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Chapter 7

Quadratic Modeling

If you kick a ball through the air enough times, you will find its pathtends to be parabolic. Before we can answer any detailed questions aboutthis situation, we need to get our hands on a precise mathematical modelfor a parabolic shaped curve. This means we seek a function y = f(x)

whose graph reproduces the path of the ball.

ground level

Figure 7.1: Possible paths for a kicked ball are parabolic.

7.1 Parabolas and Vertex Form

OK, suppose we sit down with an xy-coordinate system and draw fourrandom parabolas; let’s label them I, II, III, and IV: See Figure 7.2. Therelationship between these parabolas and the fixed coordinate systemcan vary quite a bit: The key distinction between these four curves isthat only I and IV are the graphs of functions; this follows from the ver-tical line test. A parabola that is the graph of a function is called astandard parabola. We can see that any standard parabola has threebasic features:

85

86 CHAPTER 7. QUADRATIC MODELING

x-axis

y-axis

I

II

III

IV

not graphs of functions

Figure 7.2: Relationship between a fixed coordinate system and various parabolas.

• the parabola will either open “upward” or “downward”;

• the graph will have either a “highest point” or “lowest point,” calledthe vertex;

• the parabola will be symmetric about some vertical line called theaxis of symmetry.

Our first task is to describe the mathematical model for any standardparabola. In other words, what kind of function equations y = f(x) giveus standard parabolas as their graphs? Our approach is geometric andvisual:

• Begin with one specific example, then show every other standardparabola can be obtained from it via some specific geometric ma-neuvers.

• As we perform these geometric maneuvers, we keep track of how thefunction equation for the curve is changing.

This discussion will amount to a concrete application of a more generalset of tools developed in the following section of this chapter.

−6 −4 −2 2 4 6

5

10

15

20

25

30

35

x-axis

y-axis

Figure 7.3: Graph of y = x2.

Using a graphing device, it is an easy matter to plotthe graph of y = x2 and see we are getting the parabolapictured in Figure 7.3. The basic idea is to describe howwe can manipulate this graph and obtain any standardparabola. In the end, we will see that standard parabolasare obtained as the graphs of functions having the form

y = ax2 + bx+ c,

7.1. PARABOLAS AND VERTEX FORM 87

for various constants a, b, and c, with a 6= 0. A function of this type iscalled a quadratic function and these play a central role throughout thecourse. We will divide our task into two steps.

First we show every standard parabola arises as the graph of a func-tion having the form

y = a(x− h)2 + k,

for some constants a, h, and k, with a 6= 0. This is called the vertexform of a quadratic function. Notice, if we were to algebraically expandout this equation, we could rewrite it in the y = ax2 + bx + c form. Forexample, suppose we start with the vertex form y = 2(x − 1)2 + 3, sothat a = 2, h = 1, k = 3. Then we can rewrite the equation in the formy = ax2 + bx+ c as follows:

2(x− 1)2 + 3 = 2(x2 − 2x+ 1) + 3 = 2x2 − 4x + 5,

so a = 2, b = −4, c = 5. The second step is to show any quadratic functioncan be written in vertex form; the underlying algebraic technique usedhere is called completing the square. This is a bit more involved. Forexample, if you are simply handed the quadratic function y = −3x2+6x−1,it not at all obvious why the vertex form is obtained by this equality:

−3x2 + 6x− 1 = −3(x − 1)2 + 2.

The reason behind this equality is the technique of completing the square.In the end, we will almost always be interested in the vertex form of aquadratic. This is because a great deal of qualitative information aboutthe parabolic graph can simply be “read off” from this form.

7.1.1 First Maneuver: Shifting

−6 −4 −2 2 4 6

10

30

40

20

x-axis

y-axis

Figure 7.4: Shift to the right.

Suppose we start with the graph in Figure 7.3 and hori-zontally shift it h units to the right. To be specific, con-sider the two cases h = 2 and h = 4. To visualize this,imagine making a wire model of the graph, set in on topof the curve, then slide the wire model h units to the right.What you will obtain are the two “dashed curves” in Fig-ure 7.4. We will call the process just described a hori-zontal shift. Since the “dashed curves” are no longer theoriginal parabola in Figure 7.3, the corresponding func-tion equations must have changed.

Using a graphing device, you can check that the corresponding equa-tions for the dashed graphs would be

y = (x− 2)2

= x2 − 4x + 4,


which is the plot with lowest point (2, 0) and

y = (x− 4)2

= x2 − 8x+ 16,

which is the plot with lowest point (4, 0). In general, if h is positive, thegraph of the function y = (x− h)2 is the parabola obtained by shifting thegraph of y = x2 by h units to the right.

