# Tangents, Stationary Points and Curve Sketching

> Equations of tangents and normals, increasing and decreasing, stationary points, concavity and sketching curves.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/curve-sketching/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The roller coaster designer
Priya designs roller coasters for a theme park on the Gold Coast. She needs to know three things about every curve in the track: where it climbs, where it falls, and where the riders reach the very top and the very bottom. Get the tops wrong and the cars might not make it over.

The derivative tells her all of this. Where the gradient is positive, the track climbs. Where it is negative, it falls. Where it is zero, the track is flat for an instant: a peak or a dip. That is exactly what you need to sketch any curve.

**In short:** Curve sketching uses the derivative to find a graph's shape. A curve increases where $f'(x) > 0$ and decreases where $f'(x) < 0$, stationary points are where $f'(x) = 0$, and the second derivative $f''(x)$ tells a maximum from a minimum and shows concavity.

## Starting point: Years 9 and 10

### Reading a graph

Look at any graph from left to right.
- It is **increasing** where it goes up.
- It is **decreasing** where it goes down.
- A **turning point** is where it changes direction. At the very top or bottom it is flat for an instant.

For a parabola $y = ax^2 + bx + c$, the turning point is at $x = -\dfrac{b}{2a}$. For $y = x^2 - 6x + 5$ the turning point is at $x = 3$, $y = -4$, and it is a minimum.

### Equation of a line through a point

You will often know a gradient and a point. Use $y - y_1 = m(x - x_1)$.

### Worked example: Line through a point
Gradient 4, through $(2, 3)$: $y - 3 = 4(x - 2)$, so $y = 4x - 5$.

*Diagram: Curve y equals x cubed minus 3x with a maximum at minus 1, a minimum at 1 and an inflection at the origin*
*y = x³ − 3x. The tangent is flat at the maximum and the minimum.*

### Key idea: Signs of the gradient
- Gradient positive: increasing.
- Gradient negative: decreasing.
- Gradient zero: flat. This is a **stationary point**.

### Watch out: Increasing means left to right
Always read a graph from left to right. A graph that goes down as you move right is decreasing, even if it is high up on the page.

**Mind map**

```text
Using the derivative
  Tangent and normal
    Gradient m = f'(a)
    Normal gradient -1/m
  Increasing, decreasing
    f'(x) > 0 rising
    f'(x) < 0 falling
  Stationary points
    f'(x) = 0
    Max, min, inflection
  Second derivative
    f''(x) > 0 concave up
    f''(x) < 0 concave down
    Sign change: inflection
  Sketch
    Intercepts
    Stationary points
    Ends
```

### Year 11: Tangents and normals
The gradient of the tangent at $x = a$ is $f'(a)$. The **normal** is the line at right angles to the tangent, so its gradient is $-\dfrac{1}{f'(a)}$.

**Example.** Find the tangent and normal to $y = x^2 + 1$ at $x = 2$.

Point: $y = 5$, so $(2, 5)$. Gradient: $\dfrac{dy}{dx} = 2x = 4$.

Tangent: $y - 5 = 4(x - 2)$, so $y = 4x - 3$.

Normal: gradient $-\dfrac{1}{4}$, so $y - 5 = -\dfrac{1}{4}(x - 2)$, which gives $x + 4y - 22 = 0$.

**Finding unknowns from a tangent.** If $y = ax^2 + bx$ has tangent $y = 5x - 4$ at $x = 2$, the curve and tangent share a point and a gradient: $4a + 2b = 6$ (both give $y = 6$) and $4a + b = 5$ (gradients match). So $b = 1$ and $a = 1$.

**Where a normal meets the curve again.** The normal to $y = x^2$ at $(1, 1)$ is $y = -\frac{1}{2}x + \frac{3}{2}$. Solve with the curve: $x^2 = -\frac{1}{2}x + \frac{3}{2}$, so $2x^2 + x - 3 = 0$, giving $x = 1$ (the start) or $x = -\frac{3}{2}$. It meets again at $\left(-\frac{3}{2}, \frac{9}{4}\right)$.

**Tangents through an outside point.** For tangents to $y = x^2 + 4$ through the origin, let the tangent touch at $x = a$: its gradient is $2a$ and it passes through $(a, a^2 + 4)$. Through the origin means $a^2 + 4 = 2a \times a$, so $a^2 = 4$, $a = \pm 2$, and the gradients are $\pm 4$.

### Year 11: Increasing, decreasing and stationary points
For $y = x^3 - 3x$: $\dfrac{dy}{dx} = 3x^2 - 3 = 3(x - 1)(x + 1)$.

- Stationary points where $3x^2 - 3 = 0$: $x = \pm 1$. Points $(-1, 2)$ and $(1, -2)$.
- $\dfrac{dy}{dx} < 0$ between $-1$ and $1$, so the curve decreases there.
- $\dfrac{dy}{dx} > 0$ for $x < -1$ and $x > 1$, so it increases there.

