Equation of a Tangent to a Curve

Q: How do I find the equation of a tangent in IEB Maths?

Find the point of contact from $f(x)$, the gradient from $f'(x)$ at that point, then use $y - y_1 = m(x - x_1)$.

Q: What does it mean if a tangent is horizontal?

Its gradient is 0, so you solve $f'(x) = 0$. Horizontal tangents occur at turning points and sometimes at stationary points of inflection.

The derivative at a point gives the gradient of the tangent there, and with one point and a gradient you can write the tangent's equation. Tangent questions appear in Paper 1 on their own and inside cubic graph questions, often combined with parallel lines or a given gradient.

Gradient at a point

The gradient of the curve $y = f (x)$ at $x = a$ is $f^{'} (a)$ . This is also the gradient of the tangent at that point.

Example. Find the gradient of $f (x) = x^{3} - 4 x$ at $x = 2$ .

$f^{'} (x) = 3 x^{2} - 4$

$f^{'} (2) = 12 - 4 = 8$

You can work backwards too. If you are told the gradient of the tangent is $m$ , solve $f^{'} (x) = m$ to find the $x$ -value of the point of contact. A tangent parallel to a given line has the same gradient as that line. A horizontal tangent has gradient 0.

Equation of the tangent

To find the equation of the tangent to $y = f (x)$ at $x = a$ :

Find the point of contact: $(a; f (a))$ .
Find the gradient: $m = f^{'} (a)$ .
Substitute into $y - y_{1} = m (x - x_{1})$ and make $y$ the subject.

Example. Tangent to $f (x) = x^{2} - 2 x + 3$ at $x = 3$ .

Point: $f (3) = 9 - 6 + 3 = 6$ , so $(3; 6)$
Gradient: $f^{'} (x) = 2 x - 2$ , so $f^{'} (3) = 4$
Equation: $y - 6 = 4 (x - 3)$ , so $y = 4 x - 6$

Always find $y_{1}$ from the original function $f$ , never from $f^{'}$ . That is the most common lost mark on this topic.

Key terms

Point of contact: The point where the tangent touches the curve.
Gradient of a tangent: The value of the derivative at the point of contact.
Parallel lines: Lines with equal gradients.
Perpendicular lines: Lines whose gradients multiply to $- 1$ . A normal to a curve is perpendicular to the tangent.

Common mistakes

Substituting into $f^{'} (x)$ to find the $y$ -coordinate of the point of contact.
Using the derivative expression $f^{'} (x)$ as the gradient instead of the number $f^{'} (a)$ .
Setting $f (x)$ , not $f^{'} (x)$ , equal to a given gradient.
Leaving the equation in point-gradient form when the question asks for $y = m x + c$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

routine

Calculate the gradient of the tangent to $f (x) = x^{2} - 4 x + 1$ at $x = 3$ .

Worked solution

Hidden - try the question first

Solution

$f^{'} (x) = 2 x - 4$
$f^{'} (3) = 6 - 4 = 2$

Question 2

6 marks total

routine

Determine the equation of the tangent to $y = x^{3} - 2 x$ at the point where $x = - 1$ .

Worked solution

Hidden - try the question first

Solution

$f^{'} (x) = 2 x - 4$ , so the point is $(- 1; 1)$
$\frac{d y}{d x} = 3 x^{2} - 2$
At $x = - 1$ : $m = 3 - 2 = 1$
$y - 1 = 1 (x + 1)$
$y = x + 2$

Question 3

7 marks total

complex

The tangent to $f (x) = x^{2} + x + 1$ at point $P$ is parallel to the line $y = 5 x + 4$ .

Determine the coordinates of $P$ and the equation of the tangent at $P$ .

Worked solution

Hidden - try the question first

Solution

$f^{'} (x) = 2 x - 4$
Parallel, so $f^{'} (x) = 5$
$2 x + 1 = 5 \Rightarrow x = 2$
$f (2) = 4 + 2 + 1 = 7$ , so $P (2; 7)$
$y - 7 = 5 (x - 2)$
$y = 5 x - 3$

Question 4

9 marks total

problem-solving

Given $f (x) = - x^{3} + 3 x^{2}$ .

The tangent to $f$ at the point $P (a; f (a))$ , where $a \neq = 0$ , passes through the origin.

Determine the value of $a$ and the equation of this tangent.

Worked solution

Hidden - try the question first

Solution

$f^{'} (x) = 2 x - 4$ , so the gradient at $P$ is $- 3 a^{2} + 6 a$
Tangent: $y - f (a) = f^{'} (a) (x - a)$
Through $(0; 0)$ : $0 - (- a^{3} + 3 a^{2}) = (- 3 a^{2} + 6 a) (0 - a)$
$a^{3} - 3 a^{2} = 3 a^{3} - 6 a^{2} \Rightarrow 2 a^{3} - 3 a^{2} = 0$
$a^{2} (2 a - 3) = 0$ , and $a \neq = 0$ , so $a = \frac{3}{2}$
$f (\frac{3}{2}) = - \frac{27}{8} + \frac{27}{4} = \frac{27}{8}$ and $f^{'} (\frac{3}{2}) = - \frac{27}{4} + 9 = \frac{9}{4}$
The tangent passes through the origin, so $y = \frac{9}{4} x$

FAQs

How do I find the equation of a tangent in IEB Maths?

Find the point of contact from $f (x)$ , the gradient from $f^{'} (x)$ at that point, then use $y - y_{1} = m (x - x_{1})$ .

What does it mean if a tangent is horizontal?

Its gradient is 0, so you solve $f^{'} (x) = 0$ . Horizontal tangents occur at turning points and sometimes at stationary points of inflection.

Can a tangent cut the curve somewhere else?

Yes. A tangent to a cubic often crosses the curve at another point. 'Tangent' only means it touches at the point of contact with the same gradient.

Gradient at a point

Equation of the tangent

Key terms

Common mistakes

Practice questions

FAQs

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