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]}

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