A discriminant reveals a forbidden output interval

The curve CC has equation y=2x2+43(x+λ+1)y=\frac{2x^2+4}{3(x+\lambda+1)}. It is given that CC has a vertical asymptote x=0x=0.

(i) Determine the value of λ\lambda. mark

(ii) Find the equation of the other asymptote of CC. marks

(iii) Prove, using an algebraic method, that CC cannot lie between −4a3<y<4a3-\frac{4\sqrt a}{3}<y<\frac{4\sqrt a}{3}, a<0a<0, where aa is to be determined. marks

(iv) When λ=0\lambda=0, the graph of y=2x2+43(x+λ+1)+ky=\frac{2x^2+4}{3(x+\lambda+1)}+k cuts the xx-axis at 2 distinct points where kk is a positive integer.

(a) Sketch the graph of y=2x2+43(x+λ+1)+ky=\frac{2x^2+4}{3(x+\lambda+1)}+k when λ=0\lambda=0, indicating clearly all asymptotes and the coordinates of the axial intercepts. marks

(b) Deduce the least possible integer value of kk. mark

[2009/NYJC/Prelim/I/7]

Total marks: 11

Solution

(i)

Asymptotes

The curve CC has equation y=2x2+43(x+λ+1)y=\dfrac{2x^2+4}{3(x+\lambda+1)}. It is given that CC has a vertical asymptote x=0x=0. Determine the value of λ\lambda.

Hint

  1. A vertical asymptote is where the denominator vanishes.

Answer · (i)

λ=−1\lambda=-1.

(ii)

Oblique asymptotes

Find the equation of the other asymptote of CC.

Hint

  1. With λ=−1\lambda=-1, y=2x2+43xy=\frac{2x^2+4}{3x}. Divide each term by 3x3x.

Answer · (ii)

y=2x/3y=2x/3.

(iii)

Range by discriminant

Prove, using an algebraic method, that CC cannot lie between −4a3<y<4a3-\dfrac{4\sqrt a}{3}<y<\dfrac{4\sqrt a}{3}, where aa is to be determined.

Explore this idea

Hint

  1. Let yy be a value the curve takes. Then 2x2+43x=y\frac{2x^2+4}{3x}=y must have a real solution xx.

  2. Rearrange into a quadratic in xx. A real solution exists only if the discriminant is ≥0\ge0.

  3. The heights with discriminant <0<0 are the ones the curve never reaches.

Answer · (iii)

a=2a=2; excluded outputs −42/3<y<42/3-4\sqrt2/3<y<4\sqrt2/3.

Printed a<0 is inconsistent; the real square-root answer requires a>0.

Explore the idea

Which heights does the curve never reach?

A horizontal line y = k meets the curve exactly when k is a value the curve takes.

Height of the horizontal linek = 0
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections
0

No intersection: ∣k∣<42/3|k|<4\sqrt2/3.

There are no isolated intersections at this value. A touching point counts once; hollow endpoints are excluded.

  • Y = (2x² + 4)/(3x)
  • Horizontal line y = k
  • Intersections: the values of x with y = k
  • Intersection selected in the coordinate table
Key idea

A height kk is attained exactly when 2x2−3kx+4=02x^2-3kx+4=0 has a real root. The discriminant 9k2−329k^2-32 is negative for ∣k∣<423|k|<\frac{4\sqrt2}{3}, the gap you can see between the two branches.

Ideas to try
  1. Find the two heights where the line just touches the curve.
  2. What is the smallest value on the right branch?
(iv)a

Curve sketching

When λ=0\lambda=0, the graph of y=2x2+43(x+λ+1)+ky=\dfrac{2x^2+4}{3(x+\lambda+1)}+k cuts the xx-axis at 2 distinct points, where kk is a positive integer. Sketch the graph, indicating clearly all asymptotes and the coordinates of the axial intercepts.

Explore this idea

Hint

  1. With λ=0\lambda=0: y=2x2+43(x+1)+ky=\frac{2x^2+4}{3(x+1)}+k. Divide to find the oblique asymptote.

  2. yy-intercept: put x=0x=0. xx-intercepts: solve the quadratic.

  3. Draw the two branches around x=−1x=-1.

Q18 source rational sketch
Q18 source rational sketchScroll across to see the full diagram.

Answer · (iv)a

Asymptotes x=−1,y=2x/3−2/3+kx=-1,y=2x/3-2/3+k; y-intercept (0,4/3+k)(0,4/3+k); x-intercepts ((−3k±9k2−24k−32)/4,0)((-3k\pm\sqrt{9k^2-24k-32})/4,0).

Representative diagram uses k=4.

Explore the idea

The curve lifted by k = 4

With k = 4 the x-intercepts are (−4, 0) and (−2, 0).

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Lifted curve
  • Oblique asymptote (the vertical one is x = −1)
  • X-intercepts
Ideas to try
Both intercepts are on the left of the vertical asymptote. Why?
(iv)b

Integer thresholds

Deduce the least possible integer value of kk.

Explore this idea

Hint

  1. Two distinct xx-intercepts means 2x2+3kx+(4+3k)=02x^2+3kx+(4+3k)=0 has two distinct real roots.

  2. Positive discriminant, then the least integer kk.

Answer · (iv)b

k=4k=4.

Explore the idea

How low must −k be to cut the left branch twice?

g(x) + k = 0 means g(x) = −k. Put h = −k and watch when the line y = h meets the curve twice.

Height of the horizontal linek = -3
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct x-intercepts
0

The line misses both branches.

There are no isolated intersections at this value. A touching point counts once; hollow endpoints are excluded.

  • Y = g(x)
  • Y = −k
  • X-intercepts of the lifted graph
  • Intersection selected in the coordinate table
Key idea

The curve's lower turning point is at height ≈−3.64\approx-3.64. Only when −k-k drops below that height, i.e. k≥4k\ge4 for an integer, are there two xx-intercepts.

Ideas to try
  1. Set −k=−3-k=-3: how many intersections? Then −k=−4-k=-4.
  2. The turning heights are the roots of 9k2−24k−32=09k^2-24k-32=0, with kk replaced by −k-k.