Question

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 $g(x)$.

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: \( 0 \)