Question

graph the function $f(x) = -3(3)^{x - 3}$ on the axes below. you must plot the asymptote and any two points with integer coordinates.
asymptote:
answer

Explanation:

Step1: Identify the asymptote of the exponential function

For an exponential function of the form \( f(x) = a \cdot b^{x - h} + k \), the horizontal asymptote is \( y = k \). In the given function \( f(x) = -3(3)^{x - 3} \), we can rewrite it as \( f(x) = -3(3)^{x - 3} + 0 \). So, comparing with the general form, \( k = 0 \). Thus, the horizontal asymptote is \( y = 0 \).

Step2: Find two points with integer coordinates

Point 1: Let \( x = 3 \)

Substitute \( x = 3 \) into the function \( f(x) = -3(3)^{x - 3} \).
\( f(3) = -3(3)^{3 - 3} = -3(3)^{0} = -3(1) = -3 \). So, the point is \( (3, -3) \).

Point 2: Let \( x = 4 \)

Substitute \( x = 4 \) into the function \( f(x) = -3(3)^{x - 3} \).
\( f(4) = -3(3)^{4 - 3} = -3(3)^{1} = -3(3) = -9 \). Wait, maybe a better choice. Let's take \( x = 2 \).
\( f(2) = -3(3)^{2 - 3} = -3(3)^{-1} = -3 \times \frac{1}{3} = -1 \). So, the point is \( (2, -1) \). Or \( x = 3 \) we did, let's take \( x = 1 \):
\( f(1) = -3(3)^{1 - 3} = -3(3)^{-2} = -3 \times \frac{1}{9} = -\frac{1}{3} \) (not integer). \( x = 4 \): \( f(4) = -3(3)^{1} = -9 \) (integer). \( x = 3 \): \( (3, -3) \), \( x = 4 \): \( (4, -9) \), or \( x = 2 \): \( (2, -1) \), \( x = 3 \): \( (3, -3) \).

Answer:

Asymptote: \( y = 0 \)
Two points with integer coordinates: \( (3, -3) \) and \( (4, -9) \) (or other valid integer coordinate points like \( (2, -1) \) and \( (3, -3) \))