0, -2)\) and the \(x\)-intercepts are the solutions of the equation
    \begin{equation*} 0 = x^2 - 2 \end{equation*}
    or \(\sqrt{2}\) and \(-\sqrt{2}\text{.}\) The graph is shown below.
    two vertically shifted parabolas
  2. The graph of \(y = -x^2 + 4\) opens downward and is shifted \(4\) units up, compared to the basic parabola. Its vertex is the point \((0, 4)\text{.}\) Its \(x\)-intercepts are the solutions of the equation
    \begin{equation*} 0 = -x^2 + 4 \end{equation*}
    or \(2\) and \(-2\text{.}\) You can verify both graphs with your graphing calculator.

Checkpoint 3.3.6. QuickCheck 3.

Match each equation with the description of its graph.
  1. \(y=x^2-3\)
    • i
    • ii
    • iii
    • iv
  2. \(y=-3x^2\)
    • i
    • ii
    • iii
    • iv
  3. \(y=3-x^2\)
    • i
    • ii
    • iii
    • iv
  4. \(y=\dfrac{1}{3}x^2\)
    • i
    • ii
    • iii
    • iv
  1. Wider than the basic parabola.
  2. Shifted down 3 units from the basic parabola.
  3. Opens downward, narrower than the basic parabola.
  4. Opens downward, shifted up 3 units.
Answer 1.
\(\text{ii}\)
Answer 2.
\(\text{iii}\)
Answer 3.
\(\text{iv}\)
Answer 4.
\(\text{i}\)
Solution.
  1. ii
  2. iii
  3. iv
  4. i

Checkpoint 3.3.7. Practice 2.

vertically shifted parabola
  1. Find an equation for the parabola shown above.
    \(y=\)
  2. Give the \(x\)- and \(y\)-intercepts of the graph.
    \(x\)-intercepts: Note: Use “sqrt(2)” to get \(\sqrt{2}\text{,}\) and separate different ordered pairs with a comma.
    \(y\)-intercept:
Answer 1.
\(x^{2}-5\)
Answer 2.
\(\left(\sqrt{5},0\right), \left(-\sqrt{5},0\right)\)
Answer 3.
\(\left(0,-5\right)\)
Solution.
  1. \(\displaystyle y = x^2 - 5\)
  2. \(x\)-intercepts: \((-\sqrt{5},0),(\sqrt{5},0)\text{;}\) \(y\)-intercept: \((0,-5)\)

Subsection 3.3.4 The Graph of \(y = ax^2 + bx\)

How does the linear term, \(bx\text{,}\) affect the graph?

Example 3.3.8.

Describe the graph of the equation
\begin{equation*} y = 2x^2 + 8x \end{equation*}
Solution.
The graph in the standard window is shown below. We see that the axis of symmetry for this parabola is not the \(y\)-axis: the graph is shifted to the left, compared to the basic parabola. We find the \(y\)-intercepts of the graph by setting \(y\) equal to zero:
\begin{align*} 0 \amp= 2x^2 + 8x\\ \amp= 2x(x + 4) \end{align*}
The solutions of this equation are \(0\) and \(-4\text{,}\) so the \(x\)-intercepts are the points \((0, 0)\) and \((-4, 0)\text{.}\)
shifted parabola
We can find the vertex of the graph by using the symmetry of the parabola. The \(x\)-coordinate of the vertex lies exactly half-way between the \(x\)-intercepts, so we average their \(x\)-coordinates to find:
\begin{equation*} x=\frac{1}{2}[0 + (-4)] = -2 \end{equation*}
To find the \(y\)-coordinate of the vertex, substitute \(x = -2\) into the equation for the parabola:
\begin{align*} y \amp= 2(\alert{-2})^2 + 8(\alert{-2})\\ \amp= 8 - 16 = -8 \end{align*}
Thus, the vertex is the point \((-2,-8)\text{.}\)

Checkpoint 3.3.9. Practice 3.

