**Directions:** Use the computer program ** I See Graphs** to complete the following tasks.

Click the **Standard Equations** tab.

For equation #1, type 2x + 1, and for equation #2, type x^{2}. Click the yellow **Show Tables** button. Three columns of numbers appear in a new window. the X column represents values of x. Column 1 represents values of 2x + 1. Column 2 represents x^{2}.

Compare the numbers in column 1 with the corresponding numbers in column 2. At or near the top of the lists, the numbers in column 1 are greater than the numbers in column 2. After the third number in each list, numbers in column 1 become less than the numbers in column 2. Is there a value of x for which the numbers in the two columns would be equal?

To answer that question, we need to look at values of x between integers. We need to "zoom in" on what is going on. To do that, we will change the value of **DTable. **By default, the value of **DTable **is 1. If we make that value smaller, we will be able to compare columns when the value of x is not necessarily an integer. We also notice that the columns change in relative values somewhere between 2 and 3.

Begin by double-clicking the 0 beside the **Table Minimum** button. Type 2 as the new **Table Minimum**.

Double-click the number 1 beside the **DTable **button. Type **.1 **as the new **DTable.**

The table should have been reconstructed using the new values.

Now as you compare values in column 1 to values in column 2, you should notice that column 2 values become larger than column 1 values between x = 2.4 and x = 2.5. Now we need to re-define the **Table Minimum** to be 2.4. Follow the directions above to change 2 to 2.4.

Double-click the number **.1** beside the **DTable **button. Type **.01 **as the new **DTable.**

Now we see that the two column's values "cross" between 2.41 and 2.42, so change **Table Minimum** to 2.41, and change **DTable** to **.001. **The columns now cross between

x = 2.414 and x = 2.415.

Change **Table Minimum** to the smaller of these two numbers, 2.414, and change **DTable** to **.0001**. Now we see that column 1 and column 2 are equal to 4 decimal places (5.8284) when x = 2.4142. In other words, 2x + 1 = x^{2} when x = 2.4142. Of course, this is just a decimal approximation to the actual solution to this equation. The actual solution is an irrational number, i.e., it has a non-repeating, non-terminating decimal representation.

Because this is a quadratic equation (contains x^{2}), we should expect another solution.

Change **Table Minimum **back to 0, and change **DTable **back to 1.

Click the up-arrow to scroll the columns the other direction from before.

With just a couple of clicks, values in column 1 become greater than values in column 2.

Adjust **Table Minimum **and **DTable **as we did above and find a second solution to the equation 2x + 1 = x^{2}.

1) What is the other solution to the nearest 1/10000? ___________________

Use a similar technique to find solutions to the following equations:

2) 3x - 1 = 2x^{2}

3) _{}

4) x^{3} - 5x^{2} + x - 4 = 2x - 6

5) x^{3} - 4x^{2} + 3x + 2 = 0