Evaluating
When we are evaluating a function, this means we know a value of the independent variable (an input) and we want to figure out the corresponding value of the dependent variable (the output).
Section 2.3 Using equations
So far we’ve seen examples of linear equations and exponential equations. There are lots of other kinds of equations that we might use! Whatever kind of equation we have, and however we figured it out, we can always use the equation to figure out the value of the dependent variable that’s associated with a particular value of the independent variable. We might call this process “plugging in,” but its official name is evaluating.
Just as important as evaluating a function is thinking about what our result means. Not only is this a good way to catch any calculator mistakes we might have made, but also it’s how algebra helps us make good decisions in the real world.
Opening Activity 2.3.1.
In Section 2.2, we looked at the equation
\begin{equation*}
H = 2.26\ast 1.067^{Y}
\end{equation*}
that predicted health care costs, \(H\text{,}\) measured in trillions of dollars, as a function of \(Y\text{,}\) the number of years since 2007.
-
Evaluate this equation at \(Y = 5\text{.}\)
-
Write a sentence explaining what your result means. To help you get started, here’s a template: “Our equation predicts that in years after 2007, health care costs were $ trillion.”
Subsection Braking distance
The Cadillac Escalade is a cross between a sports utility vehicle (SUV) and luxury car. Either way, it’s a big car. And it takes awhile to stop. One study showed that the 2010 Escalade traveling at 60 miles per hour takes about 144 feet to come to a complete stop from when the driver first hits the brakes. In fact, the braking distance of any car depends on how fast it is going. If someone is driving slowly they can stop in shorter distance than if they are driving fast. Which is why you should drive slowly on residential streets.
We would like to be able to calculate the braking distances at other speeds, so our two variables are
\begin{align*}
S \amp= \text{ speed (mph) } \sim \text{ indep} \\
B \amp= \text{ braking distance (feet) } \sim \text{ dep}
\end{align*}
Using the data and equations from physics, automobile analysts were able to determine that the equation relating these two variables is
\begin{equation*}
B=0.04S^2
\end{equation*}
Remember that the 0.04 written next to the \(S^2\) means they are multiplied. We might equally well have written
\begin{equation*}
B = 0.04 \ast S^2
\end{equation*}
You may be a little surprised to see the variable \(S\) squared (raised to the 2nd power) or wonder what the number 0.04 means. This equation is not something we can figure out because it relies both on the data and knowledge of the physics involved. But, we can still work with this equation to find the braking distances at any speed. (If you must know, this equation is only approximate since things like tire and road conditions are a factor, but for what we want it is good enough.)
Although in the last couple of sections we were able to find equations by generalizing examples, there are actually many different mathematical and statistical techniques for finding equations. A scientist might use lab experiments and some theory to figure it out. An economist might recognize that the equation fits a certain template because of the underlying economics. A store manager might know from years of experience that a certain equation works well to predict sales. It can be comforting to know where an equation comes from but whether we find an equation for ourselves or get it from an expert, we can use it to answer our questions and make predictions.
Now that we have an equation we should check the reported stopping distance at 60 mph. We have \(S=60\) so we substitute 60 in place of \(S\) in the equation to get
\begin{equation*}
B = 0.04 \ast 60^2 = 0.04 \times \underline{60} \wedge 2 = 144 \text{ feet} \quad \checkmark
\end{equation*}
What this means in our story is, if the car was traveling at 60 mph, it took 144 feet to stop.
Quick reminder about terminology. When we know the independent variable, like \(S=60\) and we substitute into the equation to find the dependent variable, like \(B\text{,}\) we say we evaluate the function \(B\) at \(S=60\text{.}\) You might have noticed that the 60 was underlined in the calculation above. In this book we underline the value of the independent variable when we are evaluating. That way it’s easier to see which numbers come from the equation and which number we’re plugging in. Feel free to do the same. Or not.
