Question

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

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

Explanation:

Step1: Recall exponential function form

The general form of an exponential function is \( y = a \cdot b^x \), where \( a \) is the initial value (when \( x = 0 \)) and \( b \) is the base (the common ratio between consecutive \( y \)-values).

Step2: Find the value of \( a \)

When \( x = 0 \), from the table, \( y=\frac{1}{4} \). Substitute \( x = 0 \) and \( y=\frac{1}{4} \) into \( y = a \cdot b^x \):
\( \frac{1}{4}=a\cdot b^0 \), and since \( b^0 = 1 \), we get \( a=\frac{1}{4} \).

Step3: Find the value of \( b \)

Take two consecutive points, e.g., \( (0,\frac{1}{4}) \) and \( (1,\frac{1}{2}) \). The ratio \( \frac{y_2}{y_1}=\frac{\frac{1}{2}}{\frac{1}{4}} = 2 \). So the base \( b = 2 \).

Step4: Write the equation

Substitute \( a=\frac{1}{4} \) and \( b = 2 \) into \( y = a \cdot b^x \), we get \( y=\frac{1}{4}\cdot2^x \). We can also rewrite \( \frac{1}{4}\cdot2^x \) as \( 2^{x - 2} \) (since \( \frac{1}{4}=2^{-2} \), so \( 2^{-2}\cdot2^x=2^{x - 2} \)) or check with other points. For \( x = 1 \), \( y=\frac{1}{4}\cdot2^1=\frac{1}{2} \) (matches), \( x = 2 \), \( y=\frac{1}{4}\cdot2^2 = 1 \) (matches), \( x=-1 \), \( y=\frac{1}{4}\cdot2^{-1}=\frac{1}{8} \) (matches). Another way: notice the pattern, when \( x \) increases by 1, \( y \) doubles. So starting from \( x = 0 \), \( y=\frac{1}{4} \), then \( y=\frac{1}{4}\cdot2^x \) or \( y = 2^{x - 2} \).

Answer:

\( y=\frac{1}{4}\cdot2^x \) (or \( y = 2^{x - 2} \))