−6 −4 −2 2 4 6

20

30

40

10

x-axis

y-axis

Figure 7.5: Shift to the left.

Next, if h is negative, shifting h units to the right is thesame as shifting |h| units left! On the domain −6 ≤ x ≤ 6,Figure 7.5 indicates this for the cases h = −2, − 4, using“dashed curves” for the shifted graphs and a solid linefor the graph of y = x2. Using a graphing device, we cancheck that the corresponding equations for the dashedgraphs would be

y = (x− (−2))2

= (x+ 2)2

= x2 + 4x+ 4,

which is the plot with lowest point (−2, 0) and

y = (x− (−4))2

= (x+ 4)2

= x2 + 8x+ 16,

which is the plot with lowest point (−4, 0). In general, if h is negative, thegraph of the function y = (x− h)2 gives the parabola obtained by shiftingthe graph of y = x2 by |h| units to the LEFT.

The conclusion thus far is this: Begin with the graph of y = x2 inFigure 7.3. Horizontally shifting this graph h units to the right gives anew (standard) parabola whose equation is y = (x− h)2.

−6 −4 4 6

20

30

40

−2

10

−10

2

x-axis

y-axis

Figure 7.6: Vertical shifts.

We can also imagine vertically shifting the graph inFigure 7.3. This amounts to moving the graph k unitsvertically upward. It turns out that this vertically shiftedgraph corresponds to the graph of the function y = x2 + k.We can work out a few special cases and use a graphingdevice to illustrate what all this really means.

Figure 7.6 illustrates the graphs of y = x2 + k in thecases when k = 4, 10 and k = −4, −10, leading to verticallyshifted graphs. Positive values of k lead to the upper two

“dashed curves” and negative values of k lead to the lower two “dashedcurves”; the plot of y = x2 is again the solid line. The equations givingthese graphs would be y = x2 − 10, y = x2 − 4, y = x2 + 4 and y = x2 + 10,from bottom to top dashed plot.

If we combine horizontal and vertical shifting, we end up with thegraphs of functions of the form y = (x − h)2 + k. Figure 7.7(a) illustrates

7.1. PARABOLAS AND VERTEX FORM 89

the four cases with corresponding equations y = (x±2)2±4; as an exercise,identify which equation goes with each curve.

−6 −4 4 6

20

30

40

10

−2 2

x-axis

y-axis

(a) Combined shifts.

−6 −4 4 6

−30

−20

10

20

30

−2−10

2

x-axis

y-axis

(b) Reflections.

−6 −4 −2 2 4 6

5

10

15

20

25

30

x-axis

y-axis

(c) Vertical dilations.

Figure 7.7: Shifts, reflec-tions, and dilations.

7.1.2 Second Maneuver: Reflection

Next, we can reflect any of the curves y = p(x) obtainedby horizontal or vertical shifting across the x-axis. Thisprocedure will produce a new curve which is the graph ofthe new function y = −p(x). For example, begin with thefour dashed curves in the previous figure. Here are thereflected parabolas and their equations are y = −(x±2)2±4: See Figure 7.7(b).

7.1.3 Third Maneuver: Vertical Dilation

If a is a positive number, the graph of y = ax2 is usuallycalled a vertical dilation of the graph of y = x2. There aretwo cases to distinguish here:

• If a > 1, we have a vertically expanded graph.

• If 0 < a < 1, we have a vertically compressed graph.

This is illustrated for a = 2 (upper dashed plot) and a =

1/2 (lower dashed plot): See Figure 7.7(c).

7.1.4 Conclusion

Starting with y = x2 in Figure 7.3, we can combine to-gether all three of the operations: shifting, reflection anddilation. This will lead to the graphs of functions thathave the form:

y = a(x− h)2 + k,

for some a, h and k, a 6= 0. If you think about it for awhile,it seems pretty easy to believe that any standard parabolaarises from the one in Figure 7.3 using our three geomet-ric maneuvers. In other words, what we have shown isthat any standard parabola is the graph of a quadratic equation in vertexform. Let’s summarize.

Important Fact 7.1.1. A standard parabola is the graph of a function

y = f(x) = a(x − h)2 + k, for some constants a, h, and k and a 6= 0. The

vertex of the parabola is (h, k) and the axis of symmetry is the line x = h. If

a > 0, then the parabola opens upward; if a < 0, then the parabola opens

downward.


Example 7.1.2. Describe a sequence of geometric operations leading from

the graph of y = x2 to the graph of y = f(x) = −3(x− 1)2 + 2.

reflect across x-axis

horizontal shift by h = 1

vertical dilate by 3

vertical shift by k = 2

−3(x − 1)2 + 2

(a) What do the symbols of anequation mean?

−6 −4 −2 4 6

5

15

20

2

10

−5

−10

−15

x-axis

y-axis

(b) What does the equationlook like?

