
| The Problem |
|
The
linear function given by Determine an equation whose solution is the year when there were 10,000 radio stations on the air and then solve the equation graphically, symbolically, and numerically. |
| The number 10000 in this
problem represents the value of 10000 = 214.2 (x 1950) + 2322 |
We use the Intersection-of-Graphs Method to solve this equation.
| Intersection
of Graphs Method
This method uses graphing of functions to solve an equation. STEP
1: Set STEP
2: Graph STEP 3: Locate any points of intersection. The x-values of these points correspond to the solutions to the equation. |
To solve this equation
by the intersection-of-graphs method, press
and then enter 10000 (the left side of the equation) for
and enter 214.2 (X 1950) + 2322 (the right side
of the equation) for
. On your calculator you should see:

Now select and set an appropriate Viewing Rectangle. Since we know that the x-values range from 1950 to 1995, we pick a range slightly larger than that for x, namely 1940 to 2000. We also know that Y = 10000 means that our Y range must include 10,000. Thus we select 0 to 12000.
This gives us a Viewing Rectangle of [1940, 2000, 10] by [0, 12000, 10]. To set this Viewing Rectangle, press WINDOW and then enter 1940 for Xmin, 2000 for Xmax, 10 for Xscl, 0 for Ymin, 12000 for Ymax, and 10 for Yscl.
Press GRAPH and both lines will be displayed on the screen.
Next we need to find the point of intersection. Press 2nd TRACE[CALC], then press 5 to select intersect. On the screen you should see:
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The small square near the center of the top horizontal line is the cursor mark. Use the arrow right to move the cursor so that it is very close to the point of intersection. Press ENTER, then ENTER, then ENTER. (That is 3 times). Below are the graphs you will see:
The last screen shows the Intersection X = 1985.845 Y = 10000
The solution to the equation is
X
1985.845. Thus, in 1985 there were approximately
10,000 stations on the air.
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