IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 🎯 Skill ~3 min practice

AA SL Tangent Line Equation skills

A tangent line touches a curve at one point and matches its gradient there. Three steps: differentiate to get the gradient, plug in the x-value to get the actual slope, then use the point-slope form. Same recipe for every question — once you’ve got it, you’ve got it.

The Method

y − y₁ = m(x − x₁) point-slope form · (x₁, y₁) is the touching point · m = gradient at that point

Differentiate to get f'(x) — the gradient function.
Plug in the x-value of the point to get m = f'(x₁) — that’s the actual gradient at the contact point.
Find y₁ by plugging the x-value into the original f(x), then use the point-slope formula.

What’s a tangent line?

The tangent touches the curve at exactly one point and has the same gradient there as the curve. So the slope of the tangent is the value of the derivative at that point.

The recipe — every tangent question, every time

Differentiate the curve. Find f'(x).
Find the gradient at the point: m = f'(x₁).
Find y₁ by computing f(x₁) if not given.
Substitute into y − y₁ = m(x − x₁).

Worked examples

WE 1 EASY

Find the equation of the tangent to y = x² + 3x at the point where x = 2.

step 1 — differentiate dy/dx = 2x + 3step 2 — gradient at x = 2 m = 2(2) + 3 = 7step 3 — find y₁ y₁ = (2)² + 3(2) = 10step 4 — plug into y − y₁ = m(x − x₁) y − 10 = 7(x − 2) y = 7x − 14 + 10y = 7x − 4 always find both m AND y₁ before reaching for the formula!

WE 2 MEDIUM

Find the equation of the tangent to y = x³ − 4x + 1 at the point (1, −2).

step 1 — differentiate dy/dx = 3x² − 4step 2 — gradient at x = 1 m = 3(1)² − 4 = −1step 3 — point given (no y₁ to find) (x₁, y₁) = (1, −2)step 4 — point-slope y − (−2) = −1(x − 1) y + 2 = −x + 1y = −x − 1 when point is given, skip step 3 — but always sanity-check that the point IS on the curve!

WE 3 HARD

A tangent to y = x² − 6x + 8 has gradient 4. Find the equation of this tangent.

step 1 — differentiate dy/dx = 2x − 6step 2 — find x₁ from gradient set 2x − 6 = 4 2x = 10 → x₁ = 5step 3 — find y₁ y₁ = (5)² − 6(5) + 8 = 3step 4 — point-slope y − 3 = 4(x − 5) y = 4x − 20 + 3y = 4x − 17 when gradient is given but x isn’t, set the derivative equal to the gradient first to find x!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always run all three steps in order — don’t skip ahead.

Q1 EASY Find the tangent to y = x² at x = 3. Show answer ▼Hide answer ▲

dy/dx = 2x → m = 6 y₁ = 9 y − 9 = 6(x − 3) y = 6x − 9

Q2 EASY Find the tangent to y = 2x² − 5x at x = 1. Show answer ▼Hide answer ▲

dy/dx = 4x − 5 → m = −1 y₁ = 2 − 5 = −3 y + 3 = −1(x − 1) y = −x − 2

Q3 MEDIUM Find the tangent to y = x³ − 2x + 1 at the point (2, 5). Show answer ▼Hide answer ▲

dy/dx = 3x² − 2 → m = 10 y − 5 = 10(x − 2) y = 10x − 20 + 5 y = 10x − 15

Q4 MEDIUM Find the tangent to y = 1/x at x = 2. Show answer ▼Hide answer ▲

rewrite as power first y = x⁻¹ → dy/dx = −x⁻² = −1/x² m = −1/4 y₁ = 1/2 y − 1/2 = (−1/4)(x − 2) y = (−1/4)x + 1 always rewrite roots and fractions as powers BEFORE differentiating!

Q5 HARD A tangent to y = x³ − 3x has gradient 9. Find the equation of the tangent in the first quadrant. Show answer ▼Hide answer ▲

step 1 — find x from gradient dy/dx = 3x² − 3 = 9 3x² = 12 → x² = 4 → x = ±2 step 2 — first quadrant means x > 0, so x = 2 y₁ = 2³ − 3(2) = 2 step 3 — point-slope y − 2 = 9(x − 2) y = 9x − 16 two values of x mean two tangents — pick the one matching the domain hint!

⚠ Common mistakes

Using the original function for m instead of the derivative. The gradient is f‘(x₁), not f(x₁).
Forgetting to find y₁. If only an x-value is given, you need to compute y₁ = f(x₁) too.
Sign errors in the formula. y − (−2) becomes y + 2 — the double negative trips students up.
Substituting into the wrong equation. x₁ goes in the derivative; y₁ needs the original.
Confusing tangent with normal. The normal line is perpendicular to the tangent — gradient = −1/m. They’re related but different lines.
Leaving the answer in point-slope form when y = mx + c is asked. Always simplify if the question wants explicit form.

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Want the theory?

Read the full Gradients, Tangents & Normals notes for the full geometric picture, the link to normals, and how to use tangents in optimisation problems.

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