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F(x)=2x^2 + x - 6/ x-1

User Sriram Jayaraman
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1 Answer

27 votes
27 votes

Answer:

See below

Explanation:

Zeroes:


f(x)=(2x^2+x-6)/(x-1)


0=(2x^2+x-6)/(x-1)


0=2x^2+x-6


0=(2x-3)(x+2)


x=(3)/(2),-2

Vertical and Horizontal Asymptotes:


x\\eq 1 is excluded from the domain, therefore, there's a vertical asymptote at
x=1

No horizontal asymptotes as the degree of the numerator is greater than the degree of the denominator by 1.

Oblique (Slant) Asymptote:


f(x)=(2x^2+x-6)/(x-1)=2x+3,R(-3), so the oblique/slant asymptote is
y=2x+3

F(x)=2x^2 + x - 6/ x-1-example-1
User Yousaf
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