Question

graph the function $f(x) = -5^x$ on the axes below. you must plot the asymptote and any two points with integer coordinates.
asymptote:
vertical
horizontal

Explanation:

Step1: Analyze the function type

The function \( f(x)=-5^{x} \) is an exponential function. For exponential functions of the form \( y = a\cdot b^{x}+k \), the horizontal asymptote is \( y = k \). In this case, \( a=-1 \), \( b = 5 \), and \( k = 0 \), so the horizontal asymptote is \( y=0 \) (the x - axis).

Step2: Find points with integer coordinates

• When \( x = 0 \):

Substitute \( x = 0 \) into \( f(x)=-5^{x} \), we get \( f(0)=-5^{0}=- 1 \). So the point is \( (0,-1) \).

• When \( x = 1 \):

Substitute \( x = 1 \) into \( f(x)=-5^{x} \), we get \( f(1)=-5^{1}=-5 \). So the point is \( (1, - 5) \).

• When \( x=-1 \):

Substitute \( x=-1 \) into \( f(x)=-5^{x} \), we get \( f(-1)=-5^{-1}=-\frac{1}{5}\), but this is not an integer coordinate. So we can use \( (0,-1) \) and \( (1,-5) \) or other integer - valued x. For example, when \( x = 2 \), \( f(2)=-5^{2}=-25 \), but \( (1,-5) \) and \( (0,-1) \) are simpler.

Answer:

The horizontal asymptote is \( y = 0 \). Two points with integer coordinates are \( (0,-1) \) and \( (1,-5) \) (or other valid integer - coordinate points like \( (2,-25) \), \( (- 1,-\frac{1}{5}) \) is not valid for integer coordinates, etc.). The asymptote type is Horizontal with equation \( y = 0 \).