Curve sketching with calculus: WACE Year 12 Mathematics Methods Unit 3

We sketch $f(x)=x^{3}-3x^{2}$ by systematically finding its key features before drawing.

Step 1: Find intercepts

The $y$-intercept is $f(0)=0$. Setting $f(x)=0$ gives $x^{2}(x-3)=0$, so the $x$-intercepts are at $x=0$ (a repeated root, so the curve touches the axis there) and $x=3$. Identifying the repeated root is important because the curve touches but does not cross the axis at $x=0$.

Step 2: Find stationary points

Differentiate to get $f'(x)=3x^{2}-6x=3x(x-2)$. Setting $f'(x)=0$ gives $x=0$ and $x=2$. The corresponding $y$-values are $f(0)=0$ and $f(2)=8-12=-4$, giving candidate stationary points $(0,0)$ and $(2,-4)$.

Step 3: Classify stationary points

The second derivative is $f''(x)=6x-6$. At $x=0$, $f''(0)=-6<0$, confirming a local maximum at $(0,0)$. At $x=2$, $f''(2)=6>0$, confirming a local minimum at $(2,-4)$. The sign of $f''$ tells us the concavity and therefore the nature of each turning point.

Step 4: Find the inflection point

Set $f''(x)=6x-6=0$ to get $x=1$. The sign of $f''$ changes from negative (for $x<1$) to positive (for $x>1$), confirming a genuine inflection. The $y$-value is $f(1)=1-3=-2$, so the point of inflection is $(1,-2)$.

Step 5: Describe end behaviour

Because the leading term is $x^{3}$ with a positive coefficient, $f(x)\to\infty$ as $x\to\infty$ and $f(x)\to-\infty$ as $x\to-\infty$. This confirms the overall rise-fall-rise shape between the two stationary points.

Final answer: The sketch rises from $-\infty$ to the local maximum $(0,0)$, falls through the inflection $(1,-2)$ to the local minimum $(2,-4)$, then rises through $(3,0)$ toward $+\infty$. Markers reward all labelled features and a curve whose shape matches the concavity.

Sketch $y = \dfrac{4x}{x^{2}+1}$

Intercepts. Setting $x=0$ gives $y=0$, and setting $y=0$ gives $x=0$. The only intercept is the origin.

Asymptote. As $x\to\pm\infty$, the degree of the denominator exceeds the numerator, so $y\to 0$. The horizontal asymptote is $y=0$. The denominator $x^{2}+1$ is never zero, so there is no vertical asymptote.

Stationary points. By the quotient rule,

Setting the numerator to zero, $4-4x^{2}=0$ gives $x=\pm 1$. Then $y(1)=2$ and $y(-1)=-2$.

Nature. The denominator is always positive, so the sign of $y'$ follows $4-4x^{2}$: positive for $-1<x<1$ and negative outside. So $(1,2)$ is a local maximum and $(-1,-2)$ is a local minimum.

The curve passes through the origin, rises to the maximum $(1,2)$, then decreases back toward the asymptote $y=0$, and is symmetric through the origin (an odd function). Markers reward the asymptote, both stationary points with their nature, and the symmetry observation.