Question

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

Explanation:

Step1: Analyze the logarithmic function

The general form of a logarithmic function is \( y = \log_b(x - h) + k \), where the vertical asymptote is \( x = h \). For the function \( f(x)=-\log_{5}x - 5 \), we can rewrite it as \( f(x)=-\log_{5}(x - 0)-5 \). So, comparing with the general form, \( h = 0 \).

Step2: Determine the type of asymptote

For logarithmic functions of the form \( y=\log_b(x - h)+k \), the vertical asymptote is at \( x = h \) because the logarithm is only defined for \( x - h>0 \). As \( x \) approaches \( h \) from the right, the function values tend to \( \pm\infty \). In our function, \( h = 0 \), so the vertical asymptote is \( x = 0 \) (the y - axis). Logarithmic functions do not have horizontal asymptotes in the traditional sense (they have vertical asymptotes), so the asymptote is vertical.

Step3: Find two points with integer coordinates

• When \( x = 1 \):

Substitute \( x = 1 \) into the function \( f(x)=-\log_{5}(1)-5 \). Since \( \log_{5}(1) = 0 \) (because \( 5^0=1 \)), then \( f(1)=- 0-5=-5 \). So the point is \( (1,-5) \).

• When \( x = 5 \):

Substitute \( x = 5 \) into the function \( f(x)=-\log_{5}(5)-5 \). Since \( \log_{5}(5) = 1 \) (because \( 5^1 = 5 \)), then \( f(5)=-1 - 5=-6 \). So the point is \( (5,-6) \). We could also use \( x=\frac{1}{5} \), but \( x = 1 \) and \( x = 5 \) are simpler with integer \( x \)-values.

Answer:

The asymptote is vertical (the equation of the vertical asymptote is \( x = 0 \)). Two points with integer coordinates are \( (1,-5) \) and \( (5,-6) \) (or other valid integer - coordinate points like \( ( \frac{1}{5},-4) \) etc., but \( (1,-5) \) and \( (5,-6) \) are straightforward). The type of asymptote (from the dropdown) should be selected as "Vertical".