Free Tangents & normals questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Calculus. Each question has the answer and the marking points an examiner would look for.
Generate a full Tangents & Normals worksheet →All Calculus questionsThe tangent to the curve y = e3x at the point P passes through the origin.
Find the exact coordinates of P and the equation of the tangent.
| Part | Answer and marking points | Marks |
|---|---|---|
| P(13, e), y = 3ex at x = p, gradient 3e3p A1 tangent: y − e3p = 3e3p(x − p) M1 substituting (0, 0) M1 3p = 1 ⇒ p = 13 A1 P(13, e) A1 y = 3ex A1 | 6 |
The curve C has equation y = x3 − 2x2 − 4x − 1. The point P on C has x-coordinate 1.
Find the equation of the tangent to C at P.
The tangent at P meets C again at the point Q. Find the coordinates of Q.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | y = −5x − 1 dydx = 3x2 − 4x − 4 A1 gradient at P = −5 A1 P(1, −6) (A1) y = −5x − 1 A1 | 4 |
| (b) | (0, −1) equating the curve and the tangent M1 x3 − 2x2 + x = 0 A1 (x − 1)2(x) = 0, since x = 1 is a repeated root R1 Q(0, −1) A1 | 4 |
The tangent to the curve y = ex at the point P passes through the origin.
Find the exact coordinates of P and the equation of the tangent.
| Part | Answer and marking points | Marks |
|---|---|---|
| P(1, e), y = ex at x = p, gradient ep A1 tangent: y − ep = ep(x − p) M1 substituting (0, 0) M1 p = 1 A1 P(1, e) A1 y = ex A1 | 6 |
The curve C has equation y = x3 − 2x2 + 5. The point P on C has x-coordinate −2.
Find the equation of the tangent to C at P.
The tangent at P meets C again at the point Q. Find the coordinates of Q.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | y = 20x + 29 dydx = 3x2 − 4x A1 gradient at P = 20 A1 P(−2, −11) (A1) y = 20x + 29 A1 | 4 |
| (b) | (6, 149) equating the curve and the tangent M1 x3 − 2x2 − 20x − 24 = 0 A1 (x + 2)2(x − 6) = 0, since x = −2 is a repeated root R1 Q(6, 149) A1 | 4 |
The tangent to the curve y = e2x at the point P passes through the origin.
Find the exact coordinates of P and the equation of the tangent.
| Part | Answer and marking points | Marks |
|---|---|---|
| P(12, e), y = 2ex at x = p, gradient 2e2p A1 tangent: y − e2p = 2e2p(x − p) M1 substituting (0, 0) M1 2p = 1 ⇒ p = 12 A1 P(12, e) A1 y = 2ex A1 | 6 |
The curve C has equation y = x3 − 3x2 − 3x − 3. The point P on C has x-coordinate −1.
Find the equation of the tangent to C at P.
The tangent at P meets C again at the point Q. Find the coordinates of Q.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | y = 6x + 2 dydx = 3x2 − 6x − 3 A1 gradient at P = 6 A1 P(−1, −4) (A1) y = 6x + 2 A1 | 4 |
| (b) | (5, 32) equating the curve and the tangent M1 x3 − 3x2 − 9x − 5 = 0 A1 (x + 1)2(x − 5) = 0, since x = −1 is a repeated root R1 Q(5, 32) A1 | 4 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in