Let’s use our equation to calculate the braking distance for a Cadillac Escalade traveling 30, 50, 70 or 90 miles per hour. For 30 miles per hour, we have \(S=30\text{.}\) So, we evaluate at \(S=30\) to get
\begin{equation*}
B = 0.04 \ast 30^2 = 0.04\times \underline{30} \wedge 2 = 36 \text{ feet}
\end{equation*}
At 30 mph, it takes the Cadillac Escalade 36 feet to stop. As we expected, it doesn’t take nearly as far to stop as it did at 60 mph. For the other speeds we do the same thing: evaluate at the appropriate value of \(S\text{.}\) When \(S=50\) mph we get
\begin{equation*}
B = 0.04 \ast 50^2 = 0.04\times \underline{50}\wedge2 = 100 \text{ feet}
\end{equation*}
When \(S=70\) mph we get
\begin{equation*}
B = 0.04 \ast 70^2 = 0.04\times \underline{70}\wedge2 = 196\text{ feet}
\end{equation*}
When \(S=90\) mph we get
\begin{equation*}
B = 0.04 \ast 90^2 = 0.04\times \underline{90}\wedge2 = 324 \text{ feet}
\end{equation*}
And, what does our equation tell us when the speed is 0 mph? We evaluate at \(S=0\) mph to get
\begin{equation*}
B = 0.04 \ast 0^2 = 0.04\times \underline{0} \wedge 2 = 0 \text{ feet}
\end{equation*}
Well, sure! If the car isn’t moving, then it won’t need any distance to stop. Here’s what we’ve found so far, displayed in a table.
| \(S\) | 0 | 30 | 50 | 60 | 70 | 90 |
|---|---|---|---|---|---|---|
| \(B\) | 0 | 36 | 100 | 144 | 196 | 324 |
Activity 2.3.2.
Dontrell and Kim borrowed money to buy a house on a 30-year mortgage. After \(T\) months of making payments, Dontrell and Kim will still owe $\(D\) where
\begin{equation*}
D=236{,}000-56{,}000 \ast 1.004^T
\end{equation*}
\(D\) is also known as the payoff — how much they would need to pay to settle the debt.
For example, if we evaluate at \(T = 2\text{,}\)
\begin{equation*}
D = 236{,}000-56{,}000 \ast 1.004^2 \approx $179{,}550
\end{equation*}
This means that after 2 months of making payments, Dontrell and Kim still owe about $179,550. (Notice that’s not how much they’ve paid so far!)
-
Name the variables \(T\) and \(D\text{,}\) including units and dependence.
letter = everyday words (units) ~ dep or indep = \(\qquad\qquad\qquad\qquad\qquad\) \(\displaystyle\Big(\qquad\Big)\) ~ = \(\displaystyle\Big(\qquad\Big)\) ~ -
Evaluate the equation at \(T = 12\text{.}\) What does this number mean in terms of the story?
-
What number would \(T\) be at the very beginning of the story?
-
How much did Dontrell and Kim originally borrow to buy their house?
Subsection Adding in reaction time
My neighbor Jeff happens to drive a 2010 Cadillac Escalade. The other day he almost was in an accident on the highway. Luckily no one was hurt, but he had to slam on the brakes to stop. The police report mentioned they believe it took his car 183 feet to stop. Jeff says he was not driving over the posted speed limit of 65 mph. Should we believe him?
We can see from the table that braking distance of 183 feet falls in between the 144 and 196 on our table which leads us to believe that Jeff was traveling faster than 60 mph and slower than 70 mph. We can figure out if Jeff were driving at 65 mph, then his braking distance would have been
\begin{equation*}
B = 0.04 \ast 65^2 = 0.04\times \underline{65} \wedge 2 = 169 \text{ feet}
\end{equation*}
That’s less than the 183 feet Jeff took to stop. So, it appears that Jeff was driving faster than 65 mph.
