Question

the table of ordered pairs (x, y) gives an exponential function. write an equation for the function. \\(\

$$\begin{array}{|c|c|} \\hline x & y \\\\ \\hline -1 & \\dfrac{1}{8} \\\\ \\hline 0 & \\dfrac{1}{4} \\\\ \\hline 1 & \\dfrac{1}{2} \\\\ \\hline 2 & 1 \\\\ \\hline \\end{array}$$

\\)

Explanation:

Step1: Recall exponential form

The general form of an exponential function is $y = ab^x$, where $a$ is the initial value (when $x=0$), and $b$ is the base.

Step2: Find $a$ using $x=0$

When $x=0$, $y=\frac{1}{4}$. Substitute into $y=ab^x$:
$\frac{1}{4} = ab^0$
Since $b^0=1$, we get $a = \frac{1}{4}$.

Step3: Find $b$ using another point

Use the point $(1, \frac{1}{2})$. Substitute $a=\frac{1}{4}$, $x=1$, $y=\frac{1}{2}$ into $y=ab^x$:
$\frac{1}{2} = \frac{1}{4}b^1$
Solve for $b$:
$b = \frac{1}{2} \div \frac{1}{4} = 2$

Step4: Verify with another point

Use $(2, 1)$: $y = \frac{1}{4}(2)^2 = \frac{1}{4} \times 4 = 1$, which matches. Use $(-1, \frac{1}{8})$: $y = \frac{1}{4}(2)^{-1} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$, which matches.

Answer:

$y = \frac{1}{4}(2)^x$