this is the graph of an exponential function.\nwrite an equation for the function in the form $f(x) =…

this is the graph of an exponential function.\nwrite an equation for the function in the form $f(x) = a(b)^x$.\nuse whole numbers, decimals, or simplified fractions for the values of $a$ and $b$.\n$f(x) = \\square$

this is the graph of an exponential function.\nwrite an equation for the function in the form $f(x) = a(b)^x$.\nuse whole numbers, decimals, or simplified fractions for the values of $a$ and $b$.\n$f(x) = \\square$

Answer

Explanation:

Step1: Identify the y-intercept

The exponential function is in the form ( f(x) = a(b)^x ). The y-intercept occurs when ( x = 0 ). From the graph, when ( x = 0 ), ( y=\frac{1}{2} ) (or 0.5)? Wait, no, looking at the graph, when ( x = 0 ), the point is at ( (0, \frac{1}{2}) )? Wait, no, let's check again. Wait, the graph passes through ( (0, \frac{1}{2}) )? Wait, no, maybe I made a mistake. Wait, when ( x = 0 ), ( f(0)=a(b)^0=a(1)=a ). So the y-intercept is ( a ). Looking at the graph, when ( x = 0 ), the function value is ( \frac{1}{2} )? Wait, no, the graph at ( x = 0 ) is at ( y = \frac{1}{2} )? Wait, no, let's see the grid. The y-axis: at ( x = 0 ), the point is between 0 and 1, closer to 0.5? Wait, no, maybe the graph passes through ( (0, \frac{1}{2}) ) or ( (0, 1) )? Wait, no, the graph at ( x = 0 ) is at ( y = \frac{1}{2} )? Wait, no, let's check the general form. Wait, maybe the graph is a decay function, so ( 0 < b < 1 ). Let's take another point. Let's see, when ( x = 1 ), what's the y-value? From the graph, when ( x = 1 ), ( y ) is very close to 0, but maybe let's take ( x = -1 ). Wait, no, let's start with ( x = 0 ). So ( f(0) = a = \frac{1}{2} )? Wait, no, maybe ( a = \frac{1}{2} ), and then find ( b ). Wait, let's suppose another point. Let's say when ( x = 1 ), what's the value? Wait, maybe the graph is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x )? No, that would be ( \left( \frac{1}{2} \right)^{x + 1} ). Wait, no, let's do it properly.

Wait, the standard form is ( f(x) = a b^x ). At ( x = 0 ), ( f(0) = a ). From the graph, when ( x = 0 ), ( y = \frac{1}{2} )? Wait, no, the graph at ( x = 0 ) is at ( y = \frac{1}{2} )? Wait, the grid lines: the y-axis has 0, 1, 2, etc. Wait, the graph crosses the y-axis at ( (0, \frac{1}{2}) )? Wait, no, the point at ( x = 0 ) is between 0 and 1, maybe ( \frac{1}{2} ). Then, let's take another point. Let's say when ( x = 1 ), what's the value? If ( x = 1 ), then ( f(1) = a b^1 ). If ( a = \frac{1}{2} ), and let's see the graph. Wait, maybe the correct approach is:

  1. Find ( a ): when ( x = 0 ), ( f(0) = a \cdot b^0 = a \cdot 1 = a ). So the y-intercept is ( a ). From the graph, at ( x = 0 ), the function is at ( y = \frac{1}{2} )? Wait, no, the graph at ( x = 0 ) is at ( y = \frac{1}{2} )? Wait, the grid: the y-axis has 0, 1, 2, ... So at ( x = 0 ), the point is at ( (0, \frac{1}{2}) ), so ( a = \frac{1}{2} ).

  2. Now find ( b ). Let's take another point. Let's say when ( x = 1 ), what's the y-value? From the graph, when ( x = 1 ), ( y ) is ( \frac{1}{4} )? Wait, if ( x = 1 ), ( f(1) = \frac{1}{2} \cdot b^1 ). If ( f(1) = \frac{1}{4} ), then ( \frac{1}{2} b = \frac{1}{4} ), so ( b = \frac{1}{2} ). So then the function is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ), which is ( f(x) = \left( \frac{1}{2} \right)^{x + 1} ), but in the form ( a b^x ), that's ( \frac{1}{2} \left( \frac{1}{2} \right)^x ). Wait, but let's check ( x = -1 ). If ( x = -1 ), then ( f(-1) = \frac{1}{2} \left( \frac{1}{2} \right)^{-1} = \frac{1}{2} \cdot 2 = 1 ). Does the graph pass through ( (-1, 1) )? Looking at the graph, when ( x = -1 ), the y-value is 1? Let's see the grid. At ( x = -1 ), the graph is at ( y = 1 )? Yes, because when ( x = -1 ), the point is at ( ( -1, 1) ). So let's verify:

  • When ( x = -1 ): ( f(-1) = a b^{-1} = \frac{a}{b} ). If ( a = \frac{1}{2} ) and ( b = \frac{1}{2} ), then ( \frac{\frac{1}{2}}{\frac{1}{2}} = 1 ), which matches ( ( -1, 1) ).

  • When ( x = 0 ): ( f(0) = \frac{1}{2} \left( \frac{1}{2} \right)^0 = \frac{1}{2} \cdot 1 = \frac{1}{2} ), which matches the y-intercept.

  • When ( x = 1 ): ( f(1) = \frac{1}{2} \left( \frac{1}{2} \right)^1 = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} ), which is consistent with the graph (at ( x = 1 ), y is ( \frac{1}{4} )).

So the equation is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ), or simplified as ( f(x) = \left( \frac{1}{2} \right)^{x + 1} ), but in the form ( a b^x ), it's ( \frac{1}{2} \left( \frac{1}{2} \right)^x ). Wait, but let's check again. Wait, maybe ( a = \frac{1}{2} ) and ( b = \frac{1}{2} ), so ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ).

Wait, another way: the general form of an exponential decay function is ( f(x) = a (1 - r)^x ), but here it's ( a b^x ) with ( 0 < b < 1 ). From the points ( (0, \frac{1}{2}) ) and ( (-1, 1) ), we can set up equations.

For ( x = 0 ): ( f(0) = a = \frac{1}{2} ).

For ( x = -1 ): ( f(-1) = a b^{-1} = 1 ). Substitute ( a = \frac{1}{2} ):

( \frac{1}{2} \cdot \frac{1}{b} = 1 )

Multiply both sides by ( b ): ( \frac{1}{2} = b )

So ( b = \frac{1}{2} )

Thus, the function is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ), which can be written as ( f(x) = \left( \frac{1}{2} \right)^{x + 1} ), but in the form ( a b^x ), it's ( \frac{1}{2} \left( \frac{1}{2} \right)^x ).

Wait, but let's check the graph again. At ( x = 0 ), the y-value is ( \frac{1}{2} ), at ( x = 1 ), it's ( \frac{1}{4} ), at ( x = -1 ), it's 1. So yes, that fits. So the equation is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ), or ( f(x) = \frac{1}{2} \cdot \left( \frac{1}{2} \right)^x ).

Step2: Write the equation

Now that we have ( a = \frac{1}{2} ) and ( b = \frac{1}{2} ), the function is ( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ).

Answer:

( f(x) = \frac{1}{2} \left( \frac{1}{2} \right)^x ) (or simplified as ( f(x) = \left( \frac{1}{2} \right)^{x + 1} ), but in the form ( a b^x ), it's ( \frac{1}{2} \left( \frac{1}{2} \right)^x ))