Question

what are the equations of the horizontal and vertical asymptotes of $y = \frac{1}{x - 9} - 7$?
horizontal asymptote:
vertical asymptote:
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Explanation:

Step1: Find Vertical Asymptote

For a rational function \( y=\frac{f(x)}{g(x)} \), vertical asymptotes occur where \( g(x) = 0 \) (and \( f(x)
eq0 \) there). Here, the function is \( y=\frac{1}{x - 9}-7=\frac{1-7(x - 9)}{x - 9}=\frac{1-7x + 63}{x - 9}=\frac{-7x + 64}{x - 9} \). The denominator \( x - 9 = 0 \) when \( x = 9 \). So vertical asymptote is \( x = 9 \).

Step2: Find Horizontal Asymptote

For a rational function, if the degree of numerator and denominator are equal, horizontal asymptote is ratio of leading coefficients. If degree of numerator < degree of denominator, horizontal asymptote is \( y = 0 \) (shifted if there's a vertical shift). Here, the original function can be thought of as a transformation of \( y=\frac{1}{x} \). The function \( y=\frac{1}{x - 9}-7 \) has a vertical shift of -7. For \( y=\frac{1}{x - h}+k \), horizontal asymptote is \( y = k \) (since as \( x\to\pm\infty \), \( \frac{1}{x - 9}\to0 \), so \( y\to0 - 7=-7 \)). So horizontal asymptote is \( y=-7 \).

Answer:

horizontal asymptote: \( y = -7 \)
vertical asymptote: \( x = 9 \)