Sketch the curve $y = \frac{4x^2}{9}$ showing all asymptotes - HSC - SSCE Mathematics Extension 2 - Question 3 - 2004 - Paper 1

To sketch $y = |f(x)|$, reflect any part of $f(x)$ that is below the x-axis onto the above. This gives a curve that is entirely non-negative, maintaining the shape of $f(x)$ where it is above the x-axis.

When sketching $y = (f(x))^2$, transform every point $(x, f(x))$ to $(x, (f(x))^2)$. This creates a curve that is always non-negative, with sections where $f(x)$ was negative being reflected above the x-axis, and steepening the values of $f(x)$.

For the sketch of $y = \frac{1}{\sqrt{f(x)}}$, wherever $f(x) > 0$, the plot will reflect $y$ values greater than 1 approaching infinity as $f(x)$ approaches 0, while points where $f(x)$ is negative are undefined.

To find the tangent line at $(2, -1)$, we first differentiate the implicit function using implicit differentiation. [ \frac{d}{dx}(x^2 - xy + y^3) = \frac{d}{dx}(5) \implies 2x - (y + x \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} = 0 ] Plugging in $(2, -1)$, we can solve for $\frac{dy}{dx}$, and then use the point-slope form to find the equation of the tangent line.

The area $A$ of an equilateral triangle can be found using the formula ( A = \frac{\sqrt{3}}{4}s^2 ), where $s$ is the length of a side. The length of the side at any point can be found from the intersection of the curves at that value of $x$. The calculations yield the area $A(h) = \sqrt{3}h^2$.

To find the volume $V$ of the solid, integrate the area of the cross-section along the x-axis from 0 to 2: [ V = \int_{0}^{2} \sqrt{3}h^2 ,dh = \sqrt{3} \int_{0}^{2} h^2 , dh = \sqrt{3} \left[ \frac{h^3}{3} \right]_{0}^{2} = \sqrt{3} \cdot \frac{8}{3} = \frac{8\sqrt{3}}{3} ] Thus, the volume of the solid is ( \frac{8\sqrt{3}}{3} ).