match each description of the rational function with its correct equation. each may be used exactly once…

Explanation:

Step1: Find vertical asymptotes

For a rational function $y = \frac{N(x)}{D(x)}$, vertical asymptotes occur at the values of $x$ that make $D(x)=0$. For $f(x)=\frac{777}{16x^{2}-40x + 25}=\frac{777}{(4x - 5)^{2}}$, setting $(4x - 5)^{2}=0$, we get $x=\frac{5}{4}$ (one vertical asymptote). For $g(x)=\frac{555}{16x^{2}-25}=\frac{555}{(4x + 5)(4x - 5)}$, setting $(4x + 5)(4x - 5)=0$, we get $x =-\frac{5}{4}$ and $x=\frac{5}{4}$ (two vertical asymptotes). For $h(x)=\frac{333}{16x^{2}+25}$, since $16x^{2}+25>0$ for all real - $x$ (because $16x^{2}\geq0$ and $16x^{2}+25\geq25$), there are no vertical asymptotes.

Step2: Find horizontal asymptotes

For a rational function $y=\frac{N(x)}{D(x)}$ where $N(x)=a_{n}x^{n}+\cdots+a_{0}$ and $D(x)=b_{m}x^{m}+\cdots + b_{0}$, if $n<m$, the horizontal asymptote is $y = 0$. Here, for $f(x),g(x),h(x)$, the degree of the numerator is $n = 0$ and the degree of the denominator is $m = 2$. So, the horizontal asymptote for all of them is $y = 0$.

Answer:

1. Has two vertical asymptotes and one horizontal asymptote: $g(x)=\frac{555}{16x^{2}-25}$
2. Has one vertical asymptote and one horizontal asymptote: $f(x)=\frac{777}{16x^{2}-40x + 25}$
3. Has no vertical asymptotes and one horizontal asymptote: $h(x)=\frac{333}{16x^{2}+25}$