Porisma

Worked solution

Recover parameters and check the complete curve

The curve CC has an equation y=ax2+bx+ax2−by=\dfrac{ax^2+bx+a}{x^2-b} where aa and bb are constants.

(i) Given that two of the asymptotes of CC are x=2x=2 and y=3y=3, determine the values of aa and bb.

(ii) Sketch the curve CC, giving the coordinates of the point(s) where the curve crosses the axes, stationary points and the equations of all asymptotes.

2019/CJC/Promo/2

CJC · Lecture · Example 11

Answer

(i) a=3,b=4a=3,b=4. (ii) Asymptotes x=±2,y=3x=\pm2,y=3; sole axial intercept (0,−3/4)(0,-3/4). Local minimum (−15−1614,9+1618)\left(\dfrac{-15-\sqrt{161}}4,\dfrac{9+\sqrt{161}}8\right) and local maximum (−15+1614,9−1618)\left(\dfrac{-15+\sqrt{161}}4,\dfrac{9-\sqrt{161}}8\right).

(i) Use the given vertical asymptote

Answer

b=4b=4.

The Teacher completes the Student working with

x2−b=(x−b)(x+b),b=2⇒b=4.x^2-b=(x-\sqrt b)(x+\sqrt b),\qquad\sqrt b=2\Rightarrow b=4.

Independently, a genuine vertical asymptote at 2 requires the original denominator to vanish there, so 4−b=04-b=0. After finding aa below, also check the numerator at both denominator zeros to rule out removable holes.

How did this part go?

(i) Use the quotient and retain the long-division stages

Answer

a=3a=3.

With b=4b=4, the source division has leading term aa. Multiplying back gives a(x2−4)=ax2−4aa(x^2-4)=ax^2-4a; subtracting from ax2+4x+aax^2+4x+a leaves 4x+5a4x+5a. Hence

y=a+4x+5ax2−4.y=a+\frac{4x+5a}{x^2-4}.

The remainder tends to 0 at either infinite end, giving horizontal asymptote y=ay=a. The given y=3y=3 therefore gives a=3a=3. This retains the Teacher’s full subtraction/remainder argument, not just its result.

How did this part go?

(ii) Establish the asymptotes and intercepts

Substitution gives y=(3x2+4x+3)/(x2−4)y=(3x^2+4x+3)/(x^2-4). Both source sketches label x=−2,x=2,y=3x=-2,x=2,y=3 and (0,−3/4)(0,-3/4). Independently the numerator is 7 at −2 and 23 at 2, so both vertical asymptotes are genuine. Its discriminant is 42−4(3)(3)=−20<04^2-4(3)(3)=-20<0, hence there is no xx-intercept. The yy-intercept is (0,−3/4)(0,-3/4).

How did this part go?

(ii) Check the turning point outside the source sketch

Both source sketches label only the central local maximum (−0.578,−0.461)(-0.578,-0.461). Independent differentiation finds two stationary points over the full domain:

y′=(6x+4)(x2−4)−2x(3x2+4x+3)(x2−4)2=−2(2x2+15x+8)(x2−4)2.\begin{aligned}&y'\\ & =\frac{(6x+4)(x^2-4)-2x(3x^2+4x+3)}{(x^2-4)^2}\\ & =-\frac{2(2x^2+15x+8)}{(x^2-4)^2}.\end{aligned}

The roots are x=(−15±161)/4x=(-15\pm\sqrt{161})/4. The far-left root gives a local minimum near (−6.922,2.711)(-6.922,2.711); the other gives the source-labelled maximum near (−0.578,−0.461)(-0.578,-0.461). Exact ordinates are respectively (9+161)/8(9+\sqrt{161})/8 and (9−161)/8(9-\sqrt{161})/8. Derivative signs are −→+-\to+ at the first root and +→−+\to- at the second, splitting at the poles as required.

The source drawing does not display the far-left turn or crossing of the horizontal asymptote. Independently y−3=(4x+15)/(x2−4)y-3=(4x+15)/(x^2-4) vanishes at x=−15/4x=-15/4 and tends to 0 from below as x→−∞x\to-\infty. A horizontal asymptote may be crossed. The complete reconstruction includes these features; the original source excerpt and its labelled maximum are preserved in review evidence.

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Complete rational curve with poles ±2 and horizontal 3, y-intercept −3/4 and no x-intercepts. Source-rounded central maximum (−0.578,−0.461) plus independently recovered far-left minimum (−6.922,2.711); crosses horizontal asymptote at −15/4.
Complete rational curve with poles ±2 and horizontal 3, y-intercept −3/4 and no x-intercepts. Source-rounded central maximum (−0.578,−0.461) plus independently recovered far-left minimum (−6.922,2.711); crosses horizontal asymptote at −15/4.
Complete rational curve with poles ±2 and horizontal 3, y-intercept −3/4 and no x-intercepts. Source-rounded central maximum (−0.578,−0.461) plus independently recovered far-left minimum (−6.922,2.711); crosses horizontal asymptote at −15/4.
How did this part go?

Common mistakes

The source sketch does not show every stationary point. Do not infer that a horizontal asymptote cannot be crossed.

Solution notes

Printed attribution 2019/CJC/Promo/2 is retained separately from the prompt; no marks are printed. Teacher completes part (i), while both copies already show the same part (ii) sketch. The far-left stationary point and horizontal-asymptote crossing are independently added, with original evidence retained.

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