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

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Answer 1 · 2016-11-14T18:06:35

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]}