The graph of $y=\frac{x+1}{a{{x}^{2}}+bx+c}$, where $a$, $b$ and $c$ are non-zero constants, has an asymptote at $x=-\frac{1}{2}$.
The graph also has a turning point at $\left( -2,\ -\frac{1}{9} \right)$.
Find the values of $a$, $b$ and $c$.[4]

The graph of $y=\frac{x+1}{a{{x}^{2}}+bx+c}$, where $a$, $b$ and $c$ are non-zero constants, has an asymptote at $x=-\frac{1}{2}$.

The graph also has a turning point at $\left( -2,\ -\frac{1}{9} \right)$.

Linear Equations - General

Linear Equations and Simple Inequalities

IP Math

N2024 P1 Q1

$x = -\displaystyle\frac{1}{2}$ is a root of $ax^2 + bx + c = 0$
$\Rightarrow a\left(-\displaystyle\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + c = 0$
$\displaystyle\frac{1}{4}a - \frac{1}{2}b + c = 0$ ... (1)

$x = -2, y = -\displaystyle\frac{1}{9}$
$\Rightarrow -\displaystyle\frac{1}{9} = \frac{-2+1}{a(-2)^2 + b(-2) + c}$
$\Rightarrow -\displaystyle\frac{1}{9} = -\frac{1}{4a-2b+c}$
$\Rightarrow 4a - 2b + c = 9$ ... (2)

$y = \displaystyle\frac{x+1}{ax^2+bx+c}$
Differentiate w.r.t $x$:
$y(2ax+b) + (ax^2+bx+c)\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x} = 1$

At $x = -2, y = -\displaystyle\frac{1}{9}, \frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$-\displaystyle\frac{1}{9}(2a(-2)+b) + 0 = 1$
$-4a + b = -9$ ... (3)

From GC, solving (1), (2), (3):
$a = 2, b = -1, c = -1$

N2024 P1 Q1

N2024 P1 Q1

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