Back To CourseMath 105: Precalculus Algebra
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Cat has taught a variety of subjects, including communications, mathematics, and technology. Cat has a master's degree in education and is currently working on her Ph.D.
Jill is putting together her monthly budget. Right now, she is working on her electric bill. She has a statement from her electric company that shows the kilowatts per hour that she uses for each month. She knows the rate per kilowatt per hour is 8.2 cents, but she doesn't know how much it costs her per month. How is Jill going to estimate her monthly bills if she doesn't know how much her electric bills cost her? Jill can use a linear model to write equations that will help her predict the cost of her monthly bills.
A linear model is a comparison of two values, usually x and y, and the consistent change between the values. In the opening story, Jill was analyzing two values: the amount of electricity used and the total cost of her bill. The change between these two values is the cost of each kilowatt hour. Still a little confused? We'll come back to this later.
For right now, let's talk about that consistent change, which is also known as the rate of change; in algebra, we call that the slope. You're probably familiar with the slope of a line: it describes both the length and the steepness of the line.
Okay, so back to our example. You've probably already figured out that the more electricity Jill uses, the higher her bill will be at the end of the month. Therefore, the rate of change shows us just how much that bill will increase as Jill uses more electricity.
As you can see, linear models are very applicable to real life and can be used to predict certain information that is useful for making decisions and problem solving.
Before we can figure out how much Jill's bill is going to be, we need to know the independent and dependent variables.
An independent variable is a variable that stands alone and is the input for the equation. For example, if you were to conduct an experiment and determine how much water a plant needs to survive, you might try using different amounts of water on plants in the same lighting and soil conditions. You are using water as the independent variable, or the input in this situation. In a linear equation, x represents the independent variable.
The dependent variable is the result of the independent variable, or the output. After watering your plants, you may want to measure how much each plant grew. The amount the plant grows is dependent upon how much water you gave it. In a linear equation, y represents the dependent variable.
In our example, Jill has two pieces of information that are very important to this problem: the amount of electricity and the cost of the bill. Which piece of information is the dependent variable and which piece of information is the independent variable? Well, the cost of the bill is dependent upon how much electricity Jill uses. Therefore, the cost of the bill is the dependent variable and the amount of electricity is the independent variable. We can write our linear model like this: y = .082x, where y is the cost of the bill, and x is the amount of electricity used.
You can use slope-intercept form, which is y = mx + b, to write equations for linear models. m is the slope or rate-of-change, and b is the y-intercept. Often, the y-intercept represents the starting point of the equation. For example, if Jill had a $20 per month fee for her electricity bill, then her bill would always be $20 or more. Therefore, our equation would be y = .082x + 20. The graph of our equation would look like this.
Now Jill can use a graph like this to figure out how much each month's electricity bill will be. If she uses 1,200 Kilowatts per hour in September, then her bill would be about $118.40.
Jill's husband, Danny, has a car that he wants to sell. The car is pretty average with little damage and average mileage. They have owned the car for five years. Danny wants to know how much his car will be worth now and in two years. He has developed the following formula to help in his decision making: y = -500x + 16,000. We can use what we know of linear models to understand and evaluate this equation.
Remember, the slope-intercept form of an equation is y = mx + b. We can see a similar pattern with Danny's equation: y = -500x + 16,000. First, let's examine the numbers in Danny's equation. The first number that appears in the equation is -500. This number is sitting in the m, or slope, spot in the slope-intercept equation. That tells me that -500 represents the rate of change in the scenario. That makes sense, because we know that Danny's car will likely depreciate in value as time passes. Therefore, the slope will be negative and go down, rather than up.
The next number that stands out is 16,000. This number is sitting in the b or y-intercept spot in the slope-intercept equation. That tells me that 16,000 represents the starting point for Danny's car. What kind of starting point? Well, since we are talking about cost and depreciation, I can guess that the car was originally worth $16,000.
Now, let's identify the independent and dependent variables. We already know that y represents the dependent variable, and x represents the independent variable, but what does that mean in regards to the scenario?
We already know that -500 represents rate of change in regards to time. Therefore, we can identify x as the year or time. This makes sense because x is the independent variable and time doesn't usually wait or depend on anything, much less the cost of a car.
That leaves us with y, or the dependent, variable. Since we already know that Danny wants to know the cost of the car after time, then we can deduce that y represents the depreciated cost of the car at a certain point in time.
Now we are ready to evaluate this equation, so Danny can make an informed decision about selling his car. We know that at the time of the purchase, the car was worth $16,000. That is represented on this graph as (0,16000). No time has passed: our x value is equal to zero. Don't forget that Danny has owned this car for 5 years. So how much is his car currently worth? We can use the equation y = -500x + 16,000 to find out the answer. Simply replace x with 5, and evaluate the equation.
y = -500(5) + 16,000
y = -2,500 + 16,000
y = 13,500
Now we know that Danny's car is currently worth $13,500 and our point on the graph would be (5, 13500). Danny also wants to know how much his car would be worth two years from now. That would give us the equation y = -500(7) + 16,000. You may be asking, 'where did the 7 come from?' Remember that the car is already 5 years old and our equation represents the worth of the car at the time. Therefore, we can't simply have y = -500(2) + 16,000 because that would give us how much the car was worth when it was two years old.
Now that we have the correct equation it's time to evaluate:
y = -500(7) + 16,000
y = -3,500 + 16,000
y = 12,500
Danny can predict that his car will be worth $12,500 in the next two years. Of course, this is just an example. It doesn't really represent how much your car is worth, since that depends on mileage, type of car, fuel mileage, damage, and other factors.
In summary, you can use four pieces of information to evaluate and write linear models. First, you must know the slope or the rate of change, which is the length and steepness of the line. In the last example, you saw that slope represents how much something changes over time. You also need to know the independent and dependent variables. The independent variable is the variable that stands alone or is the input variable. For example, if your linear model uses time, time will almost always be the independent variable. The dependent variable is the variable that is the result of the independent variable or the output. For example, if you need to know the total cost of something, cost is often dependent on other factors. Cost is almost always considered the dependent variable. Lastly, if your linear model has an extra fee or a starting point, that will be the y-intercept of the linear equation.
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Back To CourseMath 105: Precalculus Algebra
14 chapters | 113 lessons | 12 flashcard sets