Question

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

Explanation:

Step1: Recall exponential function form

The general form of an exponential function is \( y = ab^x \), where \( a \) and \( b \) are constants, and \( b>0, b
eq1 \).

Step2: Use \( x = 0 \) to find \( a \)

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: Use another point to find \( b \)

Let's use \( x = 1 \), \( y = 2 \). Now we know \( a=\frac{1}{4} \), substitute into \( y = ab^x \):
\( 2=\frac{1}{4}b^{1} \)
Solve for \( b \): Multiply both sides by 4, \( b = 8 \)? Wait, no, wait. Wait, when \( x = 1 \), \( y = 2 \), \( a=\frac{1}{4} \), so \( 2=\frac{1}{4}b \), so \( b = 8 \)? Wait, but let's check with \( x = 2 \), \( y = 16 \). If \( b = 8 \), then \( y=\frac{1}{4}(8)^2=\frac{1}{4}(64)=16 \), which matches. Wait, but let's check \( x=-1 \), \( y=\frac{1}{32} \). \( y=\frac{1}{4}(8)^{-1}=\frac{1}{4}\times\frac{1}{8}=\frac{1}{32} \), which matches. Wait, but earlier when I thought \( b = 8 \), but let's re - check the steps.

Wait, maybe I made a mistake in step 3. Wait, when \( x = 1 \), \( y = 2 \), \( a=\frac{1}{4} \), so \( 2=\frac{1}{4}b^{1}\), so \( b=2\times4 = 8 \). Yes, that's correct.

Wait, but let's write the function. So \( y=\frac{1}{4}\times8^{x} \). Let's verify all points:

• For \( x=-1 \): \( y=\frac{1}{4}\times8^{-1}=\frac{1}{4}\times\frac{1}{8}=\frac{1}{32} \), correct.
• For \( x = 0 \): \( y=\frac{1}{4}\times8^{0}=\frac{1}{4}\times1=\frac{1}{4} \), correct.
• For \( x = 1 \): \( y=\frac{1}{4}\times8^{1}=\frac{1}{4}\times8 = 2 \), correct.
• For \( x = 2 \): \( y=\frac{1}{4}\times8^{2}=\frac{1}{4}\times64 = 16 \), correct.

Alternatively, we can rewrite \( 8^{x}=(2^{3})^{x}=2^{3x} \), and \( \frac{1}{4}=2^{-2} \), so \( y = 2^{-2}\times2^{3x}=2^{3x - 2} \). Let's check:

• When \( x=-1 \): \( 2^{-3 - 2}=2^{-5}=\frac{1}{32} \), correct.
• When \( x = 0 \): \( 2^{0 - 2}=2^{-2}=\frac{1}{4} \), correct.
• When \( x = 1 \): \( 2^{3 - 2}=2^{1}=2 \), correct.
• When \( x = 2 \): \( 2^{6 - 2}=2^{4}=16 \), correct.

So the exponential function is \( y = 2^{3x - 2} \) or \( y=\frac{1}{4}\times8^{x} \) (since \( 8^{x}=2^{3x} \) and \( \frac{1}{4}=2^{-2} \), so \( 2^{-2}\times2^{3x}=2^{3x - 2} \)).

Answer:

The equation of the exponential function is \( y = \frac{1}{4}\times8^{x} \) (or \( y = 2^{3x - 2} \))