The diagram shows the graph $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2009 - Paper 1

Since $y = [f(x)]^2$ takes positive values of $f(x)$, sketch the graph by squaring the output of the original function.

• All points where $f(x)$ is zero will also result in $y = 0$.
• Ensure that minima become points of local minimum and zero crossings remain consistent.
• Highlight symmetry about the x-axis if it's applicable.

To sketch $y = f(x^2)$, note how this modifies the x-values:

• This transformation makes $x$ symmetric about the y-axis.
• For positive and negative $x$, compute $f(x^2)$ which is the same for both.
• Identify evenness and any notable peaks or troughs which will now correspond to $x = \pm a$, where $f(a)$ gives the same outputs.
• Draw the modified curve based on these evaluations and note any transformations.

1. Implicitly differentiate the equation:
   $2x + 2y + 2x \frac{dy}{dx} + 6y \frac{dy}{dx} = 0$

2. Factor out the derivative:
   $(2x + 2y) + (2x + 6y) \frac{dy}{dx} = 0$

3. Set the derivative to zero for horizontal tangents:
   $\frac{dy}{dx} = 0 \Rightarrow 2x + 2y = 0 \Rightarrow y = -x$

4. Substitute $y = -x$ into the original equation:
   $x^2 + 2(-x)x + 3(-x)^2 = 18 \Rightarrow 4x^2 = 18 \Rightarrow x^2 = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}$

5. Find $y$ corresponding to these $x$ values, leading to coordinates:
   $\left(\frac{3}{\sqrt{2}}, -\frac{3}{\sqrt{2}}\right)$ and $\left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)$.

1. Factor condition leads to two roots at $x = 1$:
   • From $P(1) = 0$:
     $1 + a + b + 5 = 0 \Rightarrow a + b = -6$
2. Using derivative $P'(1) = 0$ leads to:
   • Compute $P'(x)$:
     $P'(x) = 3x^2 + 2ax + b$
   • From $P'(1) = 0$:
     $3 + 2a + b = 0 \Rightarrow 2a + b = -3$
3. Solve the system of equations:
   • From $a + b = -6$ and $2a + b = -3$,
   • Subtract the two leading to $a = 3$ then substituting $b = -9$.

1. Set up the volume of revolution:
   $V = 2\pi \int_{0}^{1} x((x - 1)^2 - (x + 1)) \,dx$

2. Simplifying integrals:

   • Compute the bounds relative to the curves $y = (x - 1)^2$ and $y = x + 1$.

3. Set up the final integration:
   $V = 2\pi \int_{0}^{1} x[(x^2 - 2x + 1) - (x + 1)] \,dx$

4. Expanding and calculating the bounds provides the total volume in terms of rac{1}{2} with respect to pi as the final output.