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# Inverse Variation Models

## Identify and solve y=k/x form equations

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Inverse Variation Models

Suppose a greeting card company has found that the number of greeting cards it sells is inversely proportional to the price of each card. If it sells 500 cards per month when the price of each card is $2, how many cards would it sell per month if the price of each card were$3? How would you go about calculating this number?

### Inverse Variation Models

Previously, you learned how to write direct variation models. In direct variation, the variables changed in the same way and the graph contained the origin. But what happens when the variables change in different ways? Consider the following situation:

A group of friends rent a beach house and decide to split the cost of the rent and food. Four friends pay $170 each. Five friends pay$162 each. Six friends pay 157. If nine people were to share the expense, how much would each pay? Let’s look at this in a table. \begin{align*}n\end{align*} (number of friends) \begin{align*}t\end{align*} (share of expense) 4 170 5 162 6 157 9 ??? As the number of friends gets larger, the cost per person gets smaller. This is an example of inverse variation. An inverse variation function has the form \begin{align*}f(x)=\frac{k}{x}\end{align*}, where \begin{align*}k\end{align*} is called the constant of variation and must be a counting number and \begin{align*}x \neq 0\end{align*}. To show an inverse variation relationship, use either of the phrases: • Is inversely proportional to • Varies inversely as #### Using the inverse variation function, let's find the constant of variation of the beach house situation: Use the inverse variation equation to find \begin{align*}k\end{align*}, the constant of variation. \begin{align*}&& y &= \frac{k}{x}\\ && 170 &= \frac{k}{4}\\ \text{Solve for} \ k: && 170 \times 4 &= \frac{k}{4} \times 4\\ && k &= 680\end{align*} You can use this information to determine the amount of expense per person if nine people split the cost. \begin{align*}y &= \frac{680}{x}\\ y &= \frac{680}{9}=75.56\end{align*} If nine people split the expense, each would pay75.56.

Using a graphing calculator, look at a graph of this situation.

The graph of an inverse variation function \begin{align*}f(x)=\frac{k}{x}\end{align*} is a hyperbola. It has two branches in opposite quadrants.

If \begin{align*}k>0\end{align*}, the branches are in quadrants I and III.

If \begin{align*}k<0\end{align*}, the branches are in quadrants II and IV.

The graph appears to not cross the axes. In fact, this is true of any inverse variation equation of the form \begin{align*}y=\frac{k}{x^n}\end{align*}. These lines are called asymptotes. Because of this, an inverse variation function has a special domain and range.

\begin{align*}Domain: \ & x \neq 0\\ Range: \ & y \neq 0\end{align*}

You will investigate these excluded values later in this chapter.

#### Let's use the inverse variation function to solve the following problem:

A sound signal that has a wavelength of 34 meters has a frequency of 10 hertz. What frequency does a sound signal of 120 meters have?

The frequency, \begin{align*}f\end{align*}, of sound varies inversely with wavelength, \begin{align*}\lambda\end{align*}.

Use the inverse variation equation to find \begin{align*}k\end{align*}, the constant of variation.

\begin{align*}&& f &= \frac{k}{\lambda}\\ && 10 &= \frac{k}{34}\\ \text{Solve for} \ k: && 10 \times 34 &= \frac{k}{34} \times 34\\ && k &= 340\\ \text{Use} \ k \ \text{to answer the question:} &&\\ && f &= \frac{340}{120}=2.83 \ hertz\end{align*}

#### Now, determine if the following situations are an example of inverse variation and explain why or why not:

1. A student spends $4 a day for snacks at school. The total amount she spends each week depends on the number of days school was in session. The student can attend school 1-5 days a week, since school is never in session for more than 5 days a week. If school is in session 1 day, she spends$4. If school is in session 2 days, she spends $8. The amount of money she spends is increasing as the number of days increases. This is not an inverse variation function. Another way to think of this problem is to see that there is a constant rate involved: the student spends$4 per day on snacks. This would be a linear function, with a slope of 4. This is not an example of inverse variation.

1. A school has $200 to spend on prizes for an event. The number of prizes that can be bought depends on the price of each prize. The number of prizes that can be bought is determined by dividing by the cost of each prize. If each prize costs$10, then \begin{align*}\frac{200}{10}=20\end{align*} prizes can be bought. If each prize costs 20 then \begin{align*}\frac{200}{20}=10\end{align*} prizes can be bought. As the cost of the prizes increases, the number of prizes that can be bought decreases. The function is \begin{align*}f(x)=\frac{200}{x}\end{align*}, where \begin{align*}x\end{align*} is the price of the prizes and \begin{align*}f(x)\end{align*} represents the number of prizes that can be bought. This fits the definition of inverse variation. ### Examples #### Example 1 Earlier, you were asked to determine how many greeting cards a store could sell if they were3 a piece. You know that the number of cards sold is inversely proportional to the price per card, and that the store sold 500 cards when they were 2 each. Since you know that the number of cards sold is inversely proportional to the price, you can set up the following equation: \begin{align*}&y= number\ of\ cards\ sold\\ &k= constant\ of\ proportionality\\ &x= price\ per\ card\\ & y=\frac{k}{x}\end{align*}Next, you can solve for the constant of proportionality (k) by plugging in the data from when the cards were2 each.

\begin{align*}&y=\frac{k}{x}\\ &500=\frac{k}{2}\\ & 1000=k\\\end{align*}

In order to find the number of cards the store will sell if they are 3 each, plug x=3 and k=1000 into the equation and solve for y. \begin{align*}&y=\frac{1000}{3}\\ &y=333.33\end{align*} When you solve for y you get 333.33. You cannot sell a fraction of a card, so you have to round this to the nearest whole number. Therefore, 333 cards can be sold at3 a piece.

