Use a circle to test negative roots

y=\frac{ax^2+bx+c}{x+d}

Total marks: 8

Solution

(i)

Asymptotes and coefficients

The graph of $y=\dfrac{ax^2+bx+c}{x+d}$ , where $a,b,c,d$ are constants, has asymptotes $x=-2$ and $y=x-4$ . Write down the value of $d$ and find the values of $a$ and $b$ .

Hint

The vertical asymptote is where the denominator vanishes.
The oblique asymptote is the quotient of the division. Write $y=x-4+\frac{h}{x+2}$ and combine over $x+2$ .
Compare coefficients.

Answer · (i)

d=2,a=1,b=-2

(ii)

Substitution

Given that the curve passes through the point $(0,-4.5)$ , find the value of $c$ .

Hint

Substitute $(0,-4.5)$ into $y=\frac{x^2-2x+c}{x+2}$ .

Answer · (ii)

c=-9

(iii)

Circles and loci

By drawing a suitable curve on the same diagram, determine the range of values of $k$ such that the equation $(x-4)^2+\left(\dfrac{ax^2+bx+c}{x+d}\right)^2=k^2$ , with the values found above, has at least one negative root.

Explore this idea

Hint

Let $y=\frac{x^2-2x-9}{x+2}$ . Then the equation says $(x-4)^2+y^2=k^2$ : the point $(x,y)$ is at distance $|k|$ from $(4,0)$ .
So the curve and the circle of centre $(4,0)$ , radius $|k|$ must meet at a point with $x<0$ .
How small can that radius be? The nearest approach to $(4,0)$ for $x<0$ .

Q13 circle and curveScroll across to see the full diagram.

Answer · (iii)

|k|>\sqrt{145}/2

Explore the idea

How far is the curve from (4, 0)?

The graph shows the distance from (4, 0) to the point of the curve with abscissa x. A horizontal line at height |k| is the circle's radius: its intersections with x < 0 are the negative roots.

Height of the horizontal linek = 5

Try an example

Test boundary heights

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

Number of distinct negative roots

No negative root. At $|k|=\sqrt{145}/2$ the limiting point is at $x=0$ , which is excluded.

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

Distance from the point (4, 0) to the curve at each x
Radius |k|
Points where the circle meets the curve with x < 0: the negative roots
Intersection selected in the coordinate table

Key idea

A sum of squares $(x-4)^2+y^2$ is a squared distance. Turning the equation into 'circle meets curve' reduces the problem to finding the nearest point on the left, the $y$ -intercept.

Ideas to try

At $x=0$ the distance is exactly $\frac{\sqrt{145}}2$ . Which point of the curve is that?
Why does the count jump from 0 to 1 at that radius?

Use a circle to test negative roots

Solution

Asymptotes and coefficients

Vertical asymptote

Rewrite using the oblique asymptote

Compare coefficients

Substitution

Substitute

Circles and loci

Interpret geometrically

Which part of the curve?

Distance to the nearest point

State the range

How far is the curve from (4, 0)?