But wait a minute. The braking distance is just the time it takes from when the driver’s foot hits the brake until the car stops. That distance doesn’t take into account the driver’s reaction time — how long between when the driver thinks to stop and when the driver’s foot actually hits the brake. We have a new dependent variable:
\begin{align*}
D \amp= \text{ total stopping distance (feet) } \sim \text{ dep} \\
S \amp= \text{ speed (mph) } \sim \text{ indep}
\end{align*}
How can we include this reaction time into an equation? Suppose it takes 1 second to react. We would like to know how many feet that adds to the equation. This is something we can figure out. We know the speed and the time, so multiply them to get the distance, right? One small snag: the speed is in mph (miles per hour). We need to convert units.
\begin{align*}
1 \cancel{\text{ sec}} \ast
\frac{1 \cancel{\text{ min}}}{60 \cancel{\text{ sec}}} \ast
\frac{1 \text{ hour}}{60 \cancel{\text{ min}}} \ast
\frac{5{,}280\text{ feet}}{1 \text{ mile}} \amp =
1 \div 60 \div 60 \times 5{,}280\\
\amp= 1.4666\ldots \approx 1.47 \text{ feet per mph}
\end{align*}
added to the stopping distance. Notice the fancy fraction work with the units?
\begin{equation*}
\frac{\text{hour}\ast \text{feet}}{\text{mile}}=\frac{\text{feet}}{\quad \frac{\text{miles}}{\text{hour}}\quad~}=\frac{\text{feet}}{\text{mph}}
\end{equation*}
What all this means is that we should add 1.47 feet for every mph of speed. So \(S\) mph adds \(1.47\ast S\) feet to the stopping distance. Long story short, our new equation is
\begin{equation*}
D=0.04S^2+1.47S
\end{equation*}
Something interesting about this equation. The independent variable \(S\) appears twice; first for the braking distance and again because of the reaction time. When we evaluate the function we need to plug in the value of \(S\) in two places. Check it out. When \(S=30\) mph we have
\begin{equation*}
D = 0.04 \ast 30^2 + 1.47 \ast 30 = 0.04\times \underline{30} \wedge 2 +1.47 \times \underline{30} = 80.10 \approx 80 \text{ feet}
\end{equation*}
That’s a lot further than just the braking distance of 36 feet. When \(S=50\) mph we have
\begin{equation*}
D = 0.04 \ast 50^2 +1.47\ast 50 = 0.04\times \underline{50} \wedge 2 +1.47\times \underline{50} = 173.50 \approx 174 \text{ feet}
\end{equation*}
Again, much more than the braking distance of 100 feet. Here’s the revised table of values.
| \(S\) | 0 | 30 | 50 | 60 | 70 | 90 |
|---|---|---|---|---|---|---|
| \(D\) | 0 | 80 | 174 | 232 | 299 | 456 |
These numbers make us rethink Jeff’s assertion. Given that he stopped in 183 feet, which is much less than the 232 feet it takes to stop at 60 mph, it looks like Jeff was driving less than 60 mph. To be sure, calculate that at 65 mph, it would have taken Jeff
\begin{equation*}
D = 0.04 \ast 65^2 + 1.47 \ast 65 = 0.04\times \underline{65} \wedge 2+ 1.47\times \underline{65} =264.55 \approx 265 \text{ feet}
\end{equation*}
Again, we should believe Jeff. And, be glad nobody was hurt.
A quick glance at the graph confirms our findings.
The distance of 183 feet falls just above the unlabeled gridline for 175 and below the gridline for 200. Looking at the corresponding point on the braking distance curve, it looks like it falls around 67 mph, but looking at the corresponding point on the stopping distance curve, it looks like just over 50 mph.
Activity 2.3.3.
Let’s practice with another equation where the independent variable occurs twice.
“Rose gold” is a mix of gold and copper. We start with 2 grams of an alloy that is equal parts gold and copper and add \(A\) grams of pure gold to lighten the color. The percentage of gold in the resulting rose gold alloy, \(R\) is given by
\begin{equation*}
R = 100\left(\frac{1+A}{2+A}\right)
\end{equation*}
For example, if we add 0.8 grams of pure gold, then \(A=0.8\) and so the percentage is
\begin{equation*}
R=100\left(\frac{1+0.8}{2+0.8}\right)= 100 \times (1 + \underline{0.8})\div(2+\underline{0.8})=64.28571428\ldots \approx 64.3\%
\end{equation*}
-
Name the variables \(A\) and \(R\text{,}\) including units and dependence.
letter = everyday words (units) ~ dep or indep = \(\qquad\qquad\qquad\qquad\qquad\) \(\displaystyle\Big(\qquad\Big)\) ~ = \(\displaystyle\Big(\qquad\Big)\) ~ -
Calculate the percentage of gold in the alloy if we add 1.2 grams of pure gold.