#### Example 2

If \begin{align*}f(x)\end{align*} varies inversely with \begin{align*}x\end{align*} and \begin{align*}f(12)=5\end{align*}, find \begin{align*}f(20)\end{align*}

Since all functions of inverse variation look like:

\begin{align*}f(x)=\frac{k}{x}\end{align*}

simply substitute in \begin{align*}x\end{align*} and \begin{align*}f(x)\end{align*} and solve for \begin{align*}k\end{align*}.

\begin{align*}f(12)&=\frac{k}{12}\\ 5&=\frac{k}{12}\\ 60&=k \end{align*}

Now, use the function to find \begin{align*}f(20)\end{align*}:

\begin{align*}f(20)&=\frac{60}{20}\\ f(20)&=3 \end{align*}

### Review

1. Define inverse variation.
2. Explain three main differences between direct variation and inverse variation.

Read each statement and decide if the relationship is direct, inverse, or neither.

1. The weight of a book __________________ as the number of pages it contains.
2. The temperature outside __________________ as the time of day.
3. The amount of prize money you receive from winning the lottery __________ as the number of people who split the ticket cost.
4. The cost of a ferry ride ___________________ as the number of times you ride.
5. The area of a square _______________________ as the length of its side.
6. The height from the ground ___________________ as the number of seconds you have been on a roller coaster.
7. The time it takes to wash a car ___________________ as the number of people helping.
8. The number of tiles it takes to tile a floor ___________________ as the size of each tile.

Graph each inverse equation. State the domain and range.

1. \begin{align*}y=\frac{3}{x}\end{align*}
2. \begin{align*}y=\frac{1}{x^2}\end{align*}
3. \begin{align*}f(x)=- \frac{4}{x}\end{align*}
4. \begin{align*}y=\frac{10}{x}\end{align*}
5. \begin{align*}h(x)=-\frac{1}{x}\end{align*}
6. \begin{align*}y=\frac{1}{4x}\end{align*}
7. \begin{align*}g(x)=-\frac{2}{x^2}\end{align*}
8. \begin{align*}y=\frac{4}{x^2}\end{align*}
9. \begin{align*}y=\frac{5}{6x}\end{align*}

In 20–25, model each situation with an inverse variation equation, finding \begin{align*}k\end{align*}. Then answer the question.

1. \begin{align*}y\end{align*} varies inversely as \begin{align*}x\end{align*}. If \begin{align*}y=24\end{align*} when \begin{align*}x=3\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x=-1.5\end{align*}.
2. \begin{align*}d\end{align*} varies inversely as the cube of \begin{align*}t\end{align*}. If \begin{align*}d=-23.5\end{align*} when \begin{align*}t=3\end{align*}, find \begin{align*}d\end{align*} when \begin{align*}x=\frac{1}{4}\end{align*}.
3. If \begin{align*}z\end{align*} is inversely proportional to \begin{align*}w\end{align*} and \begin{align*}z=81\end{align*} when \begin{align*}w=9\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}z=24\end{align*}.
4. If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y=2\end{align*} when \begin{align*}x=8\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x=12\end{align*}.
5. If \begin{align*}a\end{align*} is inversely proportional to the square root of \begin{align*}b\end{align*}, and \begin{align*}a=32\end{align*} when \begin{align*}b=9\end{align*}, find \begin{align*}b\end{align*} when \begin{align*}a=6\end{align*}.
6. If \begin{align*}w\end{align*} is inversely proportional to the square of \begin{align*}u\end{align*} and \begin{align*}w=4\end{align*} when \begin{align*}u=2\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}u=8\end{align*}.
7. The law of the fulcrum states the distance from the fulcrum varies inversely as the weight of the object. Joey and Josh are on a seesaw. If Joey weighs 40 pounds and sits six feet from the fulcrum, how far would Josh have to sit to balance the seesaw? (Josh weighs 65 pounds.)
8. The intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated. A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source?
9. Ohm’s Law states that current flowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 amperes when the resistance is 20 ohms, find the resistance when the current is 5 amperes.
10. The number of tiles it takes to tile a bathroom floor varies inversely as the square of the side of the tile. If it takes 112 six-inch tiles to cover a floor, how many eight-inches tiles are needed?

Mixed Review

1. Solve and graph the solutions on a number line: \begin{align*}16 \ge -3x+5\end{align*}.
2. Graph on a coordinate plane: \begin{align*}x=\frac{7}{14}\end{align*}.
3. Simplify \begin{align*}\sqrt[3]{320}\end{align*}.
4. State the Commutative Property of Multiplication.
5. Draw the real number hierarchy and provide an example for each category.
6. Find 17.5% of 96.

To see the Review answers, open this PDF file and look for section 12.1.

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### Vocabulary Language: English Spanish

asymptote

An asymptote is a straight line to which, as the distance from the origin gets larger, a curve gets closer and closer but never intersects. The line is often shown with a dashed or dotted line.

hyperbola

A hyperbola is a curve with two distinct and similar branches. Geometrically, the curve's two branches are formed by cutting a right circular cone with a plane. Algebraically, hyperbolas come from equations of the form $\frac{x^2}{a}-\frac{y^2}{b}=\pm 1$.

Inverse Variation

Inverse variation is a relationship between two variables in which the product of the two variables is equal to a constant. As one variable increases the second variable decreases proportionally.

Constant of Proportionality

The constant of proportionality, commonly represented as $k$ is the constant ratio of two proportional quantities such as $x$ and $y$.

Direct Variation

When the dependent variable grows large or small as the independent variable does.

Joint Variation

Variables exhibit joint variation if one variable varies directly as the product of two or more other variables.