Equation of the Tangent Line

Equations of tangents and normals: the point and dy/dx at it, finding a point from a given gradient, the normal in ax + by + c = 0 form, and estimating from a graph.

In the exam for Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606, Edexcel IGCSE Maths 4MA1 · 18 real past-paper questions below

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Tangents and Normals (Higher)

A tangent touches a curve at one point and has the same gradient as the curve there. A normal goes through the same point at right angles to it.

Formula

m_{tangent} = \frac{d y}{d x}

m_{normal} = - \frac{1}{m _{tangent}}

Key idea

Every tangent or normal question ends with a straight line. You build it from a point and a gradient.

The Tangent at a Point

Three things: the point, the gradient there, then the line. The line needs a point $(x_{1}, y_{1})$ on it and a gradient $m$ .

Formula

y - y_{1} = m (x - x_{1})

Worked example

1
$y = x^{3} - 2 x, x = 2$ — The curve and the x value you are given.
2
$y = 2^{3} - 2 (2)$ — Find y first. Put x = 2 into the curve, not into dy/dx.
3
$y = 8 - 4 = 4$ — The point is (2, 4).
4
$\frac{d y}{d x} = 3 x^{2} - 2$ — Differentiate: x³ → 3x², and −2x → −2.
5
$3 (2)^{2} - 2$ — Put x = 2 into dy/dx. This gives the gradient at the point.
6
$3 (4) - 2 = 10$ — (2)² = 4, then 3 × 4 = 12, and 12 − 2 = 10. The gradient is 10.
7
$y - 4 = 10 (x - 2)$ — Put the point (2, 4) and the gradient 10 into the line formula.
8
$y - 4 = 10 x - 20$ — Multiply out the bracket: 10 × x = 10x, and 10 × (−2) = −20.
9
$y = 10 x - 16$ — Add 4 to both sides. −20 + 4 = −16.

Common mistake

Using $3 x^{2} - 2$ as the gradient. Put x = 2 in first to get a number.

Key idea

At a turning point $\frac{d y}{d x} = 0$ , so the tangent is flat. Its equation is just $y =$ the y-coordinate of the point.

Finding the Point From the Gradient

Some questions give the gradient and ask for the point. Set dy/dx equal to it.

Worked example

1
$y = 5 x^{2} - 12 x + 4$ — The curve. The gradient at G is given as 18.
2
$\frac{d y}{d x} = 10 x - 12$ — Differentiate: 5x² → 10x, and −12x → −12.
3
$10 x - 12 = 18$ — Set dy/dx equal to 18, the gradient you were given.
4
$10 x = 30$ — Add 12 to both sides.
5
$x = 3$ — Divide both sides by 10.
6
$y = 5 (3)^{2} - 12 (3) + 4$ — Put x = 3 into the curve (not dy/dx) to find y.
7
$y = 45 - 36 + 4 = 13$ — 5 × 9 = 45 and 12 × 3 = 36.
8
$G = (3, 13)$ — The point on the curve where the gradient is 18.

The Normal (Higher)

The 5-mark question: find the point and the tangent's gradient, flip the gradient, then write the line in the form $a x + b y + c = 0$ . Here the tangent at G(3, 13) has gradient 18.

Worked example

1
$m_{normal} = - \frac{1}{18}$ — The normal is at right angles, so flip the tangent's gradient and change its sign.
2
$y - 13 = - \frac{1}{18} (x - 3)$ — Line formula with the point G(3, 13) and the normal's gradient.
3
$18 (y - 13) = - (x - 3)$ — Multiply both sides by 18 to clear the fraction.
4
$18 y - 234 = - x + 3$ — Multiply out: 18 × (−13) = −234, and −(x − 3) = −x + 3 because the minus changes both signs.
5
$x + 18 y - 234 = 3$ — Add x to both sides so every x and y term is on the left.
6
$x + 18 y - 237 = 0$ — Take 3 from both sides, so −234 − 3 = −237. Everything is on one side, with whole-number a, b, c.

Common mistake

Using the tangent's gradient, 18. A line at right angles needs −1/18.

Estimating From a Graph

With only a graph, draw the tangent with a ruler, then find its gradient from two points on it.

Worked example

1
$(1, 3) and (6, 0)$ — Two points read from your tangent, far apart. The tangent touches the curve at (1, 3).
2
$up = 0 - 3 = - 3$ — Up is the difference in y. It is negative because the line goes down.
3
$across = 6 - 1 = 5$ — Across is the difference in x, in the same order.
4
$m = \frac{- 3}{5} = - 0.6$ — Gradient is up ÷ across.
5
$y = - 0.6 x + c$ — Any straight line is y = mx + c, and you now know m.
6
$3 = - 0.6 (1) + c$ — Put in the point (1, 3) to find c.
7
$c = 3 + 0.6 = 3.6$ — −0.6 × 1 = −0.6. Add 0.6 to both sides.
8
$y = - 0.6 x + 3.6$ — The tangent. Your answer only needs to be close to this.

Key idea

Draw the tangent long. Points far apart give a better estimate.

Which Method? Read the Question

Find what the question gives you: the point, the gradient or a graph.

You see	You do
Tangent at x = a	point, dy/dx at a, then the line
Gradient given	set dy/dx equal to it
Normal or perpendicular	−1 over the tangent's gradient
A drawn curve	draw the tangent, rise over run
Turning point	Stationary Points

Key idea

Check your line: put the point back in. It must fit.

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