A turning point fixes a rational curve

The curve CC has equation y=ax+bx2+cy=\frac{ax+b}{x^2+c}, where a,b,ca,b,c are constants. It is given that x=2x=2 is an asymptote of CC.

(i) State the value of cc. mark

(ii) Given also that CC has a turning point at (6,14)(6,\frac14), find the values of a,ba,b. marks

(iii) Sketch CC, stating the equations of the asymptotes and the coordinates of the turning points. [You need not indicate the coordinates of intersections with the axes.] marks

(iv) Hence find the range of values of kk for which the equation kx2−3x+10−4k=0kx^2-3x+10-4k=0 has no real root. marks

[2011/PJC/Promo/9]

Total marks: 10

Solution

(i)

Asymptotes

The curve CC has equation y=ax+bx2+cy=\dfrac{ax+b}{x^2+c}, where aa, bb, cc are constants. It is given that x=2x=2 is an asymptote of CC. State the value of cc.

Hint

  1. A vertical asymptote occurs where the denominator is zero.

Answer · (i)

c=−4c=-4.

(ii)

Turning points

Given also that CC has a turning point at (6,14)\left(6,\frac14\right), find the values of aa and bb.

Hint

  1. A turning point gives two facts: the point lies on the curve, and dydx=0\frac{dy}{dx}=0 there.

  2. Use each fact to write one linear equation in aa and bb.

  3. Solve the pair.

Answer · (ii)

a=3,b=−10a=3,b=-10.

(iii)

Curve sketching

Sketch CC, stating the equations of the asymptotes and the coordinates of the turning points. You need not indicate the coordinates of the intersections with the axes.

Explore this idea

Hint

  1. The curve is now y=3x−10x2−4y=\frac{3x-10}{x^2-4}. List its vertical and horizontal asymptotes.

  2. There are two turning points. One is given; find the other.

  3. The three branches meet asymptotes at x=±2x=\pm2.

Q9 source rational sketch
Q9 source rational sketchScroll across to see the full diagram.

Answer · (iii)

Asymptotes x=±2,y=0x=\pm2,y=0; minimum (2/3,9/4)(2/3,9/4), maximum (6,1/4)(6,1/4).

Explore the idea

Three branches, two turning points

Switch on each feature to see how it shapes the curve.

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

  • Y = (3x − 10)/(x² − 4)
  • Vertical asymptotes
  • Horizontal asymptote y = 0
Ideas to try
  1. Count the branches. Which one contains both turning-point heights?
  2. What is the range of the central branch?
(iv)

Roots from a graph

Hence find the range of values of kk for which the equation kx2−3x+10−4k=0kx^2-3x+10-4k=0 has no real root.

Explore this idea

Hint

  1. Rearrange so that one side is the curve from (iii): isolate kk.

  2. Real roots of the equation are intersections of y=ky=k with the curve.

  3. Read off the values of kk for which the horizontal line misses the curve.

Answer · (iv)

1/4<k<9/41/4<k<9/4.

Explore the idea

How many roots at each height k?

Each intersection of the line y = k with the curve is a real root of kx² − 3x + 10 − 4k = 0.

Height of the horizontal linek = 1
Try an example
Test boundary heights

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

Number of distinct intersections
0

No real roots: 14<k<94\frac14<k<\frac94.

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

  • Y = (3x − 10)/(x² − 4)
  • Horizontal line y = k
  • Intersections: the real roots
  • Intersection selected in the coordinate table
Key idea

No intersections means no real roots. The curve’s gap, 14<y<94\frac14<y<\frac94, is exactly the set of kk for which kx2−3x+10−4k=0kx^2-3x+10-4k=0 has no real root.

Ideas to try
  1. Drag k slowly from 0 up to 3. Where does the count drop to zero, and where does it come back?
  2. Use the presets to see what happens exactly at the turning-point heights.