Question

secondary math iii // module 4
rational expressions & functions - 4.2
4.2

1. $f(x) = -\frac{5}{x - 3}$

v.a.
h.a.
domain:
range:

1. $f(x) = \frac{1}{(x + 3)} - 4$

v.a.
h.a.
domain:
range:

1. write a function of the form $f(x) = \frac{a}{x - h} + k$ with a vertical asymptote at $x = -15$ and a horizontal asymptote at $y = -6$.

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Explanation:

Step1: Recall Asymptote Rules

For a function \( f(x)=\frac{a}{x - h}+k \), the vertical asymptote (V.A.) is at \( x = h \) (since the denominator cannot be zero), and the horizontal asymptote (H.A.) is at \( y = k \) (as \( x\to\pm\infty \), \( \frac{a}{x - h}\to0 \), so \( f(x)\to k \)).

Step2: Determine \( h \) and \( k \) for the Function

We need a vertical asymptote at \( x=-15 \) and horizontal asymptote at \( y = -6 \). Comparing with \( f(x)=\frac{a}{x - h}+k \), we set \( h=-15 \) (so the denominator is \( x - (-15)=x + 15 \)) and \( k=-6 \). The value of \( a \) can be any non - zero real number (let's choose \( a = 1 \) for simplicity, but any non - zero \( a \) works).

Step3: Write the Function

Substituting \( a = 1 \), \( h=-15 \), and \( k=-6 \) into the form \( f(x)=\frac{a}{x - h}+k \), we get \( f(x)=\frac{1}{x+15}-6 \). (We could also choose other non - zero values for \( a \), for example, if \( a = 2 \), the function would be \( f(x)=\frac{2}{x + 15}-6 \), but the simplest choice is \( a = 1 \).)

Answer:

A possible function is \( f(x)=\frac{1}{x + 15}-6 \) (or any function of the form \( f(x)=\frac{a}{x+15}-6 \) where \( a
eq0 \))