Use a circle to test negative roots · source reference

Printed question

The diagram below shows a sketch of the graph of $y=\frac{ax^2+bx+c}{x+d}$ where $a,b,c,d$ are constants. The asymptotes, also shown in the diagram, are $x=-2$ and $y=x-4$.

(i) Write down the value of $d$ and find the values of $a,b$. [4]

(ii) Given that the curve passes through the point $(0,-4.5)$, find the value of $c$. [1]

(iii) By drawing a suitable curve on the same diagram shown above, determine the range of values of $k$ such that the equation $(x-4)^2+(\frac{ax^2+bx+c}{x+d})^2=k^2$, where $a,b,c,d$ are the values found above, has at least one negative root. [3]

[modified 2010/RVHS/Prelim/I/7]

Printed working

(i) Vertical asymptote: $x=-d=-2\Rightarrow d=2$.

$y=x-4+\frac h{x+2}$.

$=\frac{(x-4)(x+2)+h}{x+2}$.

$=\frac{x^2-2x-8+h}{x+2}$.

$\therefore a=1,b=-2$.

(ii) $y=\frac{x^2-2x+c}{x+2}$.

At $(0,-4.5)$: $-4.5=\frac c2\Rightarrow\therefore c=-9$.

(iii) $(x-4)^2+(\frac{ax^2+bx+c}{x+d})^2=k^2$.

$(x-4)^2+(\frac{x^2-2x-9}{x+2})^2=k^2$ — (*).

The roots of equation (*) are the $x$-coordinates of the points of intersection of the 2 curves $(x-4)^2+y^2=k^2$ and $y=\frac{x^2-2x-9}{x+2}$.

$(x-4)^2+y^2=k^2$ is a circle with centre $(4,0)$, radius $|k|$.

For the 2 curves to intersect at at least one point with negative $x$-coordinate, the radius $|k|>\sqrt{4.5^2+4^2}$, the distance between the points $(4,0)$ and $(0,-4.5)$.

$k<-\frac{\sqrt{145}}2$ or $k>\frac{\sqrt{145}}2$.

Teacher notes

1. To find vertical asymptote: Put denominator $=0$. So $x+d=0\Rightarrow x=-d$.

2. To find oblique asymptote: Use long division to express $y=\text{quotient}+\text{proper fraction}$ where $y=\text{quotient}$ is the oblique asymptote.

$\therefore y=x-4+\frac h{x+2}$, $h$ constant.

For eqn (*) to have at least 1 negative root, there must be at least 1 pt of intersection between the 2 curves with negative $x$-coordinate.

For pt(s) of intersection between the 2 curves to have negative $x$-coordinate(s), the circle must cut the part of curve $y=\frac{x^2-2x-9}{x+2}$ for which $x<0$.

So the radius of circle $|k|$ must be greater than the distance between centre of circle $A(4,0)$ and $y$-intercept (pt $B$) of curve $y=\frac{x^2-2x-9}{x+2}$.

Printed answer key

(i) $d=2,a=1,b=-2$. (ii) $c=-9$. (iii) $k<-\frac{\sqrt{145}}2$ or $k>\frac{\sqrt{145}}2$; native graph above.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.