Question

graph the function $f(x) = -2\log_{4} x - 4$ on the axes below. you must plot the asymptote and any two points with integer coordinates.
asymptote:
vertical
horizontal

Explanation:

Step1: Determine the asymptote of the logarithmic function

For a logarithmic function of the form \( y = \log_b(x - h) + k \), the vertical asymptote is \( x = h \). In the function \( f(x)=- 2\log_4x - 4 \), we can rewrite it as \( f(x)=-2\log_4(x - 0)-4 \). So the vertical asymptote is \( x = 0 \) (the y - axis), and the type of asymptote is vertical.

Step2: Find two points with integer coordinates

• When \( x = 1 \):

Substitute \( x = 1 \) into the function \( f(x)=-2\log_4x - 4 \). We know that \( \log_41 = 0 \) (since \( 4^0=1 \)). Then \( f(1)=-2\times0 - 4=-4 \). So the point is \( (1, - 4) \).

• When \( x = 4 \):

Substitute \( x = 4 \) into the function. We know that \( \log_44 = 1 \) (since \( 4^1 = 4 \)). Then \( f(4)=-2\times1-4=-2 - 4=-6 \). So the point is \( (4, - 6) \). Or we can also use \( x=\frac{1}{4} \): \( \log_4\frac{1}{4}=\log_44^{-1}=- 1 \), then \( f(\frac{1}{4})=-2\times(-1)-4 = 2 - 4=-2 \), the point is \( (\frac{1}{4},-2) \), but we need integer \( x \) values, so \( x = 1 \) and \( x = 4 \) are better.

Answer:

Asymptote type: Vertical; Asymptote equation: \( x = 0 \); Two points: \( (1, - 4) \), \( (4, - 6) \) (you can plot the vertical line \( x = 0 \), the point \( (1, - 4) \) and \( (4, - 6) \) on the graph)