# TI-83 Examples

## Finding Slopes of Secant Lines

The first goal is to show one way
to do some of the problems in Section 2.1 of Stewart.
**Exercise 4** (page 71 in Stewart)

The point P(0.5,2) lies on the curve y = 1/x.

If Q is the point (x,1/x), use your calculator
to find the slope of the secant line PQ
for the following values of x:

(i) 2 (ii) 1 (iii) 0.9

### Step 1

Use the
key to call the function editor,
and enter the function y_{1} = 1/x.

On your calculator, the entry for Y_{1} in the function editor
should look like this:
________________________________
| |
| Y_{1}=1/x |
| Y_{2}= |
| Y_{3}= |
| Y_{4}= |
| Y_{5}= |
| |
| |
| |
| |
|________________________________|

### Step 2

You need to compute the slope
(f(x) - 2) / (x - 0.5),
for various values of x.

For part (i),
you first need to store 2 in the variable X.

This is done by entering 2 followed by the keystrokes
.

Use the keystrokes
to enter a " : ".

Now use the Y-variables menu to enter the formula
(y_{1}-2)/(x-0.5).

(See the
"Function Evaluation"
tutorial if you are unsure how to call the Y-variables.)

Now press the key to do the calculation.
Your screen should look like this:
________________________________
| |
| 2->X:(Y_{1}-2)/(X-.5) |
| -1 |
| |
| |
| |
| |
| |
| |
| |
|________________________________|

### Step 3

You can easily repeat the process
by using the keystrokes
to recall your previous entry.

Then you can use the cursor keys to change the input for x,
to do the remaining parts of the question.

Your screen should look like this:
________________________________
| |
| 2->X:(Y_{1}-2)/(X-.5) |
| -1 |
| 1->X:(Y_{1}-2)/(X-.5) |
| -2 |
| .9->X:(Y_{1}-2)/(X-.5) |
| -2.22222222222 |
| |
| |
| |
|________________________________|

Note: Here is the general form of the slopes
you are evaluating.

To change to a new function f(x),
change the entry Y_{1} in the function editor.

On the main screen,
change the constants a and f(a).

You are now ready to do the calculation for any value of x.
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