Tangent and normal lines | Year 11

01 / LEARN

Understand the idea.

Find both point and gradient

A line needs a point and a gradient. At input $a$ , calculate the point $(a, f (a))$ from the original function and the tangent gradient $f^{'} (a)$ from the derivative. Insert these into point–gradient form before simplifying.

The normal passes through the same point and is perpendicular to the tangent. If the tangent has finite nonzero gradient $m$ , use normal gradient $- \frac{1}{m}$ . Check that the two gradients multiply to $- 1$ ; changing the sign alone is insufficient.

y - f (a) = f^{'} (a) (x - a) .

The graph of y equals x squared on minus one to three, with y from minus two to ten. A gold tangent is y equals two x minus one, touching at one, one.

Special cases and checks

A horizontal tangent has gradient zero, so its equation is $y = f (a)$ . Its normal is vertical, with equation $x = a$ . Do not attempt to assign that normal the numerical gradient $- \frac{1}{0}$ ; a vertical line has no finite gradient.

Check both line equations by substituting the point of contact. Also check that the tangent gradient agrees with the derivative there. If a question supplies a point, verify that it lies on the curve before treating it as the contact point.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find tangent and normal to $y = x^{2}$ at x=2.

Worked solution

1
Find the point of contact.
$x = 2, y = 2^{2} = 4$
The original curve supplies the point $(2, 4)$ for both lines.
2
Find the tangent gradient.
$\frac{d y}{d x} = 2 x, m_{\tan} = 2 (2) = 4$
The derivative gives the slope at the specified input.
3
Write the tangent through the point.
$\begin{aligned} y - 4 & = 4 (x - 2) \\ y & = 4 x - 4 \end{aligned}$
Point–gradient form fixes the line of gradient 4 through $(2, 4)$ .
4
Use the perpendicular gradient for the normal.
$\begin{aligned} m_{normal} & = - \frac{1}{4} \\ y - 4 & = - \frac{1}{4} (x - 2) \\ y & = - \frac{1}{4} x + \frac{9}{2} \end{aligned}$
The negative reciprocal makes the finite gradients multiply to $- 1$ , and the normal shares the contact point.

Answer

Tangent $y = 4 x - 4$ ; normal $y = - \frac{1}{4} x + \frac{9}{2}$ .

EXAMPLE 02

Worked example 2

Find tangent and normal to $y = {(x - 2)}^{2}$ at x=2.

Worked solution

1
Calculate the point and derivative.
$\begin{aligned} y (2) & = {(2 - 2)}^{2} = 0 \\ \frac{d y}{d x} & = 2 x - 4 \end{aligned}$
Expanding the square before differentiating gives the derivative and the contact point $(2, 0)$ .
2
Find the tangent line.
$m_{\tan} = 2 (2) - 4 = 0, y = 0$
A line with zero gradient through $(2, 0)$ is horizontal.
3
Write the perpendicular normal.
$x = 2$
The perpendicular to a horizontal line is vertical and must pass through $(2, 0)$ .

Answer

Tangent $y = 0$ ; normal $x = 2$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

The tangent gradient is −3. What is the normal gradient?

Show answer & reasoning

$1 / 3$

The normal gradient is $- \frac{1}{- 3} = \frac{1}{3}$ . The product $(- 3) (\frac{1}{3}) = - 1$ confirms perpendicularity.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

Find the tangent to $y = x^{2}$ at $x = 1$ .
3 marks
Find the normal to $y = x^{2}$ at $x = 1$ .
3 marks
Find the tangent to $y = x^{3}$ at $x = 2$ .
3 marks
Find the normal to $y = 3 x^{2}$ at $x = - 1$ .
3 marks
Find tangent and normal to $y = x^{2}$ at the origin.
3 marks
Can $(1, 4)$ be a point of tangency on $y = x^{2} + 1$ ? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the tangent to $y = x^{2}$ at $x = 1$ .
3 marks
Point $(1, 1)$ , gradient 2; $y - 1 = 2 (x - 1)$ .
$y = 2 x - 1$
Find the normal to $y = x^{2}$ at $x = 1$ .
3 marks
Tangent gradient 2 gives normal gradient $- \frac{1}{2}$ .
$y - 1 = - \frac{1}{2} (x - 1)$ .
$y = - \frac{1}{2} x + \frac{3}{2}$
Find the tangent to $y = x^{3}$ at $x = 2$ .
3 marks
Point $(2, 8)$ ; gradient $3 {(2)}^{2} = 12$ .
$y - 8 = 12 (x - 2)$ .
$y = 12 x - 16$
Find the normal to $y = 3 x^{2}$ at $x = - 1$ .
3 marks
Point $(- 1, 3)$ ; tangent gradient $- 6$ ; normal gradient $\frac{1}{6}$ .
$y - 3 = \frac{1}{6} (x + 1)$
Find tangent and normal to $y = x^{2}$ at the origin.
3 marks
The derivative is 0, giving a horizontal tangent and vertical normal.
$y = 0$ and $x = 0$
Can $(1, 4)$ be a point of tangency on $y = x^{2} + 1$ ? Explain.
3 marks
$1^{2} + 1 = 2$ , so $(1, 4)$ is not on the curve.
No. The curve has $y = 2$ at $x = 1$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Find both point and gradient

Special cases and checks

See the method in action.

Worked example 1

Find the point of contact.

Find the tangent gradient.

Write the tangent through the point.

Use the perpendicular gradient for the normal.

Worked example 2

Calculate the point and derivative.

Find the tangent line.

Write the perpendicular normal.

Try it yourself.

Check your understanding

Keep practising