Tangents, gradients, and stationary points

IB Mathematics: Applications and Interpretation SL· 45 min read

1. Gradients of Tangents and the First Derivative★★☆☆☆⏱ 15 min

📘 Definition

Gradient of a Tangent

$f^{'} (a)$ or $\frac{d y}{d x}_{x = a}$

The gradient of the tangent to a curve $y = f (x)$ at $x = a$ equals the value of the first derivative evaluated at $x = a$ .

Example:

For $f (x) = x^{2}$ , gradient at $x = 2$ is $f^{'} (2) = 4$ .

The tangent to a curve at a point is the straight line that just touches the curve at that point, matching its slope exactly at that point. Differentiation gives us a gradient function that works for any point on the curve, so we only need to evaluate it at the desired $x$ -coordinate.

📐 Worked Example

Find the gradient of the tangent to $y = x^{3} - 2 x + 5$ at the point where $x = 1$ .

1
Differentiate the function to get the gradient function:
2
$\frac{dy}{dx} = 3x^2 - 2$
3
Evaluate the derivative at $x = 1$ :
4
$\frac{dy}{dx}\bigg|_{x=1} = 3(1)^2 - 2 = 1$
5
The gradient of the tangent at $x = 1$ is $1$ .

Exam tip:

Always substitute the given $x$ -value into the derivative to get a numerical gradient, don't leave it as a function of $x$ .

2. Finding the Equation of a Tangent Line★★☆☆☆⏱ 20 min

To find the full equation of a tangent, you already know the gradient from the derivative. The tangent always passes through the point of tangency $(a, f (a))$ , so you can use the point-gradient formula for a straight line.

Differentiate $f (x)$ to get the gradient function $f^{'} (x)$
Calculate gradient $m = f^{'} (a)$ at the point $x = a$
Calculate the $y$ -coordinate: $b = f (a)$ to get point $(a, b)$
Substitute into $y - b = m (x - a)$
Rearrange to $y = m x + c$ as required

📐 Worked Example

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

1
Find the derivative (gradient function):
2
$\frac{dy}{dx} = 4x - 3$
3
Calculate gradient at $x = 2$ :
4
$m = 4(2) - 3 = 5$
5
Find the $y$ -coordinate of the point of tangency:
6
$y = 2(2)^2 - 3(2) = 2, \text{ so point } = (2, 2)$
7
Substitute into point-gradient formula and rearrange:
8
$y - 2 = 5(x - 2) \implies y = 5x - 8$

3. Stationary Points: Definition and Location★★★☆☆⏱ 15 min

📘 Definition

Stationary Point

$f^{'} (x) = 0$

A point on a curve where the gradient is zero, so the tangent is horizontal, and the function is neither increasing nor decreasing.

To find stationary points, you set the first derivative equal to zero, then solve the resulting equation for $x$ , then calculate the corresponding $y$ -coordinate for each solution. For IB AI SL, you only need to classify stationary points as local maxima or local minima.

📐 Worked Example

Locate all stationary points of $y = x^{3} - 3 x + 2$ .

1
Calculate the first derivative:
2
$\frac{dy}{dx} = 3x^2 - 3$
3
Set derivative equal to zero and solve for $x$ :
4
$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1$
5
Calculate corresponding $y$ -coordinates:
6
$x=1: y = 1^3 - 3(1) + 2 = 0 \\ x=-1: y = (-1)^3 - 3(-1) + 2 = 4$
7
Stationary points are at $(1, 0)$ and $(- 1, 4)$ .

4. Classifying Stationary Points★★★☆☆⏱ 20 min

For IB AI SL, the most common method to classify stationary points is the first derivative test: you test the sign of the gradient just to the left and right of the stationary point.

For a stationary point at $x = a$ , calculate $f^{'} (a - h)$ (left) and $f^{'} (a + h)$ (right) for small $h$
Gradient changes positive $\to$ negative = local maximum
Gradient changes negative $\to$ positive = local minimum
No sign change = horizontal point of inflection

📐 Worked Example

Classify the stationary points of $y = x^{3} - 3 x + 2$ from the previous example.

1
Test the stationary point at $x = - 1$ , with $h = 0.1$ :
2
$f'(-1.1) = 3(-1.1)^2 - 3 = 0.63 > 0 \quad f'(-0.9) = 3(-0.9)^2 - 3 = -0.57 < 0$
3
Gradient changes positive to negative, so $(- 1, 4)$ is a local maximum.
4
Test the stationary point at $x = 1$ :
5
$f'(0.9) = -0.57 < 0 \quad f'(1.1) = 0.63 > 0$
6
Gradient changes negative to positive, so $(1, 0)$ is a local minimum.

5. Common Pitfalls

Wrong move:

Forgetting to calculate the $y$ -coordinate when finding the tangent equation

Why:

You only get the gradient from the derivative, you need the $y$ -coordinate to form the full equation

Correct move:

Always evaluate the original function at the given $x$ -value to get the point of tangency

Wrong move:

Setting the original function equal to zero instead of the derivative for stationary points

Why:

Stationary points require the gradient (not the function value) to be zero

Correct move:

Always set the first derivative $f^{'} (x) = 0$ , not $f (x) = 0$

Wrong move:

Mixing up the sign change rule for maxima and minima

Why:

It's easy to flip the order of signs when testing left and right of the point

Correct move:

Remember: Positive to Negative = Peak (Maximum), Negative to Positive = Valley (Minimum)

Wrong move:

Using the gradient value as the $y$ -intercept in the tangent equation

Why:

Confusing the roles of gradient and intercept when rearranging the formula

Correct move:

Always start from the point-gradient formula $y - y_{1} = m (x - x_{1})$ to avoid this error

Wrong move:

Stopping after finding $x$ -coordinates of stationary points

Why:

Exam questions almost always ask for the full coordinates of the stationary points

Correct move:

Always substitute $x$ back into the original function to get the corresponding $y$ -coordinate

6. Quick Reference Cheatsheet

Step	Finding Tangent Equation	Finding & Classifying Stationary Points
1	Differentiate $y = f (x)$ to get $\frac{d y}{d x}$	Differentiate $y = f (x)$ to get $\frac{d y}{d x}$
2	Evaluate $m = \frac{d y}{d x} ∣_{x = a}$	Set $\frac{d y}{d x} = 0$ , solve for $x$
3	Calculate $y_{1} = f (a)$ to get $(a, y_{1})$	Find $y$ -coordinate for each $x$
4	Substitute into $y - y_{1} = m (x - a)$	Test sign of derivative either side
5	Rearrange to required form	Max: + $\to$ -, Min: - $\to$ +

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.

2022 · 1
Find equation of tangent to cubic curve
2023 · 2
Find and classify stationary points of quadratic

Going deeper

Formula sheetIB AI SL Calculus Formula ReferenceRelevant for tangent and gradient formulas

What's Next

Understanding tangents and stationary points is the foundation for solving optimization problems, one of the most common applied calculus questions on the IB AI SL exam. These concepts also let you sketch accurate graphs of functions by identifying key features, which is required for many graphing and problem-solving questions. Next, you will apply these skills to solve practical real-world optimization problems, where you need to find the maximum or minimum value of a function, such as maximizing area or minimizing production cost. These skills also connect directly to graph sketching and further work with integration, the reverse of differentiation.