  1. Find the \(x\)-intercepts and the vertex of the parabola \(y = 6x - x^2\text{.}\)
    \(x\)-intercepts: Note: Use a comma to separate different points.
    Vertex:
  2. Verify your answers by graphing the function in the window
    \begin{equation*} \begin{aligned} \text{Xmin} \amp = -9.4 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10 \end{aligned} \end{equation*}
Answer 1.
\(\left(0,0\right), \left(6,0\right)\)
Answer 2.
\(\left(3,9\right)\)
Solution.
\(x\)-intercepts: \((0, 0)\) and \((6, 0)\text{;}\) vertex: \((3, 9)\)

Checkpoint 3.3.10. QuickCheck 4.

True or false.
  1. The axis of symmetry of \(y=ax^2+bx\) is shifted horizontally, compared to the basic parabola.
    • True
    • False
  2. The \(x\)-coordinate of the vertex is the average of the \(x\)-intercepts.
    • True
    • False
  3. The \(y\)-coordinate of the vertex is the same as the \(y\)-intercept.
    • True
    • False
  4. We can use extraction of roots to solve \(ax^2+bx=0\text{.}\)
    • True
    • False
Answer 1.
\(\text{True}\)
Answer 2.
\(\text{True}\)
Answer 3.
\(\text{False}\)
Answer 4.
\(\text{False}\)
Solution.
  1. True
  2. True
  3. False
  4. False

Subsection 3.3.5 A Formula for the Vertex

We can find a formula for the vertex of any parabola of the form
\begin{equation*} y = ax^2 + bx \end{equation*}
First, find the \(x\)-intercepts of the graph by setting \(y\) equal to zero and solving for \(x\text{.}\)
parabola with vertex and line of symmetry
\begin{align*} 0 \amp= ax^2 + bx \amp\amp \blert{\text{Factor.}} \\ \amp= x(ax + b) \amp\amp \blert{\text{Set each factor equal to zero.}} \\ x \amp=0 ~~~\text{or}~~~ ax + b = 0 \amp\amp \blert{\text{Solve for}~x.} \\ x \amp=0 ~~~\text{or}~~~ x =\frac{-b}{a} \end{align*}
The \(x\)-intercepts are the points \((0, 0)\) and \((\dfrac{-b}{a}, 0)\text{.}\)
Next, we find the \(x\)-coordinate of the vertex by taking the average of the two \(x\)-intercepts.
\begin{equation*} x=\frac{1}{2}\left[0+\left(\frac{-b}{a}\right)\right]=\frac{-b}{2a} \end{equation*}
Now we have a formula for the \(x\)-coordinate of the vertex.

Vertex of a Parabola.

For the graph of \(y = ax^2 + bx\text{,}\) the \(x\)-coordinate of the vertex is
\begin{equation*} \blert{x_v=\frac{-b}{2a}} \end{equation*}
We find the \(y\)-coordinate of the vertex by substituting its \(x\)-coordinate into the equation for the parabola.

Example 3.3.11.

  1. Find the vertex of the graph of \(f(x) = -1.8x^2 - 16.2x\text{.}\)
  2. Find the \(x\)-intercepts of the graph.
Solution.
  1. The \(x\)-coordinate of the vertex is
    \begin{equation*} x_v=\frac{-b}{2a}=\frac{-(-16.2)}{2(-1.8)}=-4.5 \end{equation*}
    To find the \(y\)-coordinate of the vertex, evaluate \(f(x)\) at \(x = \alert{-4.5}\text{.}\)
    \begin{equation*} y_v=-1.8(\alert{-4.5})^2 - 16.2(\alert{-4.5}) = 36.45 \end{equation*}
    The vertex is \((-4.5, 36.45)\text{.}\)
  2. To find the \(x\)-intercepts of the graph, set \(y = 0\) and solve.
    \begin{align*} -1.8x^2 - 16.2x \amp= 0\amp\amp \blert{\text{Factor.}}\\ -x(1.8x + 16.2) \amp= 0\amp\amp \blert{\text{Set each factor equal to zero.}}\\ -x =0 \hphantom{blank} 1.8x + 16.2 \amp = 0\amp\amp \blert{\text{Solve each equation.}}\\ x = 0 \hphantom{blankblankblan} x \amp= -9 \end{align*}
    The \(x\)-intercepts of the graph are \((0, 0)\) and \((-9, 0)\text{.}\) The graph is shown below.
downward opening parabola

Checkpoint 3.3.12. QuickCheck 5.

