What is the difference between a tangent and a normal?

The tangent touches the curve with the same gradient as the curve at that point. The normal passes through the same point at right angles to the tangent, so its gradient is −1/m.

What if I am only given the x-coordinate?

Substitute it into the equation of the curve to find the y-coordinate first. You need both to write the line.

How to Find the Equation of a Tangent to a Curve

Short answer

How do you find the equation of a tangent to a curve?

Find the point on the curve, differentiate and substitute x to get the gradient m, then use y − y₁ = m(x − x₁). For the normal, use the negative reciprocal gradient −1/m instead.

A tangent touches a curve at one point and has the same gradient as the curve there. So you need two things: the point, and the gradient.

The method

The point: substitute $x$ into the curve to get $y$ .
The gradient: differentiate, then substitute $x$ into $\frac{d y}{d x}$ to get $m$ .
The line: $y - y_{1} = m (x - x_{1})$ .

Example 1: tangent and normal

Find the tangent and the normal to $y = x^{2} - 4 x + 5$ at $x = 1$ .

Point: $y = 1 - 4 + 5 = 2$ , so $(1, 2)$ .

Gradient: $\frac{d y}{d x} = 2 x - 4$ , so at $x = 1$ , $m = - 2$ .

Tangent:

y - 2 = - 2 (x - 1) ⟹ y = - 2 x + 4

Normal: its gradient is $- \frac{1}{m} = \frac{1}{2}$ .

y - 2 = \frac{1}{2} (x - 1) ⟹ y = \frac{1}{2} x + \frac{3}{2}

Example 2: a cubic

Find the tangent to $y = x^{3} - 2 x^{2} + 1$ at $x = 2$ .

Point: $y = 8 - 8 + 1 = 1$ , so $(2, 1)$ .

Gradient: $\frac{d y}{d x} = 3 x^{2} - 4 x = 12 - 8 = 4$ .

y - 1 = 4 (x - 2) ⟹ y = 4 x - 7

Try it: predict the tangent, then check

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Mistakes that cost marks

Using the curve's equation for the gradient instead of $\frac{d y}{d x}$ .
Leaving $x$ in the gradient. Substitute the point's $x$ to get a number.
Using $m$ for the normal. The normal needs $- \frac{1}{m}$ .

How to find the equation of a tangent to a curve