A **table of gradients** shows the type of each point:

| $x$ | $-2$ | $-1$ | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $\dfrac{dy}{dx}$ | 9 | 0 | $-3$ | 0 | 9 |
| slope | up | flat | down | flat | up |

Up, flat, down means a **maximum** at $x = -1$. Down, flat, up means a **minimum** at $x = 1$.

**Unknown coefficients.** If $y = x^3 + px + q$ has a stationary point at $(2, 1)$, then $y' = 3x^2 + p = 0$ at $x = 2$ gives $p = -12$, and the point gives $8 - 24 + q = 1$, so $q = 17$.

**Reading $f'(x)$ in factorised form.** If $f'(x) = (x - 2)^2(x + 1)$, the stationary points are at $x = 2$ and $x = -1$. The sign of $f'$ changes at $x = -1$ (a turning point) but not at $x = 2$, because $(x - 2)^2$ never changes sign. So $x = 2$ is a horizontal point of inflection.

**Counting solutions.** $x^3 - 3x + k = 0$ means $x^3 - 3x = -k$. The curve $y = x^3 - 3x$ has a maximum of 2 and a minimum of $-2$. A horizontal line meets it exactly twice when it touches a turning point, so $-k = \pm 2$ and $k = \pm 2$.

### Year 12: The second derivative and concavity
Differentiate twice to get $f''(x)$, the rate of change of the gradient.

- $f''(x) > 0$: **concave up**, curving like a cup.
- $f''(x) < 0$: **concave down**, curving like a cap.

**Testing stationary points.** At a point where $f'(a) = 0$:
- $f''(a) > 0$ means a minimum.
- $f''(a) < 0$ means a maximum.
- $f''(a) = 0$: use a table of gradients.

A **point of inflection** is where concavity changes, so $f''(x) = 0$ **and** changes sign. For $y = x^3 - 3x$, $f''(x) = 6x$, which changes sign at $x = 0$. The point $(0, 0)$ is an inflection.

A **horizontal point of inflection** has $f'(a) = 0$ too, like $y = x^3$ at the origin.

**Sketching checklist:** domain, intercepts, stationary points and their nature, inflections, and what happens as $x \to \pm\infty$.

**Global maximum.** On a closed interval, compare the stationary values with the values at the **endpoints**. The largest is the global maximum.

**Rates in words.** "Increasing at a decreasing rate" means $f'(x) > 0$ (going up) and $f''(x) < 0$ (the gradient is shrinking). "Decreasing at an increasing rate" means $f' < 0$ and $f'' < 0$.

**When $f''(a) = 0$ at a stationary point**, use a table of gradients. For $y = x^4 - 4x^3$, $y' = 4x^2(x - 3)$ and $y'' = 12x(x - 2)$. At $x = 0$, $y'' = 0$ and $y'$ is negative on both sides, so $(0, 0)$ is a horizontal point of inflection. At $x = 3$, $y'' > 0$, a minimum.

**Unknowns from an inflection.** If $y = x^4 + ax^3$ has an inflection at $x = 1$, then $y'' = 12x^2 + 6ax = 0$ at $x = 1$ gives $a = -2$.

**Concave up everywhere.** For $f(x) = x^4 + kx^3 + 6x^2$, $f''(x) = 12x^2 + 6kx + 12$. This quadratic is never negative when its discriminant is at most 0: $36k^2 - 576 \leq 0$, so $-4 \leq k \leq 4$.

## Quick summary

- Increasing where the gradient is positive. Decreasing where it is negative.
- Stationary points are where $f'(x) = 0$.
- Tangent gradient $f'(a)$. Normal gradient $-\dfrac{1}{f'(a)}$. Use $y - y_1 = m(x - x_1)$.
- Use a table of gradients or $f''(a)$ to tell a maximum from a minimum.
- Concave up: $f'' > 0$. Concave down: $f'' < 0$. Inflection: $f''$ changes sign.

## Quiz

Try these questions from the chapter quiz.

1. Reading left to right, a graph that goes down is:
   - A. increasing
   - B. decreasing
   - C. stationary
   - D. concave up
   - Answer: B. decreasing. Down to the right means decreasing.

2. At a turning point the tangent is:
   - A. horizontal
   - B. vertical
   - C. steepest
   - D. undefined
   - Answer: A. horizontal. The curve is flat for an instant.

3. Where is the turning point of $y = x^2 - 4x + 1$?
   - A. $(4, 1)$
   - B. $(2, -3)$
   - C. $(-2, 13)$
   - D. $(2, 1)$
   - Answer: B. $(2, -3)$. $x = \dfrac{4}{2} = 2$, $y = 4 - 8 + 1 = -3$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/curve-sketching/#quiz