Sketch the graph of the curve $y = -x^3 + 3x^2 - 1$, labelling the stationary points and point of inflection - HSC - SSCE Mathematics Advanced - Question 16 - 2020 - Paper 1

Setting the derivative to zero to find stationary points:

$-3x(x - 2) = 0$

This gives us:

$x = 0 \quad \text{or} \quad x = 2$

To find the corresponding y-values:

1. For $x = 0$:

   $y = -(0)^3 + 3(0)^2 - 1 = -1$

   Thus, the stationary point is $(0, -1)$.

2. For $x = 2$:

   $y = -(2)^3 + 3(2)^2 - 1 = 3$

   Thus, the stationary point is $(2, 3)$.

Next, we need to check for points of inflection by finding the second derivative:

$\frac{d^2y}{dx^2} = -6x + 6$

Setting the second derivative to zero:

$-6x + 6 = 0$

Solving gives:

$x = 1$

To find the y-coordinate at this point of inflection:

$y = -(1)^3 + 3(1)^2 - 1 = 1$

Thus, the point of inflection is $(1, 1)$.

For the final graph:

• The stationary points $(0, -1)$ and $(2, 3)$ should be labeled on the graph.
• The point of inflection $(1, 1)$ should also be indicated.
• The graph should exhibit a local minimum at $(0, -1)$ and a local maximum at $(2, 3)$ with a change in concavity at the point of inflection $(1, 1)$.