graph the function $f(x) = -3(3)^x - 1$ on the axes below. you must plot the asymptote and any two points…

Explanation:

Step1: Find the Asymptote

For an exponential function of the form ( f(x) = a(b)^x + k ), the horizontal asymptote is ( y = k ). In the function ( f(x) = -3(3)^x - 1 ), ( k = -1 ), so the horizontal asymptote is ( y = -1 ).

Step2: Find Points with Integer Coordinates

Point 1: Let ( x = 0 )

Substitute ( x = 0 ) into ( f(x) ):
( f(0) = -3(3)^0 - 1 = -3(1) - 1 = -3 - 1 = -4 ).
So, the point is ( (0, -4) ).

Point 2: Let ( x = 1 )

Substitute ( x = 1 ) into ( f(x) ):
( f(1) = -3(3)^1 - 1 = -3(3) - 1 = -9 - 1 = -10 ).
So, the point is ( (1, -10) ). (Alternatively, ( x = -1 ): ( f(-1) = -3(3)^{-1} - 1 = -3(\frac{1}{3}) - 1 = -1 - 1 = -2 ), point ( (-1, -2) ) is also valid.)

Answer:

Asymptote: ( y = -1 )
Points: ( (0, -4) ) and ( (1, -10) ) (or other valid integer-coordinate points like ( (-1, -2) ))