(Q2) [Cambridge/9231/2007] The curve C has equation $y = \frac{ax^2 + bx + c}{x + 4},$ where $a$, $b$ and $c$ are constants. It is given that $y = 2x - 5$ is an asymptote of C.

1. Express the rational function using an identity with the given oblique asymptote and an unknown remainder.

Since $y = 2x - 5$ is an oblique asymptote of the curve, the numerator can be written in terms of the denominator and the asymptote as:
$\frac{ax^2 + bx + c}{x + 4} \equiv 2x - 5 + \frac{R}{x + 4}$
where $R$ is a constant remainder.
Multiplying by $(x + 4)$ gives:
$ax^2 + bx + c \equiv (2x - 5)(x + 4) + R$

2. Expand the right-hand side to group the terms by powers of $x$.

$ax^2 + bx + c \equiv 2x^2 + 8x - 5x - 20 + R$
$ax^2 + bx + c \equiv 2x^2 + 3x + (R - 20)$

3. Compare coefficients of $x^2$ and $x$ to find $a$ and $b$.

Comparing the coefficient of $x^2$:
$a = 2$

Comparing the coefficient of $x$:
$b = 3$

1. Determine the values of $a$ and $b$ from the given asymptote equation.

To find the oblique asymptote of $y = \frac{ax^2 + bx + c}{x + 4}$, we perform algebraic division:
$ax^2 + bx + c = (ax + b - 4a)(x + 4) + c - 4(b - 4a)$
$y = ax + (b - 4a) + \frac{c - 4(b - 4a)}{x + 4}$
Comparing the linear quotient with the given asymptote $y = 2x - 5$:
$a = 2$
$b - 4(2) = -5 \implies b = 3$
Thus, the equation of the curve is $y = \frac{2x^2 + 3x + c}{x + 4}$.

2. Find the derivative $\frac{dy}{dx}$ using the quotient rule.

$\frac{dy}{dx} = \frac{(x + 4) \frac{d}{dx}(2x^2 + 3x + c) - (2x^2 + 3x + c) \frac{d}{dx}(x + 4)}{(x + 4)^2}$
$\frac{dy}{dx} = \frac{(x + 4)(4x + 3) - (2x^2 + 3x + c)(1)}{(x + 4)^2}$

3. Apply the condition for a turning point to find $c$.

At the turning point where $x = -1$, the gradient $\frac{dy}{dx} = 0$:
$\frac{(-1 + 4)(4(-1) + 3) - (2(-1)^2 + 3(-1) + c)}{(-1 + 4)^2} = 0$
$\frac{(3)(-1) - (2 - 3 + c)}{9} = 0$
$-3 - (-1 + c) = 0$
$-2 - c = 0 \implies c = -2$

1. Substitute the values of $a$, $b$, and $c$ from previous parts and rearrange the equation of C into a quadratic equation in $x$.

Using $a=2, b=3$ and $c=-2$ from parts (i) and (ii), the equation of C is:
$y = \frac{2x^2 + 3x - 2}{x + 4}$
Multiply by $(x+4)$ and rearrange to form a quadratic equation in $x$:
$y(x + 4) = 2x^2 + 3x - 2$
$xy + 4y = 2x^2 + 3x - 2$
$2x^2 + (3 - y)x - (4y + 2) = 0$

2. Apply the discriminant condition for the curve to have no real points.

For there to be no points on C for a certain value of $y$, the quadratic equation in $x$ must have no real roots. Therefore, the discriminant must be strictly less than zero ($\Delta < 0$):
$(3 - y)^2 - 4(2)(-(4y + 2)) < 0$

3. Expand and simplify the discriminant inequality.

$9 - 6y + y^2 + 8(4y + 2) < 0$
$y^2 - 6y + 9 + 32y + 16 < 0$
$y^2 + 26y + 25 < 0$

4. Factorise and solve the quadratic inequality for $y$.

$(y + 1)(y + 25) < 0$
The roots of the corresponding quadratic equation are $y = -1$ and $y = -25$. Since the parabola defined by $Y = (y+1)(y+25)$ opens upwards, the expression is strictly negative between its roots:
$-25 < y < -1$

1. Recognise the given equation as a horizontal translation of curve $C$.

From earlier parts, the equation of $C$ is $y = \mathrm{f}(x) = \frac{2x^2 + 3x - 2}{x + 4}$.
Observe that the new equation can be rewritten by completing the denominator to match the transformation:
$y = \frac{2(x - 7)^2 + 3(x - 7) - 2}{(x - 7) + 4} = \mathrm{f}(x - 7)$
This signifies that the new curve is a translation of curve $C$ by $7$ units in the positive $x$-direction.

2. Determine the new asymptotes by translating the asymptotes of $C$.

The asymptotes of $C$ are $x = -4$ and $y = 2x - 5$.
Applying the translation (replacing $x$ with $x - 7$ for equations of lines, or simply adding $7$ to the vertical asymptote):
Vertical asymptote: $x = -4 + 7 \implies x = 3$
Oblique asymptote: $y = 2(x - 7) - 5 = 2x - 14 - 5 \implies y = 2x - 19$

3. Determine the new turning points by translating the turning points of $C$ and sketch the curve.

From part (iii), the boundary $y$-values for regions with no points are $y = -1$ and $y = -25$, which correspond to the local turning points.
Substituting these $y$-values back into $C$ (or using the derivative from earlier):
At $y = -1$, $\frac{2x^2+3x-2}{x+4} = -1 \implies 2x^2+4x+2=0 \implies x = -1$.
At $y = -25$, $\frac{2x^2+3x-2}{x+4} = -25 \implies 2x^2+28x+98=0 \implies x = -7$.
The turning points of $C$ are $(-1, -1)$ and $(-7, -25)$.
Applying the translation $(x, y) \to (x + 7, y)$, the new turning points are:
$(-1 + 7, -1) \implies (6, -1)$ (Local Minimum)
$(-7 + 7, -25) \implies (0, -25)$ (Local Maximum)

To sketch:

• Draw axes, and insert dashed lines for $x = 3$ and $y = 2x - 19$.
• Plot the turning points at $(6, -1)$ and $(0, -25)$ (noting the latter is also the $y$-intercept).
• Draw the two standard hyperbolic branches ensuring they smoothly approach the asymptotes and turn at the stated points.