Question

enter the function of the form $f(x)=a(2)^x$ that is represented by the graph.
graph of an exponential function with grid, x-axis from -4 to 4, y-axis from 0 to 32, passing through (0,4) and (2,16) approximately
enter the correct function in the box below.
show hints
$f(x)=\square$

Explanation:

Step1: Identify a point on the graph

From the graph, when \( x = 1 \), we can estimate the value of \( f(x) \). Wait, actually, looking at the y - intercept (when \( x = 0 \)), the graph passes through \( (0, 4) \)? Wait, no, let's check again. Wait, the grid: when \( x = 0 \), the graph is at \( y = 4 \)? Wait, no, let's see the coordinates. Wait, the function is \( f(x)=a(2)^x \). Let's find a point on the graph. Let's take \( x = 2 \), what's \( f(2) \)? From the graph, when \( x = 2 \), \( f(x)=16 \)? Wait, no, let's use the y - intercept. Wait, when \( x = 0 \), \( f(0)=a(2)^0=a\times1 = a \). Looking at the graph, when \( x = 0 \), the function value is 4? Wait, no, the graph at \( x = 0 \) is at \( y = 4 \)? Wait, let's check the grid. The y - axis has marks at 8, 16, 24, 32. Wait, between 0 and 8, there is a mark at 4? Wait, maybe I made a mistake. Wait, let's take \( x = 2 \), \( f(2)=16 \). Let's substitute into the function \( f(x)=a(2)^x \). So when \( x = 2 \), \( f(2)=a(2)^2 = 4a \). If \( f(2)=16 \), then \( 4a = 16 \), so \( a = 4 \)? Wait, no, wait when \( x = 1 \), what's \( f(1) \)? If \( a = 4 \), then \( f(1)=4\times2^1 = 8 \), which matches the graph (at \( x = 1 \), the graph is at 8). And when \( x = 0 \), \( f(0)=4\times2^0 = 4 \), which also matches the graph (the y - intercept is 4). Wait, let's confirm with the function form. The general form is \( f(x)=a(2)^x \). We need to find \( a \). Let's use the point \( (0, 4) \). When \( x = 0 \), \( f(0)=a(2)^0=a \). So \( a = f(0) \). From the graph, when \( x = 0 \), \( f(0)=4 \). So \( a = 4 \). Therefore, the function is \( f(x)=4(2)^x \). Wait, but let's check \( x = 2 \): \( 4\times2^2=4\times4 = 16 \), which matches the graph (at \( x = 2 \), the graph is at 16). And \( x = 1 \): \( 4\times2^1 = 8 \), which also matches. So that works.

Wait, maybe I misread the y - intercept. Let's do it properly. The function is \( f(x)=a(2)^x \). We can pick a point from the graph. Let's take \( (0, 4) \) (since when \( x = 0 \), the graph is at \( y = 4 \)). Substitute \( x = 0 \) and \( f(0)=4 \) into the function:

Step1: Substitute \( x = 0 \) into \( f(x)=a(2)^x \)

We know that for any number \( b
eq0 \), \( b^0 = 1 \). So when \( x = 0 \), \( f(0)=a\times(2)^0=a\times1=a \).

From the graph, when \( x = 0 \), \( f(0) = 4 \). So \( a = 4 \).

Step2: Write the function

Now that we know \( a = 4 \), the function is \( f(x)=4(2)^x \).

Answer:

\( f(x)=4(2)^x \)