Determine when a horizontal level has a negative input · source reference

Printed question

The curve $C$ has equation given by $y=\frac{x-a+a^2}{a(x-a)}$, where $a$ is a negative constant.

(i) Write down the equations of the asymptotes of $C$. [2]

(ii) Sketch the graph of $C$, labelling clearly the asymptotes and axial intercepts. [2]

(iii) Deduce the range of values of $k$ for which the equation $\frac{x-a+a^2}{a(x-a)}=k$ has one negative real root. [2]

(iv) State the range of values of $x$ for which the curve is concave downwards. [1]

[2012/NJC/Promo/11(a)]

Printed working

(i) $y=\frac{x-a+a^2}{a(x-a)}=\frac1a+\frac a{x-a}$.

Asymptotes are $y=\frac1a$ and $x=a$.

(ii) Printed graph below labels $x$-intercept $a-a^2$ and $y$-intercept $\frac1a-1$.

(iii) $\frac{x-a+a^2}{a(x-a)}=k$ has 1 negative real root

$\Rightarrow$ there is 1 point of intersection between curve $y=\frac{x-a+a^2}{a(x-a)}$ and line $y=k$ whose $x$-coordinate is negative.

$\Rightarrow k<\frac1a-1$ or $k>\frac1a$.

(iv) $x>a$.

Teacher notes

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

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

For eqn to have 1 negative real root, the line $y=k$ must intersect the part of curve to the left of the $y$-axis (i.e. on the negative part of the $x$-axis — 2nd and 3rd quadrants) for pt of intersection to have negative $x$-coordinate.

Region where gradient of curve is decreasing.

Printed answer key

(i) Asymptotes $y=\frac1a,x=a$. (ii) Native graph above. (iii) $k<\frac1a-1$ or $k>\frac1a$. (iv) $x>a$.

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.