Question

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

x	y
-1	\\(\frac{1}{3}\\)
0	3
1	27
2	243

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.

Step2: Find the value of \( a \)

When \( x = 0 \), \( y = 3 \). Substitute into \( y = a \cdot b^x \):
\( 3 = a \cdot b^0 \)
Since \( b^0 = 1 \), we get \( a = 3 \).

Step3: Find the value of \( b \)

Use another point, say \( x = 1 \), \( y = 27 \). Substitute \( a = 3 \), \( x = 1 \), \( y = 27 \) into \( y = 3 \cdot b^x \):
\( 27 = 3 \cdot b^1 \)
Divide both sides by 3: \( b = \frac{27}{3} = 9 \)? Wait, no, wait. Wait, let's check another point. Wait, when \( x = -1 \), \( y = \frac{1}{3} \). Let's use \( x = -1 \), \( y = \frac{1}{3} \), \( a = 3 \):
\( \frac{1}{3} = 3 \cdot b^{-1} \)
\( \frac{1}{3} = \frac{3}{b} \)
Cross - multiply: \( b = 3\times3 = 9 \)? Wait, no, wait when \( x = 1 \), \( y = 27 \), \( a = 3 \), \( y = 3\cdot b^1=27\), so \( b = 9 \)? Wait, but when \( x = 2 \), \( y = 243 \). Let's check \( 3\times9^2=3\times81 = 243 \), which matches. And when \( x=-1 \), \( 3\times9^{-1}=3\times\frac{1}{9}=\frac{1}{3} \), which also matches. Wait, but wait, 9 is \( 3^2 \), but let's re - check. Wait, maybe I made a mistake. Wait, let's see the ratio between consecutive \( y \)-values. From \( x = 0 \) to \( x = 1 \), \( y \) goes from 3 to 27, the ratio is \( \frac{27}{3}=9 \). From \( x = 1 \) to \( x = 2 \), \( \frac{243}{27}=9 \). From \( x=-1 \) to \( x = 0 \), \( \frac{3}{\frac{1}{3}} = 9 \). So the common ratio \( b = 9 \)? Wait, but 9 is \( 3^2 \), but the general form is \( y=a\cdot b^x \), with \( a = 3 \), \( b = 9 \). Wait, but let's write \( 9 \) as \( 3^2 \), so \( y = 3\cdot(3^2)^x=3\cdot3^{2x}=3^{2x + 1} \), but no, the general form is \( y=a\cdot b^x \). Wait, maybe I miscalculated \( b \). Wait, when \( x = 1 \), \( y = 27 \), \( a = 3 \), so \( 27=3\times b^1\), so \( b = 9 \). And when \( x=-1 \), \( y=\frac{1}{3}=3\times b^{-1}\), so \( b = 9 \), which is correct.

Wait, but let's re - express. Wait, the general form is \( y=a\cdot b^x \). We found \( a = 3 \) (from \( x = 0 \)), and \( b = 9 \) (from the ratio of \( y \)-values). So the equation is \( y = 3\cdot9^x \). But wait, 9 is \( 3^2 \), so \( y = 3\cdot(3^2)^x=3^{2x + 1} \), but in the form \( y=a\cdot b^x \), it's \( y = 3\cdot9^x \). Let's verify all points:

• For \( x=-1 \): \( y = 3\times9^{-1}=3\times\frac{1}{9}=\frac{1}{3} \) (correct)
• For \( x = 0 \): \( y = 3\times9^{0}=3\times1 = 3 \) (correct)
• For \( x = 1 \): \( y = 3\times9^{1}=27 \) (correct)
• For \( x = 2 \): \( y = 3\times9^{2}=3\times81 = 243 \) (correct)

Wait, but another way: maybe the base is 9? Or did I mis - see the numbers? Wait, the table is \( x=-1,y = 1/3 \); \( x = 0,y = 3 \); \( x = 1,y = 27 \); \( x = 2,y = 243 \). Let's check the ratio between \( y \)-values: \( 3\div(1/3)=9 \), \( 27\div3 = 9 \), \( 243\div27=9 \). So the common ratio (base) \( b = 9 \), and the initial value \( a = 3 \) (when \( x = 0 \)). So the equation is \( y = 3\cdot9^x \). But 9 is \( 3^2 \), so \( y = 3\cdot(3^2)^x=3^{2x + 1} \), but in the form \( y=a\cdot b^x \), \( a = 3 \), \( b = 9 \).

Wait, but maybe I made a mistake in the base. Let's check with \( x = 1 \), \( y = 27 \), \( a = 3 \), so \( b=\frac{y}{a}\) when \( x = 1 \), so \( b = 9 \). Yes, that's correct.

Answer:

\( y = 3\cdot9^x \) (or \( y = 3\times9^{x} \))