-
Evaluate at \(A = 2\text{.}\) What does your result mean in the story?
Subsection New kinds of equations
Before we leave, let’s learn a little vocabulary about a few more kinds of equations. Our first equation
\begin{equation*}
B = 0.04 \ast S^2
\end{equation*}
is a power equation because the independent variable is being raised to a power, \(n=2\text{,}\) and then scaled by a proportionality constant, \(k=0.04\text{.}\) Any power equation fits this template.
Power Equation Template
\begin{equation*}
\text{dep} = k \ast \text{indep}^{n}
\end{equation*}
Our second equation
\begin{equation*}
D=0.04S^2 + 1.47S
\end{equation*}
is a polynomial equation because it includes both a linear term and powers. The exercises introduce more polynomial equations. Polynomials can have any powers, but in this equation the highest power happens to be 2. This type of polynomial equation has a special name. It is a quadratic equation. Any quadratic equation fits this template.
Quadratic Equation Template
\begin{equation*}
\text{dep} = a \ast \text{indep}^2 + b \ast\text{indep} + c
\end{equation*}
For our equation \(a=0.04\text{,}\) \(b=1.47\text{,}\) and the mysteriously missing \(c=0\text{.}\)
Do you know …
-
What it means to “evaluate” a function?
-
Why some numbers are underlined in our calculation?
-
How to evaluate an function when the independent variable occurs more than once?
-
How to generate a table or graph from an equation?
-
What graphs of different types of functions look like?
-
What a power, polynomial, or quadratic equation look like?
If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.
Exercises Exercises
Exercises 1-4 are available in a separate workbook format.
1.
Dontrell and Kim borrowed money to buy a house on a 30-year mortgage.
After \(T\) months of making payments, Dontrell and Kim will still owe $\(D\) where
Aside
\begin{equation*}
D=236{,}000-56{,}000 \ast 1.004^T
\end{equation*}
\(D\) is also known as the payoff (how much they would need to pay to settle the debt).
(a)
| letter | = | everyday words | (units) | ~ | dep or indep |
|---|---|---|---|---|---|
| = | \(\qquad\qquad\qquad\qquad\qquad\) | \(\displaystyle\Big(\qquad\Big)\) | ~ | ||
| = | \(\displaystyle\Big(\qquad\Big)\) | ~ |
(b)
How much did Dontrell and Kim originally borrow to buy their house? What value of \(T\) did you evaluate at to answer the question?
(c)
Evaluate the equation at \(T=12\text{.}\) What does this number mean in terms of Dontrell and Kim’s mortgage?
(d)
After making half the payments, how much money will Dontrell and Kim still owe on the house?
Will they have paid more or less (or exactly) half of the amount they originally borrowed?
Aside
(e)
The very last month they do not actually pay the regular monthly payment,
just whatever balance is left on the loan. How much will that be?
Aside
2.
“Rose gold” is a mix of gold and copper.
We start with 2 grams of an alloy that is equal parts gold and copper and add \(A\) grams of pure gold to lighten the color. The percentage of gold in the resulting rose gold alloy, \(R\) is given by
Aside
\begin{equation*}
R = 100\left(\frac{1+A}{2+A}\right)
\end{equation*}
For example, if we add 0.8 grams of pure gold, then \(A=0.8\) and so the percentage is
\begin{equation*}
R = 100\left(\frac{1+0.8}{2+0.8}\right)
= 100 \times (1 + \underline{0.8})\div(2+\underline{0.8})
=64.28571428\ldots \approx 64.3\%
\end{equation*}
(a)
| letter | = | everyday words | (units) | ~ | dep or indep |
|---|---|---|---|---|---|
| = | \(\qquad\qquad\qquad\qquad\qquad\) | \(\displaystyle\Big(\qquad\Big)\) | ~ | ||
| = | \(\displaystyle\Big(\qquad\Big)\) | ~ |
(b)
Calculate the percentage of gold in the alloy if we add 1.2 grams of pure gold.
