Equations of tangents and normals

IB Mathematics Analysis and Approaches SL· 12 min read

1. Calculating the gradient of the tangent at a point★★☆☆☆⏱ 8 min

The gradient of a tangent to a curve $y = f (x)$ at any point $x$ is exactly equal to the value of the first derivative $f^{'} (x)$ evaluated at that $x$ coordinate. This core principle underpins all tangent and normal calculations in IB Math AA SL.

📘 Definition

Tangent Gradient

$m_{t} = f^{'} (a)$

The slope of the line that touches the curve $y = f (x)$ at $x = a$ , matching the curve's instantaneous rate of change at that point.

📐 Worked Example

Find the gradient of the tangent to the curve $y = 3 x^{2} - 4 x + 1$ at the point where $x = 2$ .

1
Step 1: Differentiate the function with respect to $x$
$f'(x) = 6x - 4$
2
Step 2: Substitute $x = 2$ into the derivative
$f'(2) = 6(2) - 4 = 12 - 4 = 8$
3
Final tangent gradient at $x = 2$ is 8

Exam tip:

IB exam questions almost always give you the x-coordinate of the point of contact, not the full (x,y) pair, so remember to calculate the corresponding y value using the original $f (x)$ function.

2. Deriving the full equation of a tangent line★★★☆☆⏱ 10 min

Once you have the tangent gradient $m_{t}$ at the point of contact $(a, f (a))$ , you use the standard straight line point-gradient formula $y - y_{1} = m (x - x_{1})$ to get the full equation of the tangent. IB mark schemes almost always award 1 mark for correctly substituting the point and gradient into this formula.

y - f(a) = m_t (x - a)

📐 Worked Example

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

1
Step 1: Calculate the y-coordinate of the point of contact
$f(1) = (1)^3 - 2(1) = -1$
2
Step 2: Differentiate to find the gradient function
$f'(x) = 3x^2 - 2$
3
Step 3: Evaluate the derivative at $x = 1$
$m_t = f'(1) = 3(1)^2 - 2 = 1$
4
Step 4: Substitute $(1, - 1)$ and $m_{t} = 1$ into the point-gradient formula
$y - (-1) = 1(x - 1) \implies y + 1 = x - 1 \implies y = x - 2$

✓ Quick check

Test your understanding before moving on

What is the tangent to $y = 2 x^{2}$ at $x = 3$ ?
- y=12x - 18
- y=12x
- y=6x - 18
- y=12x + 18
Reveal answer
y=12x - 18 —
Correct: $f (3) = 18$ , $f^{'} (3) = 12$ , so $y - 18 = 12 (x - 3)$ simplifies to $y = 12 x - 18$ .

3. Calculating the equation of the normal line★★★☆☆⏱ 10 min

The normal line is perpendicular to the tangent at the point of contact. For two perpendicular lines with non-zero gradients, the product of their gradients is equal to -1. This means the normal gradient $m_{n}$ is the negative reciprocal of the tangent gradient $m_{t}$ .

m_n = -\frac{1}{m_t}, \quad m_t \neq 0

📐 Worked Example

Find the equation of the normal to the curve $y = 4 x + 1/ x$ at the point where $x = 1$ .

1
Step 1: Find the y-coordinate of the point
$f(1) = 4(1) + 1/1 = 5$
2
Step 2: Differentiate the function
$f'(x) = 4 - \frac{1}{x^2}$
3
Step 3: Calculate tangent gradient at $x = 1$
$m_t = 4 - 1 = 3$
4
Step 4: Take negative reciprocal to get normal gradient
$m_n = -\frac{1}{3}$
5
Step 5: Substitute $(1, 5)$ and $m_{n} = - 1/3$ into point-gradient formula
$y - 5 = -\frac{1}{3}(x - 1) \implies 3y - 15 = -x + 1 \implies x + 3y = 16$

4. Solving for unknown points of contact★★★★☆⏱ 12 min

Higher mark IB AA SL questions will give you the equation of a tangent or normal, and ask you to find the unknown x-coordinate of the point of contact. This requires setting the derivative equal to the given line gradient, then solving the resulting algebraic equation.

📐 Worked Example

The line $y = 6 x + c$ is tangent to the curve $y = 2 x^{2} - 4 x + 7$ . Find the value of $c$ .

1
Step 1: The tangent gradient is 6, so set derivative equal to 6
$f'(x) = 4x - 4 = 6$
2
Step 2: Solve for x
$4x = 10 \implies x = 2.5$
3
Step 3: Find the corresponding y value on the curve
$f(2.5) = 2(2.5)^2 -4(2.5) +7 = 12.5 - 10 +7 = 9.5$
4
Step 4: Substitute $(2.5, 9.5)$ into the tangent line equation to solve for c
$9.5 = 6(2.5) + c \implies 9.5 = 15 + c \implies c = -5.5$

5. Common Pitfalls

Wrong move:

Using the original function $f (x)$ instead of the derivative to calculate the tangent gradient

Why:

The original function returns y-values, not the slope of the curve at that point

Correct move:

Always differentiate $f (x)$ first before substituting the x coordinate to get the gradient

Wrong move:

Calculating the normal gradient as the reciprocal of the tangent, forgetting the negative sign

Why:

Parallel lines have equal gradients, not perpendicular ones

Correct move:

Always flip the sign after taking the reciprocal of $m_{t}$ to get $m_{n}$

Wrong move:

Using only the x-coordinate as the point of contact, forgetting to calculate the corresponding y value

Why:

The point-gradient formula requires both $x_{1}$ and $y_{1}$ to derive the correct line equation

Correct move:

Evaluate $f (a)$ on the original curve to get the full $(a, f (a))$ coordinate before building the line equation

Wrong move:

Simplifying the line equation incorrectly, mixing up signs when rearranging

Why:

IB mark schemes deduct 1 mark for sign errors in final line equations

Correct move:

Double check your rearranged equation by substituting the point of contact back in to confirm it satisfies the equation

Wrong move:

Trying to calculate the normal gradient when the tangent gradient is zero

Why:

A horizontal tangent ( $m_{t} = 0$ ) has a vertical normal, which has an undefined gradient

Correct move:

Write the normal line as $x = a$ , the vertical line passing through the point of contact

6. Quick Reference Cheatsheet

Quantity	Formula
Tangent gradient at $x = a$	$m_{t} = f^{'} (a)$
Normal gradient at $x = a$	$m_{n} = - 1/ m_{t}$
Equation of tangent	$y - f (a) = m_{t} (x - a)$
Equation of normal	$y - f (a) = m_{n} (x - a)$
Vertical normal (when $m_{t} = 0$ )	$x = a$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · Paper 1
Find tangent to quadratic function
2023 · Paper 2
Find normal to cubic function
2022 · Paper 1
Find tangent-curve second intersection

What's Next

Mastering tangent and normal equations is a foundational skill for all upcoming calculus topics in IB Math AA SL, including rates of change, optimization, and kinematics. You will next practice applying these differentiation skills to find maximum and minimum points on curves, and solve real-world optimization problems that are worth up to 8 marks on Paper 2. You can also review past exam-style questions that combine tangents with quadratic and cubic curve properties to reinforce your understanding before your mock exams.