Figure 7.8: Interpreting anequation.

Solution. To begin with, we can make some initial conclu-sions about the specific shifts, reflections and dilationsinvolved, based on looking at the vertex form of the equa-tion. In addition, by Fact 7.1.1, we know that the vertexof the graph of y = f(x) is (1, 2), the line x = 1 is a verticalaxis of symmetry and the parabola opens downward.

We need to be a little careful about the order in whichwe apply the four operations highlighted. We will illus-trate a procedure that works. The full explanation for thesuccess of our procedure involves function compositionsand we will return to that at the end of Chapter 8. Theorder in which we will apply our geometric maneuvers isas follows:

horizontal shift ⇒ vertical dilate

⇒ reflect

⇒ vertical shift

Figure 7.8(b) illustrates the four curves obtained by ap-plying these successive steps, in this order. As a refer-ence, we include the graph of y = x2 as a “dashed curve”:

• A horizontal shift by h = 1 yields the graph ofy = (x− 1)2; this is the fat parabola opening upward

with vertex (1, 0).

• A dilation by a = 3 yields the graph of y = 3(x −

1)2; this is the skinny parabola opening upward withvertex (1, 0).

• A reflection yields the graph of y = −3(x − 1)2; this isthe downward opening parabola with vertex (1, 0).

• A vertical shift by k = 2 yields the graph of y = −3(x−

1)2 + 2; this is the downward opening parabola withvertex (1, 2).

7.2 Completing the Square

By now it is pretty clear we can say a lot about the graph of a quadraticfunction which is in vertex form. We need a procedure for rewriting agiven quadratic function in vertex form. Let’s first look at an example.

7.2. COMPLETING THE SQUARE 91

Example 7.2.1. Find the vertex form of the quadratic function y = −3x2 +

6x− 1.

Solution. Since our goal is to put the function in vertex form, we can writedown what this means, then try to solve for the unknown constants. Ourfirst step would be to write

−3x2 + 6x− 1 = a(x− h)2 + k,

for some constants a, h, k. Now, expand the right hand side of thisequation and factor out coefficients of x and x2:

−3x2 + 6x− 1 = a(x− h)2 + k

−3x2 + 6x− 1 = a(x2 − 2xh + h2) + k

−3x2 + 6x− 1 = ax2 − 2xah+ ah2 + k

(−3)x2 + (6)x+ (−1) = (a)x2 + (−2ah)x+ (ah2 + k).

If this is an equation, then it must be the case that the coefficients oflike powers of x match up on the two sides of the equation in Figure 7.9.Now we have three equations and three unknowns (the a, h, k) and we

(−3)︸︷︷︸ x2 +

︷︸︸︷(6) x + (−1)︸︷︷︸ = (a)︸︷︷︸ x

2 +︷︸︸︷(−2ah) x + (ah2 + k)︸︷︷︸

Equal

Equal

Equal

Figure 7.9: Balancing the coefficients.

can proceed to solve for these:

−3 = a

6 = −2ah

−1 = ah2 + k

The first equation just hands us the value of a = −3. Next, we can plugthis value of a into the second equation, giving us

6 = −2ah

= −2(−3)h

= 6h,


so h = 1. Finally, plug the now known values of a and h into the thirdequation:

−1 = ah2 + k

= −3(12) + k

= −3+ k,

so k = 2. Our conclusion is then

−3x2 + 6x− 1 = −3(x− 1)2 + 2.

Notice, this is the quadratic we studied in Example 7.1.2 on page 90.

The procedure used in the preceding example will always work torewrite a quadratic function in vertex form. We refer to this as completingthe square.

Example 7.2.2. Describe the relationship between the graphs of y = x2

and y = f(x) = −4x2 + 5x + 2.

−6 −4 −2 4 6

−20

10

20

2

−10

x-axis

y-axis

Figure 7.10: Maneuvering

y = x2.

Solution. We will go through the algebra to complete thesquare, then interpret what this all means in terms ofgraphical maneuvers. We have

−4x2 + 5x+ 2 = a(x− h)2 + k

(−4)x2 + 5x+ 2 = ax2 + (−2ah)x+ (ah2 + k).

This gives us three equations:

−4 = a

5 = −2ah

2 = ah2 + k.

We conclude that a = −4, h = 58= 0.625 and k = 57

16= 3.562. So, this tells

us that we can obtain the graph of y = f(x) from that of y = x2 by thesesteps:

• Horizontally shifting by h = 0.625 units gives y = (x− 0.625)2.

• Vertically dilate by the factor a = 4 gives y = 4(x − 0.625)2.

• Reflecting across the x-axis gives y = −4(x − 0.625)2.

