How do you identify the oblique asymptote of $f(x) = (2x^2+3x+8)/(x+3)$ ?

Question

3455 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-02T14:44:50

Oblique Asymptote is $y=2x-3$
Vertical Asymptote is $x=-3$

from the given:

$f(x)=(2x^2+3x+8)/(x+3)$

perform long division so that the result is

$(2x^2+3x+8)/(x+3)=2x-3+17/(x+3)$

Notice the part of the quotient

$2x-3$

equate this to $y$ like as follows

$y=2x-3$ this is the line which is the Oblique Asymptote

And the divisor $x+3$ be equated to zero and that is the Vertical asymptote

$x+3=0$ or $x=-3$

You can see the lines $x=-3$ and $y=2x-3$ and the graph of
$f(x)=(2x^2+3x+8)/(x+3)$
graph{(y-(2x^2+3x+8)/(x+3))(y-2x+3)=0[-60,60,-30,30]}

God bless...I hope the explanation is useful..

How do you identify the oblique asymptote of f(x) = (2x^2+3x+8)/(x+3)f(x) = (2x^2+3x+8)/(x+3)?