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Worked solution

Link the discriminant interval to the missing horizontal levels

C has equation y=(x2+9x+16)/(x+2)y=(x^2+9x+16)/(x+2).

(i) Find its asymptotes.

(ii) Find, in exact form, all k for which k=(x2+9x+16)/(x+2)k=(x^2+9x+16)/(x+2) has no real roots.

(iii) Sketch C with asymptotes, stationary points and x-/y-axis intersections labelled.

[2009/IJC/II/1]

CJC · Discussion · Question 3

Answer

Asymptotes x=−2,y=x+7; no roots for 5−22<k<5+225-2\sqrt2<k<5+2\sqrt2; all intercepts/turns shown.

(i) Perform the full division

Subtract x(x+2)=x2+2xx(x+2)=x^2+2x from the numerator to leave 7 x+16. Subtract 7(x+2)=7x+147(x+2)=7x+14 to leave 2. Hence y=x+7+2/(x+2)y=x+7+2/(x+2), with asymptotes y=x+7 and x=−2.

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(ii) Form a quadratic in x

For x≠−2, kx+2k=x2+9x+16kx+2k=x^2+9x+16, hence x2+(9−k)x+(16−2k)=0x^2+(9-k)x+(16-2k)=0. Its discriminant must be negative: (9−k)2−4(1)(16−2k)<0(9-k)^2-4(1)(16-2k)<0, so 81−18k+k2−64+8k<081-18k+k^2-64+8k<0, or k2−10k+17<0k^2-10k+17<0. The excluded x=−2 cannot be a spurious quadratic root because substitution gives 2, not 0.

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(ii) Solve the boundary quadratic exactly

k=[10±100−4(1)(17)]/2=(10±32)/2=(10±42)/2=5±22k=[10\pm\sqrt{100-4(1)(17)}]/2=(10\pm\sqrt{32})/2=(10\pm4\sqrt2)/2=5\pm2\sqrt2. The upward-opening quadratic is negative between these roots. Therefore 5−22<k<5+225-2\sqrt2<k<5+2\sqrt2, as in the source number-line sketch.

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(iii) Find intercepts and stationary points

The y-intercept is (0,8); x-intercepts are ((−9±17)/2,0)((-9\pm\sqrt{17})/2,0), approximately (−6.56,0),(−2.44,0). Since y′=1−2/(x+2)2y'=1-2/(x+2)^2, stationary points are (−2−2,5−22)(-2-\sqrt2,5-2\sqrt2), local maximum≈(−3.41,2.17), and (−2+2,5+22)(-2+\sqrt2,5+2\sqrt2), local minimum≈(−0.586,7.83). Both asymptotes and both branches are retained. The exact coordinates supplement the source’s rounded graph labels.

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Compare your sketch

Expand diagram

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First panel shows k²−10k+17 below zero between 5−2√2 and 5+2√2. Second shows y=x+7+2/(x+2), both branches, the two rounded x-intercepts, y-intercept 8, both stationary points and asymptotes.
First panel shows k²−10k+17 below zero between 5−2√2 and 5+2√2. Second shows y=x+7+2/(x+2), both branches, the two rounded x-intercepts, y-intercept 8, both stationary points and asymptotes.
First panel shows k²−10k+17 below zero between 5−2√2 and 5+2√2. Second shows y=x+7+2/(x+2), both branches, the two rounded x-intercepts, y-intercept 8, both stationary points and asymptotes.

Common mistakes

Keep both discriminant endpoints excluded: at an endpoint the horizontal line touches a stationary point.

Solution notes

The missing horizontal levels lie strictly between the two stationary output values.

Exact intercept and stationary coordinates are given in the working alongside the rounded graph labels.

Read the hint