• Vertically shifting by k = 3.562 gives y = f(x) = −4(x− 0.625)2 + 3.562.

Example 7.2.3. A drainage canal has a cross-section in the shape of a

parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the

top. If the water depth in the ditch is 5 feet, how wide is the surface of the

water in the ditch?

7.3. INTERPRETING THE VERTEX 93

��

��

centerline20 feet

10 feet

Figure 7.11: A drainagecanal.

Solution. Impose an xy-coordinate system so that theparabolic cross-section of the canal is symmetric aboutthe y-axis and its vertex is the origin. The vertex form ofany such parabola is y = f(x) = ax2, for some a > 0; thisis because (h, k) = (0, 0) is the vertex and the parabolaopens upward! The dimension information given tells usthat the points (10, 10) and (−10, 10) are on the graph off(x). Plugging into the expression for f, we conclude that 10 = 100a, soa = 0.1 and f(x) = (0.1)x2. Finally, if the water is 5 feet deep, we mustsolve the equation: 5 = (0.1)x2, leading to x = ±

√50 = ±7.07. Conclude the

surface of the water is 14.14 feet wide when the water is 5 feet deep.

7.3 Interpreting the Vertex

minimum

value f

(

−b

2a

)

−b

2a

−b

2a

vertex

vertex

maximum

value f

(

−b

2a

)

Figure 7.12: The vertex as the extremum of the quadratic function.

If we begin with a quadratic function y = f(x) = ax2+bx+c, we know thegraph will be a parabola. Graphically, the vertex will correspond to eitherthe “highest point” or “lowest point” on the graph. If a > 0, the vertexis the lowest point on the graph; if a < 0, the vertex is the highest pointon the graph. The maximum or minimum value of the function is thesecond coordinate of the vertex and the value of the variable x for whichthis extreme value is achieved is the first coordinate of the vertex. Aswe know, it is easy to read off the vertex coordinates when a quadraticfunction is written in vertex form. If instead we are given a quadraticfunction y = ax2 + bx + c, we can use the technique of completing thesquare and arrive at a formula for the coordinates of the vertex in termsof a, b, and c. We summarize this below and label the two situations(upward or downward opening parabola) in the Figure 7.12. Keep in


mind, it is always possible to obtain this formula by simply completingthe square.

Important Fact 7.3.1. In applications involving a quadratic function

f(x) = ax2 + bx+ c,

the vertex has coordinates P =(

−b2a, f(

−b2a

))

. The second coordinate of the

vertex will detect the maximum or minimum value of f(x); this is often a

key step in problem solving.

Example 7.3.2. Discuss the graph of the quadratic function y = f(x) =

−2x2 + 11x− 4.

−4 −2 2 4 8

−80

−60

−40

6

−20

Figure 7.13: Sketching y =

f(x).

Solution. We need to place the equation y = f(x) in vertexform. We can simply compute a = −2, h = −b

2a= 11

4and

k = f( 114) = 89

8, using Fact 7.3.1:

f(x) = −2x2 + 11x− 4

= −2

(

x−

(

11

4

))2

+89

8.

This means that the graph of f(x) is a parabola openingdownward with vertex

(

114, 898

)

and axis x = 114; see Fig-

ure 7.13.

7.4 Quadratic Modeling Problems

The real importance of quadratic functions stems from the connectionwith motion problems. Imagine one of the three kicked ball scenarios inFigure 7.1 and impose a coordinate system with the kicker located at theorigin. We can study the motion of the ball in two ways:

• Regard time t as the important variable and try to find a functiony(t) which describes the height of the ball t seconds after the ballis kicked; this would just be the y-coordinate of the ball at time t.The function y(t) is a quadratic function. If we had this function inhand, we could determine when the ball hits the ground by solvingthe equation 0 = y(t), but we would not be able to determine wherethe ball hits the ground.

• A second approach is to forget about the time variable and simplytry to find a function y = f(x) whose graph models the exact path ofthe ball. In particular, we could find where the ball hits the groundby solving 0 = f(x), but we would not be able to determine when theball hits the ground.

7.4. QUADRATIC MODELING PROBLEMS 95

path of kicked ball

ground level

cliff

(a) What it looks likephysically.

−2 1 2 3 5

−100

−50

50

4−1

(b) What it looks likegraphically.

Figure 7.14: Different viewsof the ball’s trajectory.

Example 7.4.1. Figure 7.14(a) shows a ball is located on

the edge of a cliff. The ball is kicked and its height (in feet)

above the level ground is given by the function s = y(t) =

−16t2 + 48t + 50, where t represents seconds elapsed after

kicking the ball. What is the maximum height of the ball

and when is this height achieved? When does the ball hit

the ground? How high is the cliff?

