How do you find the slant asymptote of #y = (3x^2 + 2x - 3 )/( x - 1)#?

Question

Theodore Amidon · Accepted Answer

The slant asymptote is #y=3x+5#

The degree of the numerator #># the degree of the denominator, we expect a slant asymptote.

Just do a long division

#color(white)(aaaa)##3x^2+2x-3##color(white)(aaaa)##∣##x-1#

#color(white)(aaaa)##3x^2-3x##color(white)(aaaaaaa)##∣##3x+5#

#color(white)(aaaaaa)##0+5x-3#

#color(white)(aaaaaaaa)##+5x-5#

#color(white)(aaaaaaaaaaa)##0+2#

#y=(3x^2+2x-3)/(x-1)=3x+5+2/(x-1)#

Therefore, the slant asymptote is #y=3x+5#

graph{(y-(3x^2+2x-3)/(x-1))(y-3x-5)=0 [-32.04, 32.92, -9.17, 23.3]}

Anna Allen · Answer

undefined To find the slant asymptote of the rational function $ y = \frac{{3x^2 + 2x - 3}}{{x - 1}} $, you need to perform polynomial long division to divide the numerator $ 3x^2 + 2x - 3 $ by the denominator $ x - 1 $. The quotient of this division will give you the equation of the slant asymptote.

Performing polynomial long division:

$$ \begin{array}{r|l}
3x^2 + 2x - 3 & x - 1 \
\end{array} $$

First, divide $ 3x^2 $ by $ x $, which gives $ 3x $. Then, multiply $ x - 1 $ by $ 3x $ to get $ 3x^2 - 3x $, and subtract this from $ 3x^2 + 2x - 3 $, which gives $ 5x - 3 $.

Next, divide $ 5x $ by $ x $, which gives $ 5 $. Multiply $ x - 1 $ by $ 5 $ to get $ 5x - 5 $, and subtract this from $ 5x - 3 $, which gives $ 2 $.

Since the degree of the remainder (constant term) is less than the degree of the divisor (linear term), the slant asymptote exists. Therefore, the slant asymptote of the function is $ y = 3x + 5 $.