Tangent and normal lines | Year 11

01 / LEARN

The idea, step by step

Find both point and gradient

At x=a, a tangent passes through (a,f(a)) with gradient f′(a). Find the height using the original function and the gradient using its derivative. Then use point–gradient form.

The normal is perpendicular to the tangent. For nonzero tangent gradient m, its gradient is −1/m. Simply changing the sign of m does not give the normal.

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

Special cases and checks

If the tangent gradient is zero, the tangent is y=f(a) and the normal is the vertical line x=a. A vertical line has no finite numerical gradient.

Substitute the tangent point into both equations as a check. For finite perpendicular slopes their product should be −1. Also check that any supplied point really lies on the curve before using it as a point of tangency.

A common trap

Using the derivative value as a coordinate.

Instead: Use f(a) for the height and f′(a) for the gradient.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

The point is (2,4), and $d y / d x = 2 x$ gives gradient 4.
Tangent: $y - 4 = 4 (x - 2)$ .
Normal: $y - 4 = - \frac{1}{4} (x - 2)$ .

Tangent

y = 4 x - 4

; normal

y = - \frac{1}{4} x + \frac{9}{2}

Example 2

Worked example 2

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

The point is (2,0). The derivative $2 (x - 2)$ is 0 there.
The tangent is horizontal; its perpendicular normal is vertical.

Tangent

y = 0

; normal

x = 2

Check your understanding

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

Show answer

$1 / 3$

The negative reciprocal is 1/3, and $(- 3) (1 / 3) = - 1$ .

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF 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 −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 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 ↗

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