Solution. The function y(t) is a quadratic function witha negative leading coefficient, so its graph in the ts-coordinate system will be a downward opening parabola.We use a graphing device to get the picture in Fig-ure 7.14(b).

The vertex is the highest point on the graph, whichcan be found by writing y(t) in standard form usingFact 7.3.1:

y(t) = −16t2 + 48t+ 50

= −16

(

t−3

2

)2

+ 86.

The vertex of the graph of y(t) is(

32, 86)

, so the maximumheight of the ball above the level ground is 86 feet, occur-ing at time t = 3

2.

The ball hits the ground when its height above the ground is zero;using the quadratic formula:

y(t) = −16t2 + 48t+ 50

= 0

t =−48±

√

(48)2 − 4 (−16) (50)

2 · 16= 3.818 sec or − 0.818 sec

Conclude the ball hits the ground after 3.818 seconds. Finally, the heightof the cliff is the height of the ball zero seconds after release; i.e., y(0) = 50feet is the height of the cliff.

Here are two items to consider carefully:

1. The graph of y(t) is NOT the path followed by the ball! Finding theactual path of the ball is not possible unless additional informationis given. Can you see why?

2. The function y(t) is defined for all t; however, in the context of theproblem, there is no physical meaning when t < 0.

CAUTION!!!

!!!


The next example illustrates how we must be very careful to link thequestion being asked with an appropriate function.

Example 7.4.2. A hot air balloon takes off from the edge of a mountain

lake. Impose a coordinate system as pictured in Figure 7.15 and assume

that the path of the balloon follows the graph of y = f(x) = − 22500x2+ 4

5x. The

land rises at a constant incline from the lake at the rate of 2 vertical feet for

each 20 horizontal feet. What is the maximum height of the balloon above

lake level? What is the maximum height of the balloon above ground level?

Where does the balloon land on the ground? Where is the balloon 50 feet

above the ground?

100

200

500 1000

height above lake (ft)

lake

balloon

balloon path

(ft)

Figure 7.15: Visualizing.

Solution. In the coordinate system indicated, the origin isthe takeoff point and the graph of y = f(x) is the path ofthe balloon. Since f(x) is a quadratic function with a neg-ative leading coefficient, its graph will be a parabola whichopens downward. The difficulty with this problem is thatat any instant during the balloon’s flight, the “height ofthe balloon above the ground” and the “height of the bal-loon above the lake level” are different! The picture in

Figure 7.16 highlights this difference; consequently, two different func-tions will be needed to study these two different quantities.

200

100

500 1000

AA

lake

BB

height of balloon above lake level

height of balloon above ground level

height of ground above lake level

Figure 7.16: The height of the balloon y as a function of x.

The function y = f(x) keeps track of the height of the balloon abovelake level at a given x location on the horizontal axis. The line ℓ with slopem = 2

20= 1

10passing through the origin models the ground level. This says

that the function

y =1

10x

keeps track of the height of the ground above lake level at a given x

location on the horizontal axis.We can determine the maximum height of the balloon above lake level

by analyzing the parabolic graph of y = f(x). Putting f(x) in vertex form,via Fact 7.3.1,

f(x) = −2

2500(x− 500)2 + 200.


The vertex of the graph of y = f(x) is (500, 200). This just tells us that themaximum height of the balloon above lake level is 200 feet. To find thelanding point, we need to solve the system of equations

{y = − 2

2500x2 + 4

5x

y = 110x

}.

As usual, plugging the second equation into the first and solving for x,we get

1

10x = −

2

2500x2 +

4

5x

x2 = 875x

x2 − 875x = 0

x(x− 875) = 0

takeoff point

vertex (high point above lake)

landing point

horizontal distancefrom launch (feet)

height above lake level (feet)

200

100

500 1000

Figure 7.17: Locating the takeoff and landing points.

From the algebra, we see there are two solutions: x = 0 or x = 875;these correspond to the takeoff and landing points of the balloon, whichare the two places the flight path and ground coincide. (Notice, if we haddivided out x from the last equation, we would only get one solution; thetricky point is that we can’t divide by zero!) The balloon lands at theposition where x = 875 and to find the y coordinate of this landing pointwe plug x = 875 into our function for the balloon height above lake level:y = f(875) = 87.5 feet. So, the landing point has coordinates (875,87.5).

Next, we want to study the height of the balloon above the ground. Lety = g(x) be the function which represents the height of the balloon above


the ground when the horizontal coordinate is x. We find

g(x) =

{height of the balloonabove lake level withhorizontal coordinate x

}

−

{elevation of groundabove lake level withhorizontal coordinate x

}

= f(x) − g(x)

=

(

−2

2500x2 +

4

5x

)

︸︷︷︸(balloon above lake level)

−︸︷︷︸minus

(

1

10x

)

︸︷︷︸(ground above lake level)

= −2

2500(x − 437.5)2 + 153.12,

Notice that g(x) itself is a NEW quadratic function with a negative leadingcoefficient, so the graph of y = g(x) will be a downward opening parabola.The vertex of this parabola will be (437.5, 153.12), so the highest elevationof the balloon above the ground is 153.12 feet.

