Porisma

Worked solution

Separate polynomial parts from vanishing remainders

By carrying out long division to simplify the rational function (if applicable), find the equation(s) of the horizontal asymptote(s) or oblique asymptote(s) for the following curves.

(a) y=1/(3−x)y=1/(3-x).

(b) y=(2x+1)/(x−2)y=(2x+1)/(x-2).

(c) y=x2/(x−1)y=x^2/(x-1).

(d) y=(x2+2x−3)/(x+2)y=(x^2+2x-3)/(x+2).

CJC · Lecture · Self-Practice 3

(a) Identify the part approached at infinity

The source leaves y=1/(3−x)y=1/(3-x) unchanged and gives horizontal asymptote y=0y=0. Independently it tends to zero as x→±∞x\to\pm\infty.

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(b) Identify the part approached at infinity

The printed division is y=2+5/(x−2)y=2+5/(x-2), so horizontal asymptote y=2y=2. Independently 2x+1=2(x−2)+52x+1=2(x-2)+5.

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(c) Identify the part approached at infinity

The printed division is y=x+1+1/(x−1)y=x+1+1/(x-1), so oblique asymptote y=x+1y=x+1. Independently x2=(x+1)(x−1)+1x^2=(x+1)(x-1)+1 and the remainder quotient tends to zero.

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(d) Identify the part approached at infinity

The printed division is y=x−3/(x+2)y=x-3/(x+2), so oblique asymptote y=xy=x. Independently x2+2x−3=x(x+2)−3x^2+2x-3=x(x+2)-3.

How did this part go?

Common mistakes

Compare degrees and divide when needed; retain the original excluded inputs even after rewriting.

Solution notes

Both original copies visually compared. Independent checks and qualifications are identified in the working. No printed marks or external attribution.

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