Question

the table shows somedata for an exponential function. x: -2, -1, 0, 1, 2. P(x): 2,500, 500, 100, 20, 4. which function represents the relation shown in the table

Answer 1

To find the exponential function that fits the data provided, we will look for a function of the form:

\[
P(x) = a \cdot b^x
\]

We have the following data points from the table:

- \( P(-2) = 2500 \)
- \( P(-1) = 500 \)
- \( P(0) = 100 \)
- \( P(1) = 20 \)
- \( P(2) = 4 \)

We'll use two of the points to create equations to solve for \( a \) and \( b \).

From \( P(0) = 100 \):
\[
P(0) = a \cdot b^0 = a = 100
\]

Now substituting \( a = 100 \) into the other points:

1. For \( P(-1) = 500 \):
\[
100 \cdot b^{-1} = 500 \implies b^{-1} = \frac{500}{100} = 5 \implies b = \frac{1}{5}
\]

2. For \( P(-2) = 2500 \):
\[
100 \cdot b^{-2} = 2500 \implies b^{-2} = \frac{2500}{100} = 25 \implies b^2 = \frac{1}{25} \implies b = \frac{1}{5}
\]

Thus, we have \( a = 100 \) and \( b = \frac{1}{5} \).

Now, we can write the function:
\[
P(x) = 100 \left( \frac{1}{5} \right)^x
\]

To verify, we can calculate \( P(1) \) and \( P(2) \):

- For \( P(1) \):
\[
P(1) = 100 \left( \frac{1}{5} \right)^1 = 100 \cdot \frac{1}{5} = 20
\]

- For \( P(2) \):
\[
P(2) = 100 \left( \frac{1}{5} \right)^2 = 100 \cdot \frac{1}{25} = 4
\]

Both values are correct according to the data from the table. Thus, the exponential function that represents the relation shown in the table is:

\[
P(x) = 100 \left( \frac{1}{5} \right)^x
\]