We can now sketch the graph of g(x) and the horizontal line deter-mined by y = 50 in a common coordinate system, as below. Findingwhere the balloon is 50 feet above the ground amounts to finding wherethese two graphs intersect. We need to now solve the system of equations

{y = − 2

2500x2 + 7

10x

y = 50

}.

Plug the second equation into the first and apply the quadratic for-mula to get x = 796.54 or 78.46. This tells us the two possible x coor-dinates when the balloon is 50 feet above the ground. In terms of theoriginal coordinate system imposed, the two places where the balloon is50 feet above the ground are (78.46, 57.85) and (796.54, 129.6).

7.4.1 How many points determine a parabola?

We all recall from elementary geometry that two distinct points in theplane will uniquely determine a line; in fact, we used this to derive equa-tions for lines in the plane. We could then ask if there is a similar char-acterization of parabolas.

Important Fact 7.4.3. Let P = (x1, y1), Q = (x2, y2) and R = (x3, y3) be three

distinct non-collinear points in the plane such that the x-coordinates are all

different. Then there exists a unique standard parabola passing through

these three points. This parabola is the graph of a quadratic function y =

f(x) = ax2 + bx + c and we can find these coefficients by simultaneously


graph of height above ground function

vertex (high point above ground)

line y = 50

places where 50 feet above ground

100

500 1000

Figure 7.18: Finding heights above the ground.

solving the system of three equations and three unknowns obtained by

assuming P,Q and R are points on the graph of y = f(x):

ax21 + bx1 + c = y1

ax22 + bx2 + c = y2

ax23 + bx3 + c = y3.

.

Example 7.4.4. Assume the value of a particular house in Seattle has

increased in value according to a quadratic function y = v(x), where the

units of y are in dollars and x represents the number of years the property

has been owned. Suppose the house was purchased on January 1, 1970

and valued at $50,000. In 1980, the value of the house on January 1

was $80,000. Finally, on January 1, 1990 the value was $200,000. Findthe value function v(x), determine the value on January 1, 1996 and find

when the house will be valued at $1,000,000.

Solution. The goal is to explicitly find the value of the function y = v(x).We are going to work in a xy-coordinate system in which the first co-ordinate of any point represents time and the second coordinate repre-sents value. We need to decide what kind of units will be used. Thex-variable, which represents time, will denote the number of years thehouse is owned. For the y-variable, which represents value, we coulduse dollars. But, instead, we will follow a typical practice in real estateand use the units of K, where K = $1,000. For example, a house valued


at $235,600 would be worth 235.6 K. These will be the units we use, whichessentially saves us from drowning in a sea of zeros!

We are given three pieces of information about the value of a particularhouse. This leads to three points in our coordinate system: P = (0, 50),Q = (10, 80) and R = (20, 200). If we plot these points, they do not lieon a common line, so we know there is a unique quadratic functionv(x) = ax2 + bx+ c whose graph (which will be a parabola) passes throughthese three points. In order to find the coefficients a, b, and c, we needto solve the system of equations:

a02 + b0+ c = 50

a(10)2 + b(10) + c = 80

a(20)2 + b(20) + c = 200

,

which is equivalent to the system

c = 50

100a+ 10b+ c = 80

400a+ 20b+ c = 200

.

Plugging c = 50 into the second two equations gives the system{100a+ 10b = 30

400a+ 20b = 150

}(7.1)

Solve the first equation for a, obtaining a = 30−10b100

, then plug this intothe second equation to get:

400

(

30− 10b

100

)

+ 20b = 150

120− 40b+ 20b = 150

b = −3

2.

Now, plug b = −32into the first equation of Equation 7.1 to get 100a +

10(

−32

)

= 30; i.e., a = 920. We conclude that

y = v(x) =9

20x2 −

3

2x + 50,

keeping in mind the units here are K.To find the value of the house on January 1, 1996, we simply note this

is after x = 26 years of ownership. Plugging in, we get y = v(26) = 920(26)2−

3226 + 50 = 315.2; i.e., the value of the house is $315,200. To find when

the house will be worth $1,000,000, we note that $1,000,000 = 1,000K andneed to solve the equation

1000 = v(x) =9

20x2 −

3

2x + 50

0 =9

20x2 −

3

2x − 950.

