In many linear application problems, you needed to write the given information as ordered pairs and then find the equation that passes through the ordered pairs. Here is a similar example that use p = mt + b instead of y = mx + b.
Problem The percent of births to teenage mothers that are out of wedlock can be approximated by a linear function. In 1960, the percentage was 15% and in 1996 the percentage was 76%.
Use this information to find a linear model for the percentage of births as a linear function of the number of years since 1950.
Solution Since the problem statement specifies a linear function of the number of years since 1950, the input to this function is years since 1950 and the output is the percentage. The information in the problem can be written as ordered pairs (10, 15) and (46, 76).
Define the variables for these quantities as
t: years since 1950
p: percentage of births to teenage mothers out of wedlock
This means the form of the linear function is p = mt + b. The slope of a line passing through these points is
The slope is written as a fraction so no rounding occurs. Writing this as a decimal and rounding the amount would lead to a line that does not pass through the points. With this slope, we know the equation of the line is . To find the value of b, substitute one of the ordered pairs into the equation and solve for b.
Linear models have inputs and outputs. In applications, the input or output may be specified and you must solve for the other variable. To make sure you put in the value for the correct variable, pay careful attention to what each variable represents.
Once the resulting equation is solved, you can check it with the Method of Intersection. In this method, each side of the equation is graphed to find the point of intersection. The horizontal value of the point of intersection should match the algebraic solution if everything has been done correctly.
Example An antique is originally purchased for $20,000. The value V after t months is $latex \displaystyle V=1500t+20,000$ dollars. After how many months will the value be $36,000?
Solution To find when the value will be $36,000, set V equal to 36,000 and solve for t. Verify the algebraic solution by graphing each side of the resulting equation. Since the linear model starts at 20,000 and increases to 36,000, an appropriate window is [0,12] by [0, 40000]. Other windows like [0,20] by [10000, 40000] would also work well.
The solution to an equation may be checked by graphing each side of the equation with a graphing utility. The solution to the equation should correspond to the input value at the point of intersection.
In each of the problems below, students have solved an equation algebraically and then checked the solution on a graph.
Note that in each case the students have included an exact answer such as a fraction as well as an approximate answer. The approximate answer is appropriate for comparing to the point of intersection which is also an approximate solution.
A number of you have asked me about alternatives to a TI graphing calculator. In addition to Excel, there are many online graphing tools available. If you have an Android phone or IPhone, there are a huge number of free apps that are available. I have had a hard time finding just one that does everything we need. Most will graph formulas, but may or may not graph data and fit data to lines. None of them work identically to a graphing calculator which makes them difficult to support. But luckily there is another option!
Another option is the website WolframAlpha (http://www.wolframalpha.com/). This website is the Internet’s leading computational engine. It can do just about anything. The trouble is knowing how to use it to do just about anything.
When you go to this website, you’ll see a box in which you can enter commands to help you do mathematics. The list of things you can do is HUGE. It is best to show some examples to get you started.
For instance, suppose you want to graph the function y = -5.686x + 676.173 in a window from x = 0 to x = 25. Enter the text you see below into WolframAlpha followed by Enter.
WolframAlpha makes a nice graph. If you want to graph two formulas simultaneously, add another formula with the word “and” as you see below.
You can see that these lines are going to meet. To see this point of intersection, extend the graph by modifying the input to WolframAlpha as you see below.
We can find this point of intersection by modifying the input to WolframAlpha with the command “intersections”:
Not only does it give a decimal…it also gives the exact answer in terms of fractions!
Often you are interested in evaluating the model above at a particular point. We can do this by clicking on the formula above. This will make WolframAlpha graph the formula by itself. Now add “where x = 20” on the command line and you will see:
This gives the same output as TRACE on a graphing calculator. I’ll continue to make posts about WolframAlpha so those of you who do not have access to a graphing calculator have an alternative way to make graphs, calculate values, ect.