IB DP Mathematics exam preparation becomes more manageable when you understand why a method works instead of memorizing a list of steps. This worked example is suitable for students studying Mathematics: Analysis and Approaches (AA), including those preparing for calculus questions.
Exam-style question: find the tangent to a curve
The curve is y = x² − 4x + 1. Find the equation of the tangent at x = 3.
Step 1: Differentiate to find the gradient
Differentiate y = x² − 4x + 1 with respect to x:
dy/dx = 2x − 4.
At x = 3, the gradient of the tangent is m = 2(3) − 4 = 2.
Step 2: Find the point on the curve
Substitute x = 3 into the original equation: y = 3² − 4(3) + 1 = 9 − 12 + 1 = −2. The tangent passes through (3, −2).
Step 3: Write the tangent equation
Using the point-gradient form, y − y₁ = m(x − x₁):
y + 2 = 2(x − 3), so y = 2x − 8.
How to avoid common mistakes
- Find the gradient from the derivative, not from the original function.
- Use the original equation to calculate the y-coordinate of the point of tangency.
- Show the method clearly: derivative → gradient → point → equation.
- Check your answer: when x = 3, your tangent must give y = −2.
Try this yourself
Find the equation of the tangent to y = x² + 2x − 5 at x = 1.
Reveal the answer
dy/dx = 2x + 2, so m = 4 at x = 1. The point is (1, −2). Thus y + 2 = 4(x − 1), giving y = 4x − 6.
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