Porisma

Worked solution

Practise rational-curve features

Sketch the graph of the following curves, indicating clearly axial intercept(s), asymptote(s) and stationary point(s), if any:

(i) y=13−xy=\dfrac1{3-x}.

(ii) y=2x+1x−2y=\dfrac{2x+1}{x-2}.

(iii) y=x2x−1y=\dfrac{x^2}{x-1}.

(iv) y=x2+2x−3x+2y=\dfrac{x^2+2x-3}{x+2}.

CJC · Lecture · Self-Practice 4

Answer

(i) Intercept (0,1/3)(0,1/3); asymptotes x=3,y=0x=3,y=0; no stationary points. (ii) Intercepts (−1/2,0),(0,−1/2)(-1/2,0),(0,-1/2); asymptotes x=2,y=2x=2,y=2; no stationary points. (iii) Intercept/local maximum (0,0)(0,0); local minimum (2,4)(2,4); asymptotes x=1,y=x+1x=1,y=x+1. (iv) Intercepts (−3,0),(1,0),(0,−3/2)(-3,0),(1,0),(0,-3/2); asymptotes x=−2,y=xx=-2,y=x; no stationary points.

(i) Keep the negative reciprocal orientation

The source supplies the ask but no printed answer for this part. Independently 1/(3−x)=−1/(x−3)1/(3-x)=-1/(x-3), so x=3x=3 is a genuine pole and y=0y=0 is the horizontal asymptote. Substituting 0 gives (0,1/3)(0,1/3), and the nonzero numerator gives no xx-intercept. Since y′=1/(3−x)2>0y'=1/(3-x)^2>0 on both branches, there are no stationary points. The left branch is positive and the right branch negative.

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Increasing reciprocal branches, pole x3, asymptote y0, y-intercept 1/3, no x-intercept or stationary points.
Increasing reciprocal branches, pole x3, asymptote y0, y-intercept 1/3, no x-intercept or stationary points.
Increasing reciprocal branches, pole x3, asymptote y0, y-intercept 1/3, no x-intercept or stationary points.
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(ii) Separate constant and remainder

No answer is printed for this part. Independent division gives

2x+1x−2=2+5x−2.\frac{2x+1}{x-2}=2+\frac5{x-2}.

Thus the asymptotes are x=2x=2 and y=2y=2. Intercepts are (−1/2,0)(-1/2,0) and (0,−1/2)(0,-1/2). The derivative −5/(x−2)2-5/(x-2)^2 is negative on both branches, so there are no stationary points. The numerator at 2 is 5, so there is no removable cancellation.

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Decreasing rational branches with asymptotes x2,y2, intercepts (−1/2,0),(0,−1/2), no stationary points.
Decreasing rational branches with asymptotes x2,y2, intercepts (−1/2,0),(0,−1/2), no stationary points.
Decreasing rational branches with asymptotes x2,y2, intercepts (−1/2,0),(0,−1/2), no stationary points.
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(iii) Check the two printed stationary points

Answer

Local maximum (0,0)(0,0); local minimum (2,4)(2,4).

Both copies print Max.: (0,0)(0,0); Min.: (2,4)(2,4). Independent working completes all requested features:

x2x−1=x+1+1x−1,y′=1−1(x−1)2=x(x−2)(x−1)2.\frac{x^2}{x-1}=x+1+\frac1{x-1},\qquad y'=1-\frac1{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}.

The denominator excludes 1 and its numerator there is 1: vertical asymptote x=1x=1, slant asymptote y=x+1y=x+1. The only axial intercept is (0,0)(0,0); its double zero gives contact. The derivative changes +→−+\to- at 0 and −→+-\to+ at 2, verifying the printed local maximum and local minimum. They belong to different continuous branches and are not global extrema of the whole domain.

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Pole x=1 and slant y=x+1. Left branch touches the origin at its local maximum; right branch has local minimum (2,4).
Pole x=1 and slant y=x+1. Left branch touches the origin at its local maximum; right branch has local minimum (2,4).
Pole x=1 and slant y=x+1. Left branch touches the origin at its local maximum; right branch has local minimum (2,4).
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(iv) Use the remainder to exclude stationary points

No answer is printed for this part. Independently

x2+2x−3x+2=x−3x+2,x2+2x−3=(x+3)(x−1).\frac{x^2+2x-3}{x+2}=x-\frac3{x+2},\qquad x^2+2x-3=(x+3)(x-1).

Asymptotes are x=−2x=-2 and y=xy=x, with no cancellation at −2. Intercepts are (−3,0),(1,0),(0,−3/2)(-3,0),(1,0),(0,-3/2). Since y′=1+3/(x+2)2>0y'=1+3/(x+2)^2>0, neither branch has a stationary point.

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Increasing rational branches with pole x−2, slant yx, and intercepts (−3,0),(1,0),(0,−3/2).
Increasing rational branches with pole x−2, slant yx, and intercepts (−3,0),(1,0),(0,−3/2).
Increasing rational branches with pole x−2, slant yx, and intercepts (−3,0),(1,0),(0,−3/2).
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Common mistakes

Do not describe an extremum on one branch as the global maximum or minimum of a disconnected curve.

Solution notes

Student and Teacher PDF 23 are identical for this practice. Only the two turning-point coordinates for (iii) are printed answers. All other answers, derivations and plots here are independently authored and checked. No printed marks or external attribution.

Read the hint