SL 2.8—Reciprocal and simple rational functions, equations of asymptotes

The function is $f(x)=\frac{4x-5}{x+3}$, which is a simple rational function of the form $\frac{ax+b}{cx+d}$ with $a=4$, $b=-5$, $c=1$, $d=3$.

Set the denominator equal to zero: $x+3=0\Rightarrow x=-3$. This is the vertical asymptote.

Since numerator and denominator have equal degree, divide the leading coefficients: $y=\frac{a}{c}=\frac{4}{1}=4$. This is the horizontal asymptote.

The asymptotes are $x=-3$ and $y=4$.

We need the equations of both the vertical and horizontal asymptotes of $f(x)=\frac{4x-5}{x+3}$.

The vertical asymptote comes from $x+3=0$, giving $x=-3$, not $x=3$. So options "$x=3$ and $y=4$" and "$x=3$ and $y=-4$" are eliminated.

The horizontal asymptote is $y=\frac{4}{1}=4$, not $y=-5$. So "$x=-3$ and $y=-5$" is eliminated — the $-5$ comes from the numerator's constant term, not the asymptote.

The only remaining option is "$x=-3$ and $y=4$", which matches both asymptotes correctly.

A vertical asymptote occurs where the denominator equals zero. For $f(x)=\frac{7}{mx-14}$, set $mx-14=0$.

We are told the vertical asymptote is at $x=2$. Substitute $x=2$: $m(2)-14=0$.

$2m=14\Rightarrow m=7$.

With $m=7$: denominator $=7x-14=7(x-2)$, which equals zero at $x=2$. ✓

We need the value of $m$ such that $f(x)=\frac{7}{mx-14}$ has a vertical asymptote at $x=2$.

If $m=-7$: denominator $=-7x-14=0\Rightarrow x=-2$. The asymptote would be at $x=-2$, not $x=2$. Eliminated.

If $m=2$: denominator $=2x-14=0\Rightarrow x=7$. The asymptote would be at $x=7$, not $x=2$. Eliminated.

If $m=14$: denominator $=14x-14=0\Rightarrow x=1$. The asymptote would be at $x=1$, not $x=2$. Eliminated.

Testing $m=7$: $7x-14=0\Rightarrow x=2$ ✓. The correct answer is $m=7$.