(c)
Fill in all the missing values in this table.
| \(A\) | 0 | 0.4 | 0.8 | 1.2 | 1.6 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(R\) | 50.0 | \(\fillinmath{\displaystyle\int\int}\) | 64.3 | \(\fillinmath{\displaystyle\int\int}\) | 72.2 | \(\fillinmath{\displaystyle\int\int}\) | 80.0 | \(\fillinmath{\displaystyle\int\int}\) |
(d)
Graph the function.
(e)
What do you think happens to the percentage of gold as we add more and more pure gold? Try adding 10 grams, and then try adding 100 grams to check.
3.
Monty hopes to grow orchids but they are fragile plants.
He will consider his greenhouse a success if at least nine of the ten orchids survive. Assuming the orchids each survive at rate \(S\text{,}\) the probability his greenhouse is a success, \(P\text{,}\) is given by
Aside
\begin{equation*}
P= 10S^9-9S^{10}
\end{equation*}
(a)
| letter | = | everyday words | (units) | ~ | dep or indep |
|---|---|---|---|---|---|
| = | \(\qquad\qquad\qquad\qquad\qquad\) | \(\displaystyle\Big(\qquad\Big)\) | ~ | ||
| = | \(\displaystyle\Big(\qquad\Big)\) | ~ |
(b)
If the orchids are perfect (\(S=1\)), what is the probability of a successful greenhouse? Explain how this answer makes sense in the story.
(c)
If the orchids are complete duds (\(S=0\)), what is the probability of a successful greenhouse? Explain how this answer makes sense in the story.
(d)
Make a table comparing the probability of a successful greenhouse if the orchids each survive at rate S = 0, 0.5, 0.8, 0.9, 0.95, or 1.
(e)
Draw a graph of the function.
4.
Valerie plans to do a charity walk to raise money for breast cancer research, in honor of her aunt. Valerie’s friends have pledged a total of $93 per mile.
(a)
If Valerie walks all 50 miles of the event, how much money will she raise?
(b)
She might have to stop sooner, however. Name variables and write an equation showing how the money Valerie raises is a function of how far she is able to walk.
| letter | = | everyday words | (units) | ~ | dep or indep |
|---|---|---|---|---|---|
| = | \(\qquad\qquad\qquad\qquad\qquad\) | \(\displaystyle\Big(\qquad\Big)\) | ~ | ||
| = | \(\displaystyle\Big(\qquad\Big)\) | ~ |
(c)
How long will it take Valerie to walk the full 50 miles if she is able to keep a pace (speed) of 3.2 miles per hour? Write your answer in H:MM format.
(d)
Name the new variables and write a new equation showing how the time it takes Valerie to walk the full 50 miles depends on her pace.
| letter | = | everyday words | (units) | ~ | dep or indep |
|---|---|---|---|---|---|
| = | \(\qquad\qquad\qquad\qquad\qquad\) | \(\displaystyle\Big(\qquad\Big)\) | ~ | ||
| = | \(\displaystyle\Big(\qquad\Big)\) | ~ |
(e)
Good news:
Valerie walked the full 50 miles at a pace of 3.2 miles per hour. Way to go, girl! How much money did she raise each hour?
Aside
5.
The 2002 Chevrolet Tahoe 4WD will take about 158.1 feet to stop when traveling at 60 mph in normal highway conditions.
Let \(S\) be the speed at which the vehicle is traveling, in miles per hour (mph), and \(D\) the distance it takes to stop, in feet. This information together with a little physics gives the equation
Aside
\begin{equation*}
D = 0.0439S^2 + 1.47S
\end{equation*}
(a)
According to this equation, how many feet does it take to stop the Tahoe when traveling 80 miles per hour?
(b)
In driver’s training classes they teach the “two-second rule” for safety: you should follow no closer than two seconds behind the car in front of you. If you are traveling 80 miles per hour, how many feet can you travel in two seconds? Hint: convert to feet per second, then multiply by two seconds.
