Question

this table shows several values of an exponential function.
x | f(x)
-4 | -2/125
-3 | -2/25
-2 | -2/5
-1 | -2
write an equation for the function in the form f(x) = a(b)^x.
use whole numbers, decimals, or simplified fractions for the values of a and b.
f(x) =

Explanation:

Step1: Recall the exponential function form

The general form of an exponential function is \( f(x) = a(b)^x \). We need to find the values of \( a \) and \( b \). Let's use the given points to set up equations. Let's take the point \( x = -1 \), \( f(-1) = -2 \). Substitute into the function: \( -2 = a(b)^{-1} \), which can be rewritten as \( -2 = \frac{a}{b} \), so \( a = -2b \) (Equation 1).

Step2: Use another point, say \( x = -2 \), \( f(-2) = -\frac{2}{5} \)

Substitute into \( f(x) = a(b)^x \): \( -\frac{2}{5} = a(b)^{-2} = \frac{a}{b^2} \). Now substitute \( a = -2b \) from Equation 1 into this equation: \( -\frac{2}{5} = \frac{-2b}{b^2} \). Simplify the right - hand side: \( \frac{-2b}{b^2}=\frac{-2}{b} \) (assuming \( b eq0 \)). So we have the equation \( -\frac{2}{5}=\frac{-2}{b} \).

Step3: Solve for \( b \)

Cross - multiply the equation \( -\frac{2}{5}=\frac{-2}{b} \): \( -2\times b=-2\times5 \). Simplify both sides: \( -2b=-10 \). Divide both sides by - 2: \( b = 5 \).

Step4: Solve for \( a \)

Substitute \( b = 5 \) into Equation 1 (\( a=-2b \)): \( a=-2\times5=-10 \)? Wait, no, wait. Wait, when \( x=-1 \), \( f(-1)=-2=a(b)^{-1}=\frac{a}{b} \). If \( b = 5 \), then \( \frac{a}{5}=-2 \), so \( a=-2\times5=-10 \)? Wait, but let's check with another point. Let's take \( x=-2 \), \( f(-2)=-\frac{2}{5} \). If \( a=-10 \) and \( b = 5 \), then \( f(-2)=-10\times(5)^{-2}=-10\times\frac{1}{25}=-\frac{10}{25}=-\frac{2}{5} \), which matches. Let's check \( x=-3 \): \( f(-3)=-10\times(5)^{-3}=-10\times\frac{1}{125}=-\frac{10}{125}=-\frac{2}{25} \), which matches. And \( x = - 4 \): \( f(-4)=-10\times(5)^{-4}=-10\times\frac{1}{625}=-\frac{10}{625}=-\frac{2}{125} \), which also matches. Wait, but wait, earlier when I thought \( a=-2b \), let's re - examine. Wait, when \( x=-1 \), \( f(x)=a(b)^x \), so \( f(-1)=a(b)^{-1}=\frac{a}{b}=-2 \), so \( a=-2b \). Then when we used \( x = - 2 \), \( f(-2)=a(b)^{-2}=\frac{a}{b^2}=-\frac{2}{5} \). Substituting \( a=-2b \) gives \( \frac{-2b}{b^2}=\frac{-2}{b}=-\frac{2}{5} \), so \( \frac{2}{b}=\frac{2}{5} \), so \( b = 5 \). Then \( a=-2b=-10 \)? Wait, but let's check the function \( f(x)=-10(5)^x \). When \( x=-1 \), \( f(-1)=-10\times5^{-1}=-10\times\frac{1}{5}=-2 \), correct. When \( x=-2 \), \( f(-2)=-10\times5^{-2}=-10\times\frac{1}{25}=-\frac{2}{5} \), correct. When \( x=-3 \), \( f(-3)=-10\times5^{-3}=-10\times\frac{1}{125}=-\frac{2}{25} \), correct. When \( x=-4 \), \( f(-4)=-10\times5^{-4}=-10\times\frac{1}{625}=-\frac{2}{125} \), correct. Wait, but another way: Let's consider the ratio of consecutive terms. For an exponential function, the ratio of \( f(x) \) values when \( x \) increases by 1 should be \( b \). Let's take \( x=-4 \) to \( x=-3 \): \( \frac{f(-3)}{f(-4)}=\frac{-\frac{2}{25}}{-\frac{2}{125}}=\frac{125}{25}=5 \). From \( x=-3 \) to \( x=-2 \): \( \frac{f(-2)}{f(-3)}=\frac{-\frac{2}{5}}{-\frac{2}{25}}=\frac{25}{5}=5 \). From \( x=-2 \) to \( x=-1 \): \( \frac{f(-1)}{f(-2)}=\frac{-2}{-\frac{2}{5}} = 5 \). So the common ratio \( b = 5 \). Now, to find \( a \), we can use the formula \( f(x)=a(b)^x \). Let's use \( x=-1 \), \( f(-1)=-2=a(5)^{-1}=\frac{a}{5} \), so \( a=-2\times5=-10 \). Wait, but let's write the function: \( f(x)=-10(5)^x \)? Wait, no, wait, when \( x = 0 \), what would \( f(0) \) be? \( f(0)=-10(5)^0=-10 \). But let's check with the points. Wait, maybe I made a mistake. Wait, let's take \( x=-1 \), \( f(-1)=-2 \). If \( b = 5 \), then \( f(x)=a(5)^x \). So \( f(-1)=a(5)^{-1}=\frac{a}{5}=-2 \), so \( a=-10 \). Then \( f(-2)=-10(5)^{-2}=-10\times\frac{1}{25}=-\frac{2}{5} \), which is correct. \( f(-3)=-10(5)^{-3}=-10\times\frac{1}{125}=-\frac{2}{25} \), correct. \( f(-4)=-10(5)^{-4}=-10\times\frac{1}{625}=-\frac{2}{125} \), correct. So the function is \( f(x)=-10(5)^x \)? Wait, no, wait, wait a second. Wait, the general form is \( f(x)=a(b)^x \). Let's check the ratio again. From \( x=-4 \) to \( x=-3 \): \( f(-3)/f(-4)=(-\frac{2}{25})/(-\frac{2}{125}) = 5 \). From \( x=-3 \) to \( x=-2 \): \( (-\frac{2}{5})/(-\frac{2}{25}) = 5 \). From \( x=-2 \) to \( x=-1 \): \( (-2)/(-\frac{2}{5}) = 5 \). So the base \( b = 5 \). Now, to find \( a \), we can use the formula \( f(x)=a(b)^x \). Let's use \( x=-1 \): \( -2=a(5)^{-1}\implies a=-2\times5=-10 \). So the function is \( f(x)=-10(5)^x \)? Wait, but let's write it as \( f(x)=-10\times5^{x} \). Wait, but let's check with \( x=-1 \): \( -10\times5^{-1}=-10\times\frac{1}{5}=-2 \), correct. \( x=-2 \): \( -10\times5^{-2}=-10\times\frac{1}{25}=-\frac{2}{5} \), correct. So that's the function.

Answer:

\( f(x)=-10(5)^{x} \)