SOLUTION: I am trying to find the slant asymptote of the function {{{ f(x)=(-4*x^2+6*x+3)/(6*x+9)}}}

Algebra ->  Rational-functions -> SOLUTION: I am trying to find the slant asymptote of the function {{{ f(x)=(-4*x^2+6*x+3)/(6*x+9)}}}       Log On


   

Question 732048: I am trying to find the slant asymptote of the function +f%28x%29=%28-4%2Ax%5E2%2B6%2Ax%2B3%29%2F%286%2Ax%2B9%29

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Oblique or Slant Asymptote
An oblique or a slant asymptote is an asymptote
that is neither vertical or horizontal.
If the degree of the numerator is one more than
the degree of the denominator, then the graph of
the rational function will have a slant asymptote.
The slant asymptote is the quotient part of the answer you get when you divide the numerator by the denominator.You may have 0 or 1 slant asymptote, but no more than that.
You draw a slant asymptote on the graph by putting a dashed horizontal (left and right) line going through y+=+mx+%2B+b.
Applying long division to this problem we get:
f%28x%29=%28-4x%5E2%2B6x%2B3%29%2F%286x%2B9%29

..................-2x%2F3%2B2
(6x%2B9)|-4x%5E2%2B6x%2B3
...........-4x%5E2%2B6x
________________________
.....................12x%2B3
.....................12x%2B18
___________________
............................-15
Answer:-%282%2F3%29x%2B2+-15%2F%286x%2B9%29
The equation for the slant asymptote is the quotient part of the answer which would be y=-%282%2F3%29x%2B2
find two points and draw the slant asymptote:
if x=0 => y=2
if x=3 => y=0