parabola
parabola
parabola
parabola
Match each equation with its graph.
  1. \(y=x^2-2\)
    • i
    • II
    • III
    • IV
  2. \(y=-x^2-2x\)
    • I
    • II
    • III
    • IV
  3. \(y=x^2+2x\)
    • I
    • II
    • III
    • IV
  4. \(y=-2x^2\)
    • I
    • II
    • III
    • IV
Answer 1.
\(\text{IV}\)
Answer 2.
\(\text{I}\)
Answer 3.
\(\text{II}\)
Answer 4.
\(\text{III}\)
Solution.
  1. IV
  2. I
  3. II
  4. III

Checkpoint 3.3.13. Practice 4.

Find the intercepts and the vertex of the graph of \(y = 6.4x-3.6x^2\text{.}\)
Intercepts: Note: Use a comma to separate different points.
Vertex:
Answer 1.
\(\left(0,0\right), \left(1.8,0\right)\)
Answer 2.
\(\left(0.9,2.916\right)\)
Solution.
\((0,0)\text{,}\)\((1.8,0)\text{,}\) \((0.9,2.916)\)

Exercises 3.3.6 Problem Set 3.3

Warm Up

Exercise Group.
For problems 1 and 2, evaluate.
1.
\(~2x^2-3x-1~\text{,}\) for \(~x=-2\)
Answer.
\(13\)
2.
\(~3+4x-3x^2~\text{,}\) for \(~x=-3\)
Exercise Group.
For problems 3 and 4, which technique would you use to solve the equation, extracting roots or factoring? Then solve the equation.
3.
  1. \(\displaystyle 3x^2-15=0\)
  2. \(\displaystyle 3x^2-15x=0\)
  3. \(\displaystyle (2x-3)^2=9\)
  4. \(\displaystyle 2x^2-3x=9\)
4.
  1. \(\displaystyle 20x-2x^2\)
  2. \(\displaystyle 20=2x^2\)
  3. \(\displaystyle 4x^2=2+2x\)
  4. \(\displaystyle 4(x+2)^2-1=0\)
Answer.
  1. factoring; \(0,10\)
  2. extraction of roots; \(\pm \sqrt{10}\)
  3. factoring; \(\dfrac{-1}{2}, 1\)
  4. extraction of roots; \(\dfrac{-5}{2}, \dfrac{-5}{2}\)
Exercise Group.
For Problems 5 and 6, factor the right side of the formula.
5.
\(A=\dfrac{1}{2}bh + \dfrac{1}{2}h^2\)
6.
\(S=2 \pi R^2+2 \pi RH\)
Answer.
\(2 \pi R(R+H)\)
Exercise Group.
For Problems 7 and 8, solve the equation.
7.
\(-9x=81x^2\)
8.
\(0=-140x-4x^2\)
Answer.
\(0, -35\)

Skills Practice

9.
Match each equation with its graph. In each equation, \(k \gt 0\text{.}\)
  1. \(\displaystyle y=x^2+k\)
  2. \(\displaystyle y=x^2+kx \)
  3. \(\displaystyle y=kx^2\)
  4. \(\displaystyle y=kx \)
  5. \(\displaystyle y=x+k\)
  6. \(\displaystyle y=x^2-k \)
parabolas
Answer.
  1. ii
  2. iv
  3. i
  4. iii
  5. vi
  6. v
10.
Match each equation with its graph. In each equation, \(k \gt 0\text{.}\)
  1. \(\displaystyle y=-kx\)
  2. \(\displaystyle y=-kx^2 \)
  3. \(\displaystyle y=k-x^2\)
  4. \(\displaystyle y=x-k \)
  5. \(\displaystyle y=k-x\)
  6. \(\displaystyle y=x^2-kx \)
parabolas
Exercise Group.
For Problems 11 and 12:
  1. Describe what each graph will look like compared to the basic parabola.
  2. Sketch the graph by hand and label the coordinates of three points on the graph.
11.
  1. \(\displaystyle y=2x^2\)
  2. \(\displaystyle y=\dfrac{1}{2}x^2\)
  3. \(\displaystyle y=-x^2\)
Answer.
    1. Narrower
    2. graph y=2x-squared
    1. Wider
    2. graph y=2x-squared
    1. Opens downward
    2. graph y=neg x-squared
12.
  1. \(\displaystyle y=x^2+2\)
  2. \(\displaystyle y=x^2-9\)
  3. \(\displaystyle y=100-x^2\)
Exercise Group.
For Problems 13–16, find the \(x\)-intercepts and the vertex of the graph. Then sketch the graph by hand.
13.
\(y=x^2-4x\)
Answer.
\((0,0), (4,0); (2,-4)\)
parabola
14.
\(y=x^2+2x\)
15.
\(y=3x^2+6x\)
Answer.
\((0,0), (-2,0); (-1,-3)\)
parabola
16.
\(y=-2x^2+5x\)
17.
\(y=40x-2x^2\)
Answer.
\((0,0), (20,0); (10,200)\)
parabola
18.
\(y=144-12x^2\)

