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Worked solution

Recover the coefficients from a vertical and an oblique asymptote

C1C_1 has equation y=(ax2+bx)/(x+c)y=(ax^2+bx)/(x+c).

(i) It has vertical asymptote x=2 and an oblique asymptote through (0,10) and (−2,0). Find a,b,c.

(ii) Hence sketch C1C_1, clearly stating all asymptotes and stationary points.

[2014/NJC/MYE/7(part)]

CJC · Discussion · Question 4

Answer

a=5,b=0,c=−2; asymptotes x=2,y=5 x+10; stationary points(0,0) and (4,40).

(i) Determine the denominator shift

The vertical asymptote gives x=−c=2, so c=−2.

How did this part go?

(i) Find the quotient and remainder

ax2+bx−(ax2−2ax)=(b+2a)xax^2+bx-(ax^2-2ax)=(b+2a)x. Subtract [(b+2a)x−2(b+2a)][(b+2a)x-2(b+2a)] to leave 2(b+2a)2(b+2a). Thus (ax2+bx)/(x−2)=ax+b+2a+(2b+4a)/(x−2)(ax^2+bx)/(x-2)=ax+b+2a+(2b+4a)/(x-2), with oblique asymptote y=ax+b+2 a.

How did this part go?

(i) Substitute the two given points

At (0,10), b+2 a=10. At (−2,0), −2 a+b+2 a=0, so b=0. Hence a=5 and c=−2.

How did this part go?

(ii) Identify the curve and both stationary points

Now y=5x2/(x−2)=5x+10+20/(x−2)y=5x^2/(x-2)=5x+10+20/(x-2). The asymptotes are x=2 and y=5 x+10. The source sketch labels (0,0) and (4,40); differentiation gives y′=5x(x−4)/(x−2)2y'=5x(x-4)/(x-2)^2, confirming a local maximum (0,0) and local minimum (4,40). The oblique line’s axis crossings (−2,0),(0,10)are shown separately from the curve intercept (0,0).

How did this part go?

Compare your sketch

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The graph y=5x²/(x−2)has both branches, local maximum at (0,0), minimum at (4,40), vertical asymptote x=2 and oblique asymptote y=5x+10 passing through (−2,0) and (0,10).
The graph y=5x²/(x−2)has both branches, local maximum at (0,0), minimum at (4,40), vertical asymptote x=2 and oblique asymptote y=5x+10 passing through (−2,0) and (0,10).
The graph y=5x²/(x−2)has both branches, local maximum at (0,0), minimum at (4,40), vertical asymptote x=2 and oblique asymptote y=5x+10 passing through (−2,0) and (0,10).

Common mistakes

Points supplied on an asymptote are not necessarily points on the curve.

Solution notes

Use the two given points on the oblique asymptote to recover its slope and intercept.

The curve’s stationary points are (0,0) and (4,40); the asymptote’s crossings are different points.

Read the hint