Tangents, Stationary Points and Curve Sketching

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 $y = a x^{2} + b x + c$ , the turning point is at $x = - \frac{b}{2 a}$ . For $y = x^{2} - 6 x + 5$ 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 $y - y_{1} = m (x - x_{1})$ .

y = x³ − 3x. The tangent is flat at the maximum and the minimum.

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 11Tangents 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 $- \frac{1}{f ^{'} ( a )}$ .

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

Point: $y = 5$ , so $(2, 5)$ . Gradient: $\frac{d y}{d x} = 2 x = 4$ .

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

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

Finding unknowns from a tangent. If $y = a x^{2} + b x$ has tangent $y = 5 x - 4$ at $x = 2$ , the curve and tangent share a point and a gradient: $4 a + 2 b = 6$ (both give $y = 6$ ) and $4 a + b = 5$ (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 $2 x^{2} + x - 3 = 0$ , giving $x = 1$ (the start) or $x = - \frac{3}{2}$ . It meets again at $(- \frac{3}{2}, \frac{9}{4})$ .

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 $2 a$ and it passes through $(a, a^{2} + 4)$ . Through the origin means $a^{2} + 4 = 2 a \times a$ , so $a^{2} = 4$ , $a = \pm 2$ , and the gradients are $\pm 4$ .

Year 11Increasing, decreasing and stationary points

For $y = x^{3} - 3 x$ : $\frac{d y}{d x} = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$ .

Stationary points where $3 x^{2} - 3 = 0$ : $x = \pm 1$ . Points $(- 1, 2)$ and $(1, - 2)$ .
$\frac{d y}{d x} < 0$ between $- 1$ and $1$ , so the curve decreases there.
$\frac{d y}{d x} > 0$ 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
$\frac{d y}{d x}$	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 $y = x^{3} + p x + q$ has a stationary point at $(2, 1)$ , then $y^{'} = 3 x^{2} + p = 0$ 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. $x^{3} - 3 x + k = 0$ means $x^{3} - 3 x = - k$ . The curve $y = x^{3} - 3 x$ 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 12The 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 $y = x^{3} - 3 x$ , $f^{''} (x) = 6 x$ , 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 $y = x^{4} - 4 x^{3}$ , $y^{'} = 4 x^{2} (x - 3)$ and $y^{''} = 12 x (x - 2)$ . 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 $y = x^{4} + a x^{3}$ has an inflection at $x = 1$ , then $y^{''} = 12 x^{2} + 6 a x = 0$ at $x = 1$ gives $a = - 2$ .

Concave up everywhere. For $f (x) = x^{4} + k x^{3} + 6 x^{2}$ , $f^{''} (x) = 12 x^{2} + 6 k x + 12$ . This quadratic is never negative when its discriminant is at most 0: $36 k^{2} - 576 \leq 0$ , 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 $- \frac{1}{f ^{'} ( a )}$ . 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

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Try 3 questions now

Reading left to right, a graph that goes down is:
1. A increasing
2. B decreasing
3. C stationary
4. D concave up
Show answer
B: decreasing
Down to the right means decreasing.
At a turning point the tangent is:
1. A horizontal
2. B vertical
3. C steepest
4. D undefined
Show answer
A: horizontal
The curve is flat for an instant.
Where is the turning point of $y = x^{2} - 4 x + 1$ ?
1. A $(4, 1)$
2. B $(2, - 3)$
3. C $(- 2, 13)$
4. D $(2, 1)$
Show answer
B: $(2, - 3)$
$x = \frac{4}{2} = 2$ , $y = 4 - 8 + 1 = - 3$ .