Applications

19.
Commercial fishermen rely on a steady supply of fish in their area. To avoid overfishing, they adjust their harvest to the size of the population. The formula
\begin{equation*} R=0.4x-0.001x^2 \end{equation*}
gives the annual rate of growth, in tons per year, of a fish population of biomass \(x\) tons.
  1. Find the vertex of the graph. What does it tell us about the fish population?
  2. Find the \(x\)-intercepts of the graph. What do they tell us about the fish population?
  3. Sketch the graph for \(0 \le x \le 5000\text{.}\)
    grid
  4. For what values of \(x\) does the fish population decrease rather than increase? Suggest a reason why the population might decrease.
Answer.
  1. \(\displaystyle (2000, 400)\)
  2. \(\displaystyle (0,0), (4000,0)\)
  3. parabola
  4. \(\displaystyle x \gt 4000\)
20.
After it lands on Earth, the distance the space shuttle travels is given by
\begin{equation*} d=vT+\dfrac{v^2}{2a} \end{equation*}
where \(v\) is the shuttle’s velocity in ft/sec at touchdown, \(T\) is the pilot’s reaction time before the brakes are applied, and \(a\) is the shuttle’s deceleration.
  1. Suppose that for a particular landing, \(T=0.5\) seconds and \(a=12\) ft/sec. Write the equation for \(d\) in terms of \(v\text{,}\) using these values.
  2. Find the coordinates of the vertex and the horizontal intercepts of the graph of your equation. Use these points to help you graph the equation.
  3. Explain the meaning of the vertex and the intercepts, if any, in this context. As \(v\) increases, what happens to \(d\text{?}\)
  4. The runway at Edwards Air Force base is 15,000 feet long. Graph your equation again in an appropriate (larger) window, and use it to estimate the answer to the question: What is the maximum velocity the shuttle can have at touchdown and still stop on the runway?
Exercise Group.
For problems 21–24, graph the set of equations in the standard window on your calculator. Describe how the parameters (constants) \(a\text{,}\)\(b\text{,}\) and \(c\) in each equation transform the graph, compared to the basic parabola.
21.
  1. \(\displaystyle y=x^2\)
  2. \(\displaystyle y=3x^2\)
  3. \(\displaystyle y=\dfrac{1}{4}x^2\)
  4. \(\displaystyle y=-2x^2\)
Answer.
  1. basic parabola
  2. narrower
  3. wider
  4. narrower and reflected about \(y\)-axis
22.
  1. \(\displaystyle y=x^2\)
  2. \(\displaystyle y=x^2+1\)
  3. \(\displaystyle y=x^2+3\)
  4. \(\displaystyle y=x^2-6\)
23.
  1. \(\displaystyle y=x^2-4x\)
  2. \(\displaystyle y=x^2+4x\)
  3. \(\displaystyle y=4x-x^2\)
  4. \(\displaystyle y=-x^2-4x\)
Answer.
  1. \(x\)-intercepts at 0 and 4
  2. \(x\)-intercepts at 0 and \(-4\)
  3. reflection of (a) about \(y\)-axis
  4. reflection of (b) about \(y\)-axis
24.
  1. \(\displaystyle y=x^2-4x\)
  2. \(\displaystyle y=2x^2-8x\)
  3. \(\displaystyle y=\dfrac{1}{2}x^2-2x\)
  4. \(\displaystyle y=-x^2+4x\)
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