How do you find vertical, horizontal and oblique asymptotes for $y =(3/x)+2$ ?

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Answer 1 · 2017-02-22T07:15:56

Vertical asymptote: $x = 0$
Horizontal asymptote: $y = 2$
Oblique asymptote: none

Find a common denominator for the function:
$y = 3/x +(2x)/x = (3+2x)/x = (2x+3)/x$

Rational Function: $(N(x))/(D(x))$ , when $N(x) = 0$ gives x-intercepts,
when $D(x) = 0$ you find vertical asymptotes.

Vertical asymptote at $x = 0$

When $(N(x))/(D(x)) =( a_nx^n+....)/(b_mx^m+....)$ where $n, m$ are the degrees of the polynomials.

If $n=m$ horizontal asymptote is at $y=a_n/b_m$

$n = m = 1$ , so Horizontal asymptote: $y = 2/1; y = 2$

To have an oblique or slant asymptote, $m+1 = n$ which is not the case, so there is no oblique asymptote.

How do you find vertical, horizontal and oblique asymptotes for y =(3/x)+2y =(3/x)+2?