7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL? 101

By the quadratic formula,

x =

32±√

(

−32

)2− 4

(

920

)

(−950)

2(

920

)

=1.5±

√1712.25

0.9= 47.64 or − 44.31.

Because x represents time, we can ignore the negative solution and sothe value of the house will be $1,000,000 after approximately 47.64 yearsof ownership.

7.5 What’s Needed to Build a Quadratic Model?

Back in Fact 4.7.1 on page 44, we highlighted the information required todetermine a linear model. We now describe the quadratic model analog.

Important Facts 7.5.1. A quadratic model is completely determined by:

1. Three distinct non-collinear points, or

2. The vertex and one other point on the graph.

The first approach is just Fact 7.4.3. The second approach is basedon the vertex form of a quadratic function. The idea is that we know anyquadratic function f(x) has the form

f(x) = a(x− h)2 + k,

where (h, k) is the vertex. If we are given h and k, together with anotherpoint (x0, y0) on the graph, then plugging in gives this equation:

y0 = a(x0 − h)2 + k.

The only unknown in this equation is a, which we can solve for usingalgebra. A couple of the exercises will depend upon these observations.

7.6 Summary

• A quadratic function is one of the form

f(x) = ax2 + bx+ c.

where a 6= 0.


• The graph of a quadratic function is a parabola which is symmetricabout the vertical line through the highest (or lowest) point on thegraph. This highest (or lowest) point is known as the vertex of thegraph; its location is given by (h,k) where

h = −b

2aand k = f(h).

• If a > 0, then the vertex is the lowest (or minimum) point on thegraph, and the parabola ”opens upward”. If a < 0, then the vertexis the highest (or maximum) point on the graph, and the parabola”opens downward”.

• Every quadratic function can be expressed in the form

f(x) = a(x− h)2 + k

where (h,k) is the vertex of the function’s graph.

7.7. EXERCISES 103

7.7 Exercises

Problem 7.1. Write the following quadraticfunctions in vertex form, find the vertex, theaxis of symmetry and sketch a rough graph.

(a) f(x) = 2x2 − 16x + 41.

(b) f(x) = 3x2 − 15x − 77.

(c) f(x) = x2 − 37x+ 13.

(d) f(x) = 2x2.

(e) f(x) = 1100x2.

Problem 7.2. In each case, find a quadraticfunction whose graph passes through thegiven points:

(a) (0,0), (1,1) and (3, − 1).

(b) (−1,1), (1, − 2) and (3,4).

(c) (2,1), (3,2) and (5,1).

(d) (0,1), (1,1) and (1,3).

Problem 7.3. (a) Sketch the graph of thefunction f(x) = x2 − 3x+ 4 on the interval−3 ≤ x ≤ 5. What is the maximum valueof f(x) on that interval? What is the min-imum value of f(x) on that interval?

(b) Sketch the graph of the function f(x) =

x2−3x+4 on the interval 2 ≤ x ≤ 7. Whatis the maximum value of f(x) on that in-terval? What is the minimum value off(x) on that interval?

(c) Sketch the graph of the function g(x) =

−(x + 3)2 + 3 on the interval 0 ≤ x ≤ 4.What is the maximum value of g(x) onthat interval? What is the minimumvalue of g(x) on that interval?

Problem 7.4. If the graph of the quadraticfunction f(x) = x2 + dx + 3d has its vertex onthe x-axis, what are the possible values of d?What if f(x) = x2 + 3dx − d2 + 1 ?

Problem 7.5. The initial price of buzz.comstock is $10 per share. After 20 days the stockprice is $20 per share and after 40 days theprice is $25 per share. Assume that while theprice of the stock is not zero it can be modeledby a quadratic function.

(a) Find the multipart function s(t) givingthe stock price after t days. If you buy1000 shares after 30 days, what is thecost?

(b) To maximize profit, when should you sellshares? How much will the profit be onyour 1000 shares purchased in (a)?

Problem 7.6. Sketch the graph ofy = x2 − 2x − 3. Label the coordinates of thex and y intercepts of the graph. In thesame coordinate system, sketch the graph ofy = |x2 − 2x − 3|, give the multipart rule andlabel the x and y intercepts of the graph.

Problem 7.7. A hot air balloon takes off fromthe edge of a plateau. Impose a coordinate sys-tem as pictured below and assume that thepath the balloon follows is the graph of thequadratic function y = f(x) = − 4

2500x2 + 4

5x.

The land drops at a constant incline from theplateau at the rate of 1 vertical foot for each5 horizontal feet. Answer the following ques-tions:

takeoffhorizontal distancefrom launch (feet)

ground incline

height above plateau (feet)

balloon

(a) What is the maximum height of the bal-loon above plateau level?

(b) What is the maximum height of the bal-loon above ground level?

(c) Where does the balloon land on theground?

