an exponential function $f(x) = -4(2)^x$ is reflected across the $x$-axis to create $g(x)$. complete the…

an exponential function $f(x) = -4(2)^x$ is reflected across the $x$-axis to create $g(x)$. complete the steps explaining how to graph $g(x)$. enter your answers in the boxes. step 1: plot the initial value of $g(x)$ at $(0, \\square)$. step 2: plot two other points, one at $(1, 8)$ and the other at $(-1, \\square)$. step 3: note that the horizontal asymptote of $g(x)$ is $y = \\square$. step 4: draw the curve that includes these points and approaches the asymptote of $g(x)$.

an exponential function $f(x) = -4(2)^x$ is reflected across the $x$-axis to create $g(x)$. complete the steps explaining how to graph $g(x)$. enter your answers in the boxes. step 1: plot the initial value of $g(x)$ at $(0, \\square)$. step 2: plot two other points, one at $(1, 8)$ and the other at $(-1, \\square)$. step 3: note that the horizontal asymptote of $g(x)$ is $y = \\square$. step 4: draw the curve that includes these points and approaches the asymptote of $g(x)$.

Answer

Explanation:

Step1: Find reflection rule

Reflecting ( f(x) = -4(2)^x ) over ( x )-axis: ( g(x) = -f(x) = 4(2)^x ).

Step2: Find ( g(0) )

Substitute ( x = 0 ) into ( g(x) ): ( g(0) = 4(2)^0 = 4(1) = 4 ). So initial point is ( (0, 4) ).

Step3: Find ( g(-1) )

Substitute ( x = -1 ) into ( g(x) ): ( g(-1) = 4(2)^{-1} = 4 \times \frac{1}{2} = 2 ). So point is ( (-1, 2) ).

Step4: Find horizontal asymptote

For exponential ( g(x) = 4(2)^x ), horizontal asymptote is ( y = 0 ) (since ( \lim_{x\to -\infty} 4(2)^x = 0 )).

Answer:

Step 1: ( 4 )
Step 2: ( 2 )
Step 3: ( 0 )