Tangents, stationary points and curve sketching: VCE Math Methods Unit 3

What this dot point is asking

VCAA wants you to use the first and second derivatives to extract information from a function. Specifically: find the equation of a tangent or normal line at a given point, locate stationary points by solving $f'(x) = 0$, classify them as local maximum, minimum or stationary point of inflection, locate points of inflection (concavity changes), and combine these with intercepts and end behaviour to sketch the curve.

Tangents and normals

Tangent line

At the point $(a, f(a))$ on $y = f(x)$, the tangent line has slope $f'(a)$ and passes through $(a, f(a))$. In point-slope form:

Normal line

The normal is perpendicular to the tangent at the same point, so its slope is $-\frac{1}{f'(a)}$ (provided $f'(a) \neq 0$):

If $f'(a) = 0$, the tangent is horizontal and the normal is vertical, with equation $x = a$.

Standard pattern

Three steps every time.

1. Compute $f(a)$ to get the y-coordinate of the point.
2. Compute $f'(a)$ to get the tangent slope.
3. Substitute into the point-slope form. For the normal, use the negative reciprocal slope.

Stationary points

A stationary point is where the tangent line is horizontal: $f'(x) = 0$.

To find them, solve $f'(x) = 0$. To classify each one, apply one of two tests.

First derivative test

Check the sign of $f'(x)$ just before and just after the stationary point.

• Sign change positive to negative: local maximum.
• Sign change negative to positive: local minimum.
• No sign change: stationary point of inflection (the function has a flat spot but does not turn).

Second derivative test

Evaluate $f''(x)$ at the stationary point.

• $f''(x) < 0$: local maximum (curve is concave down).
• $f''(x) > 0$: local minimum (curve is concave up).
• $f''(x) = 0$: test is inconclusive; fall back on the first derivative test.

The second derivative test is usually faster on Paper 1 if $f''$ is easy to compute.

Points of inflection

A point of inflection is where the concavity of the curve changes: $f''$ changes sign. To find candidates, solve $f''(x) = 0$. Then check that $f''$ actually changes sign either side (not all roots of $f''$ are inflection points; e.g. $f(x) = x^4$ has $f''(0) = 0$ but no inflection at $x = 0$).

If the inflection point is also a stationary point (so $f'(a) = 0$ and $f''$ changes sign at $a$), it is called a stationary point of inflection.

Curve sketching

To sketch $y = f(x)$ in full, identify and mark:

1. Domain. Where $f$ is defined.
2. x-intercepts. Solve $f(x) = 0$.
3. y-intercept. Compute $f(0)$.
4. Stationary points. Solve $f'(x) = 0$ and classify.
5. Points of inflection. Where $f''$ changes sign.
6. Asymptotes. Vertical (where $f$ is undefined), horizontal (limit as $x \to \pm \infty$).
7. End behaviour. Direction as $x \to \pm \infty$.

A typical Paper 1 sketch question asks for axis intercepts, stationary points and end behaviour. Section B of Paper 2 may also ask for asymptotes or inflection points.

Worked example

Sketch $f(x) = x^3 - 6 x^2 + 9 x$ on $\mathbb{R}$.

Intercepts. y-intercept: $f(0) = 0$. x-intercepts: factor $x^3 - 6 x^2 + 9 x = x(x^2 - 6 x + 9) = x(x - 3)^2$. Zeros at $x = 0$ (single root) and $x = 3$ (double root).

Stationary points. $f'(x) = 3 x^2 - 12 x + 9 = 3(x^2 - 4 x + 3) = 3 (x - 1)(x - 3)$. Stationary at $x = 1$ and $x = 3$.

$f''(x) = 6 x - 12$. At $x = 1$: $f''(1) = -6 < 0$, local max. $f(1) = 1 - 6 + 9 = 4$. Point: $(1, 4)$.

At $x = 3$: $f''(3) = 6 > 0$, local min. $f(3) = 27 - 54 + 27 = 0$. Point: $(3, 0)$ (matches the double root touching the x-axis).

Point of inflection. $f''(x) = 0$ at $x = 2$. $f(2) = 8 - 24 + 18 = 2$. Inflection at $(2, 2)$.

End behaviour. Cubic with positive leading coefficient: $f \to -\infty$ as $x \to -\infty$, $f \to +\infty$ as $x \to +\infty$.

The sketch rises from bottom left, crosses at $(0, 0)$, peaks at $(1, 4)$, passes through inflection $(2, 2)$, touches the x-axis at $(3, 0)$, then rises to the top right.

Examples in context

Example 1. Tangent to a cost curve. A firm's average cost is $C(x) = x^2 - 8x + 20$ for output $x$. The tangent at $x = 5$ has slope $C'(5) = 2(5) - 8 = 2$ and passes through $(5, C(5)) = (5, 5)$, so the tangent is $y - 5 = 2(x - 5)$, i.e. $y = 2x - 5$. The positive slope shows average cost is rising at this output level.

Example 2. Profit turning point. Profit is $P(x) = -x^3 + 9x^2 - 15x$ (thousand dollars). Setting $P'(x) = -3x^2 + 18x - 15 = -3(x - 1)(x - 5) = 0$ gives $x = 1$ and $x = 5$. Since $P''(x) = -6x + 18$, $P''(5) = -12 < 0$, so $x = 5$ is a local maximum with profit $P(5) = -125 + 225 - 75 = \$25{,}000$.

Try this

Q1. Find the equation of the tangent to $y = x^2 + 1$ at $x = 2$. [3 marks]

• Cue. Point $(2, 5)$, slope $f'(2) = 4$; tangent $y - 5 = 4(x - 2)$, i.e. $y = 4x - 3$.

Q2. Find and classify the stationary points of $f(x) = x^3 - 12x$. [4 marks]

• Cue. $f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$; $x = -2$ (max, $f'' = -12$), $x = 2$ (min, $f'' = 12$).

Q3. Find the point of inflection of $f(x) = x^3 - 3x^2 + 2$. [3 marks]

• Cue. $f''(x) = 6x - 6 = 0 \Rightarrow x = 1$; $f(1) = 0$, so inflection at $(1, 0)$.