(d) Where is the balloon 50 feet above theground?

Problem 7.8. (a) Suppose f(x) = 3x2 − 2.Does the point (1,2) lie on the graph ofy = f(x)? Why or why not?

(b) If b is a constant, where does the liney = 1 + 2b intersect the graph of y =

x2 + bx+ b?


(c) If a is a constant, where does theline y = 1− a2 intersect the graph ofy = x2 − 2ax+ 1?

(d) Where does the graph of y = −2x2 + 3x+ 10

intersect the graph of y = x2 + x− 10?

Problem 7.9. Sylvia has an apple orchard.One season, her 100 trees yielded 140 applesper tree. She wants to increase her productionby adding more trees to the orchard. However,she knows that for every 10 additional treesshe plants, she will lose 4 apples per tree (i.e.,the yield per tree will decrease by 4 apples).How many trees should she have in the or-chard to maximize her production of apples?

Problem 7.10. Rosalie is organizing a circusperformance to raise money for a charity. Sheis trying to decide how much to charge for tick-ets. From past experience, she knows that thenumber of people who will attend is a linearfunction of the price per ticket. If she charges5 dollars, 1200 people will attend. If shecharges 7 dollars, 970 people will attend. Howmuch should she charge per ticket to make themost money?

Problem 7.11. A Norman window is a rectan-gle with a semicircle on top. Suppose that theperimeter of a particular Norman window is tobe 24 feet. What should its dimensions be inorder to maximize the area of the window and,therefore, allow in as much light as possible?

Problem 7.12. Jun has 300 meters of fenc-ing to make a rectangular enclosure. She alsowants to use some fencing to split the enclo-sure into two parts with a fence parallel to twoof the sides. What dimensions should the en-closure have to have the maximum possiblearea?

Problem 7.13. You have $6000 with whichto build a rectangular enclosure with fencing.The fencing material costs $20 per meter. Youalso want to have two parititions across thewidth of the enclosure, so that there will bethree separated spaces in the enclosure. Thematerial for the partitions costs $15 per meter.What is the maximum area you can achieve forthe enclosure?

Problem 7.14. Steve likes to entertain friendsat parties with “wire tricks.” Suppose he takes

a piece of wire 60 inches long and cuts it intotwo pieces. Steve takes the first piece of wireand bends it into the shape of a perfect circle.He then proceeds to bend the second piece ofwire into the shape of a perfect square. Whereshould Steve cut the wire so that the total areaof the circle and square combined is as smallas possible? What is this minimal area? Whatshould Steve do if he wants the combined areato be as large as possible?

Problem 7.15. Two particles are moving in thexy-plane. The move along straight lines at con-stant speed. At time t, particle A’s position isgiven by

x = t+ 2, y =1

2t− 3

and particle B’s position is given by

x = 12− 2t, y = 6−1

3t.

(a) Find the equation of the line along whichparticle A moves. Sketch this line, andlabel A’s starting point and direction ofmotion.

(b) Find the equation of the line along whichparticle B moves. Sketch this line on thesame axes, and label B’s starting pointand direction of motion.

(c) Find the time (i.e., the value of t) atwhich the distance between A and B isminimal. Find the locations of particlesA and B at this time, and label them onyour graph.

Problem 7.16. Sven starts walking due southat 5 feet per second from a point 120 feet northof an intersection. At the same time Rudyardstarts walking due east at 4 feet per secondfrom a point 150 feet west of the intersection.

(a) Write an expression for the distance be-tween Sven and Rudyard t seconds afterthey start walking.

(b) When are Sven and Rudyard closest?What is the minimum distance betweenthem?

Problem 7.17. After a vigorous soccer match,Tina and Michael decide to have a glass of

7.7. EXERCISES 105

their favorite refreshment. They each run ina straight line along the indicated paths at aspeed of 10 ft/sec. Parametrize the motion ofTina and Michael individually. Find when andwhere Tina and Michael are closest to one an-other; also compute this minimum distance.

Michael

Tina

(400,50)

(200,300)

soy milk(−50,275)

beet juice

Problem 7.18. Consider the equation: αx2 +

2α2x + 1 = 0. Find the values of x that makethis equation true (your answer will involve α).

Find values of α that make this equation true(your answer will involve x).

Problem 7.19. For each of the following equa-tions, find the value(s) of the constant α so thatthe equation has exactly one solution, and de-termine the solution for each value.

(a) αx2 + x+ 1 = 0

(b) x2 + αx+ 1 = 0

(c) x2 + x+ α = 0

(d) x2 + αx+ 4α + 1 = 0

Problem 7.20. (a) Solve for t

s = 2(t − 1)2 + 1

(b) Solve for x

y = x2 + 2x + 3

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