# Equations of tangents and normals

> IB Mathematics Analysis and Approaches SL · IB Math AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-equations-of-tangents-and-normals/

We cover using first derivatives to find tangent gradients, derive full tangent and normal line equations, and solve common IB exam problems.

**Prerequisites:** [Basic differentiation of polynomial functions](https://www.owlsprep.com/study/ib-math-aa-sl-u5-basic-differentiation/); [Equation of a straight line (y = mx + c)](https://www.owlsprep.com/study/ib-math-aa-sl-u2-linear-functions/)

## Learning objectives

- Calculate the gradient of a curve at a given point using first derivatives
- Derive the full equation of a tangent line to a function at a specified coordinate
- Derive the full equation of a normal line perpendicular to the tangent at the point of contact
- Solve exam-style problems where tangents or normals meet the curve at a second point

## Calculating the gradient of the tangent at a point

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.

**Tangent Gradient** — 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.

*Notation:* $m_t = f'(a)$

**Worked example:** Find the gradient of the tangent to the curve $y = 3x^2 - 4x + 1$ 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.

## Deriving the full equation of a tangent line

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 $y = x^3 - 2x$ 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

   $$y - (-1) = 1(x - 1) \implies y + 1 = x - 1 \implies y = x - 2$$

**Check your understanding**

Test your understanding before moving on

1. What is the tangent to $y=2x^2$ at $x=3$?

   - y=12x - 18
   - y=12x
   - y=6x - 18
   - y=12x + 18

   *Why:* Correct: $f(3)=18$, $f'(3)=12$, so $y-18=12(x-3)$ simplifies to $y=12x-18$.

## Calculating the equation of the normal line

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 $m_t$.

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

> **Perpendicular Gradient Rule**
>
> Negative reciprocal for perpendicular: flip the fraction, flip the sign

**Worked example:** Find the equation of the normal to the curve $y = 4x + 1/x$ 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

   $$y - 5 = -\frac{1}{3}(x - 1) \implies 3y - 15 = -x + 1 \implies x + 3y = 16$$

**Exam command terms**

IB exam command terms for this topic have specific expectations:

- **Find the tangent** — You must present the full simplified linear equation, not just the gradient

- **Show that the normal is...** — You must explicitly state the negative reciprocal step to earn full marks

## Solving for unknown points of contact

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 $y = 6x + c$ is tangent to the curve $y = 2x^2 - 4x + 7$. 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

   $$9.5 = 6(2.5) + c \implies 9.5 = 15 + c \implies c = -5.5$$

## Common pitfalls

- **Wrong:** Using the original function $f(x)$ instead of the derivative to calculate the tangent gradient
  - Why it fails: The original function returns y-values, not the slope of the curve at that point
  - Correct: Always differentiate $f(x)$ first before substituting the x coordinate to get the gradient
- **Wrong:** Calculating the normal gradient as the reciprocal of the tangent, forgetting the negative sign
  - Why it fails: Parallel lines have equal gradients, not perpendicular ones
  - Correct: Always flip the sign after taking the reciprocal of $m_t$ to get $m_n$
- **Wrong:** Using only the x-coordinate as the point of contact, forgetting to calculate the corresponding y value
  - Why it fails: The point-gradient formula requires both $x_1$ and $y_1$ to derive the correct line equation
  - Correct: Evaluate $f(a)$ on the original curve to get the full $(a, f(a))$ coordinate before building the line equation
- **Wrong:** Simplifying the line equation incorrectly, mixing up signs when rearranging
  - Why it fails: IB mark schemes deduct 1 mark for sign errors in final line equations
  - Correct: Double check your rearranged equation by substituting the point of contact back in to confirm it satisfies the equation
- **Wrong:** Trying to calculate the normal gradient when the tangent gradient is zero
  - Why it fails: A horizontal tangent ($m_t=0$) has a vertical normal, which has an undefined gradient
  - Correct: Write the normal line as $x = a$, the vertical line passing through the point of contact

## 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 $m_t=0$) | $x = a$ |

## 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.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-math-aa-sl-u5-equations-of-tangents-and-normals/