Tangent or normal

At a given point

In 13 of 20 recent exams
2% of the marks
Paper 1 or 2
1 to 5 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 1–5 marks. Very common: usually straight after "find $f'(x)$ ", or as an early part of a Section B question.

Spot it in the exam

Exam phrases to spot (7)

“find the equation of the tangent to the curve at the point where $x = 2$ ”
“the normal to the graph of $f$ at $P$ ”
“give your answer in the form $ax + by + d = 0$ , where $a, b, d \in \mathbb{Z}$ ”
“write down the gradient of the tangent at $P$ ”
“find the gradient of the curve at …”
“show that the normal at $P$ is …”
“find the angle that the tangent at $P$ makes with the positive x-axis”

In disguise (6)

the point is described by a feature: "where the curve crosses the x-axis", "the y-intercept". Find the point first.
"the line through $P$ perpendicular to the tangent at $P$ ": that is the normal
$f'(x)$ and one value of $f$ are given, but not $f$ itself: no integration is needed, because the tangent only uses $f(a)$ and $f'(a)$
the tangent's equation is given and the normal is asked: the point comes from the line. If the tangent at $x = 2$ is $y = 3x - 1$ , the point is $(2, 5)$ and the normal is $y = -\frac{1}{3}x + \frac{17}{3}$ .
"show that the curves meet at right angles": find where they meet, then show that the two gradients there multiply to $-1$
the curve is the graph of $y = f'(x)$ , not of $f$ : every step moves up one level, so its gradient at $x = a$ is $f''(a)$

Looks like it, but isn't (5)

only a gradient, or a point off the curve, is known: Find the point of contact
the line is already known and meets something else: The tangent or normal meets something
the curve has unknown constants: Find unknown constants
"the line … is a tangent to the parabola", with an unknown in the line and no point: Two graphs
a line at right angles to (or parallel to) another line, through a given point: the gradient is read from the line's equation, with no derivative: Equation of a straight line

Worked example

A normal in integer form

Paper 1 (no calculator)

Question

Let $f(x) = 3x^2 - \dfrac{8}{x}$ , for $x \ne 0$ .
(a) Find $f'(x)$ . [2]
(b) Find the equation of the normal to the graph of $f$ at the point where $x = 2$ . Give your answer in the form $ax + by + d = 0$ , where $a, b, d \in \mathbb{Z}$ . [5]

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See it move: Drag the point, the tangent follows

At $x = 0.80$ : the point is $(0.80, -8.08)$ , the gradient is $f'(0.80) = 17.3$ , and the tangent is $y = 17.3x - 21.92$ .
The gradient is a number, $f'(a)$ , read at the point: here $f'(x) = 6x + \dfrac{8}{x^{2}}$ . It is the height of the lower graph and the rise of the triangle.
The normal is at right angles to the tangent: gradient $-\frac{1}{f'(a)}$ , here $-0.06$ . The axes have different scales, so the right angle looks squashed.
Drag to $x = 2$ : $f(2) = 8$ , $f'(2) = 14$ , the normal's gradient is $-\frac{1}{14}$ , and the normal is $x + 14y - 114 = 0$ .

Steps

The point:
$x = a$ and $y = f(a)$ , from the curve. If the point is described by a feature, solve for it first.
The gradient:
$m = f'(a)$ . On the GDC gives it (gradient at a point); keep the full value (keep full-accuracy values).
Normal:
gradient $-\frac{1}{m}$ . If $m = 0$ , the tangent is $y = f(a)$ and the normal is $x = a$ .
The line:
$y - f(a) = m(x - a)$ , then write it as $y = \ldots$ , or in the form asked for (for $ax + by + d = 0$ with integers, clear the fractions).
Check:
the point satisfies your line.

Common mistakes

Using $f'(x)$ , with $x$ still in it, as the gradient.
The gradient is the number $f'(a)$ .
The wrong point
$(a, f'(a))$ is not on the curve; the point is $(a, f(a))$ . For a composite, $h(1) = f(g(1))$ , not $f(1)$ .
The normal's gradient
it is $-\frac{1}{f'(a)}$ , not $-f'(a)$ and not $\frac{1}{f'(a)}$ .

More mistakes (4)

A zero gradient
if $f'(a) = 0$ , the normal is the vertical line $x = a$ . For example, $y = x^2 - 4x$ at $x = 2$ has tangent $y = -4$ and normal $x = 2$ .
Writing " $L = \ldots$ " or "tangent $= \ldots$ "
a line is an equation, $y = \ldots$ . Otherwise the final mark is lost.
The integer form
" $a, b, d \in \mathbb{Z}$ " means no fractions: multiply through. Any integer multiple is accepted.
Rounding on too early
keep the full gradient and $y$ -value; round only the final coefficients.

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