(c)
Compare your results from parts (a) and (b) to decide if the “two-second rule” is adequate for safety at 80 miles per hour. That is, if you are following two seconds behind the car in front of you and calamity strikes that car, will you be able to stop before hitting it?
(d)
Is the “two-second rule” adequate at 50 mph?
6.
Mom always said to sit close to the lamp when I was reading.
The intensity of light \(L\text{,}\) measured in percentage (%) that you see from a lamp depends on your distance from the lamp, \(F\) feet as described by the formula
Aside
\begin{equation*}
L=\frac{100}{F^2}
\end{equation*}
(a)
Calculate the intensity when I sit 1 foot, 2 feet, or 3 feet away. Record your answers in a table.
(b)
The other day I was sitting 27 inches from the lamp. What was the intensity of the light there?
(c)
Calculate the rate of change of light intensity between 2 feet and 27 inches. What does that tell you in terms of the story?
(d)
Draw a graph illustrating the function.
7.
Urban community gardens are catching on.
What was once an abandoned lot down the block is now a thriving 10′\(\times\)25′ vegetable and berry garden for the neighborhood. (Remember ′ stands for “feet,” so the garden is 10 feet wide and 25 feet long.) One neighbor volunteered to donate gravel to make a path around the garden. The path will be 3 inches deep and the same width all around.
Aside
(a)
The other measurements are in feet, so convert the depth of the path (3 inches) into feet also.
(b)
Suppose for the moment that the path will be 4 feet wide. Calculate the area of the path. Here’s one way to do it: first, find the area of the outer rectangle, then subtract the area of the garden itself.
Hint: the length of that outer rectangle includes the 25 feet of garden plus the width of the path on each side. Same for the width of that outer rectangle.
(c)
Figure out how much gravel they would need (in cubic feet) for a 4 foot wide path by multiplying your answers to (a) and (b).
(d)
Actually, they aren’t sure how wide the path should be and how much gravel they can get. Let’s write \(W\) = width of path (feet) and \(G\) = amount of gravel (cubic feet). It turns out that
\begin{equation*}
G = W^2 + 17.5W
\end{equation*}
Check that when you evaluate at \(W=4\) you get the same answer as in (c).
(e)
How many cubic feet of gravel would they need to make the path 2 feet wide? 3 feet wide? 42 inches?
8.
One measure of the diversity of our news source is the count of the number of different daily newspapers in circulation. A reasonable equation estimating this count over the past century is
\begin{equation*}
N = -0.0021T^3+0.34T^2-20T+ 2{,}226
\end{equation*}
where \(N\) is the number of daily newspapers in circulation in the United States \(T\) years after 1900.
(a)
Based on this equation, how many daily newspapers were in circulation in 1920? In 1955? In 1995? In 2010?
(b)
During which period were the number of newspapers in circulation dropping faster: 1900 to 1920 or 1995 to 2010?
(c)
Draw a graph illustrating the dependence.
9.
Mrs. Weber’s cooking class came up with the equation
\begin{equation*}
M = 1.2F^2+4F+7
\end{equation*}
to approximate the grilling time of a piece of fish depending on its thickness.
Here \(M\) is the number of minutes to grill the fish and \(F\) is the thickness of the fish (in inches).
Aside
(a)
(b)
What do the answers you found say about cooking fish?
(c)
Draw a graph showing how the cooking time depends on the thickness of the fish.
(d)
Your graph should show that the function is increasing. Explain how that makes sense in terms of the story.
10.
Wynter has a pretty decent job.
He is paid a salary of $780 per week but his hours vary week-to-week. Even though Wynter is not paid by the hour, he can figure out what his hourly wage would be depending on the number of hours he works. For example, in a week where Wynter works 40 hours, he’s earning the equivalent of $19.50/hr because
Aside
\begin{equation*}
\frac{\$780}{40 \text{ hours}} = 780 \div 40 =\$19.50\text{/hour}
\end{equation*}
(a)
What’s Wynter’s equivalent hourly wage in a week when he works 50 hours? 60 hours?
(b)
Name the variables and write an equation relating them.
(c)
Explain why this function is decreasing.
