IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Gradients, Tangents & Normals

The gradient of a curve at a point is the gradient of its tangent there — which is just f′(x) evaluated at that point. Once you have the gradient and the point, you can write the tangent’s equation directly using y − y₁ = m(x − x₁). The normal is the line perpendicular to the tangent at the same point — its gradient is −1/m.

📘 What you need to know

Gradient at a point: substitute the x-value into f′(x) to get the tangent gradient m.
Tangent equation: y − y₁ = f′(x₁)·(x − x₁) — point-slope form.
Normal gradient: −1/f′(x₁) — the negative reciprocal of the tangent gradient.
Normal equation: y − y₁ = (−1/f′(x₁))·(x − x₁).
The point must be ON the curve: find y₁ = f(x₁) by substituting into the original function.
Tangent ⊥ normal: their gradients multiply to give −1 (a quick sanity check).
GDC tool: d/dx evaluates f′(x₁) numerically — handy for messy expressions.
Form requested: rearrange to y = mx + c or ax + by + c = 0 by clearing fractions.

Finding the gradient at a point

Differentiate the function to get f′(x), then substitute the x-value at the point of interest. That single number is the gradient — both of the curve and of its tangent — at that point.

Gradient at x = x₁ gradient at x₁ = f′(x₁)

Reading exam wording: “gradient of the curve” and “gradient of the tangent” mean exactly the same thing at any point. The tangent is the local linear copy of the curve at that point.

Equation of the tangent

Tangent at (x₁, y₁) y − y₁ = f′(x₁)·(x − x₁)

You need both the gradient (from f′) and the point (from f). The point-slope form above is the cleanest starting expression — rearrange it after, not before. To find y₁, just plug x₁ into the original f(x).

Equation of the normal

The tangent has gradient m = f′(x₁); the normal is perpendicular to it, so its gradient is −1/m. Both pass through the point P.

Normal at (x₁, y₁) y − y₁ = −1f′(x₁)·(x − x₁)

If tangent gradient = 0

tangent: y = y₁; normal: x = x₁

horizontal tangent → vertical normal

If tangent is vertical (f′ undefined)

tangent: x = x₁; normal: y = y₁

vertical tangent → horizontal normal

🧭 Recipe — find tangent or normal at a point

Find y₁ = f(x₁) — the y-coordinate of the point on the curve.
Differentiate f to get f′(x); substitute x₁ to get the tangent gradient m.
For tangent: write y − y₁ = m(x − x₁).
For normal: replace m with −1/m; same point-slope form.
Rearrange to y = mx + c or ax + by + c = 0 by clearing fractions if needed.

Worked examples

WE 1

Find the gradient at a given point

The curve y = x³ − 4x² + 7. Find the gradient of the curve at x = 3.

Differentiate dy/dx = 3x² − 8x Substitute x = 3 dy/dx = 3(9) − 8(3) = 27 − 24 = 3 Gradient at x = 3 is 3 always plug into f′, not f — gradients come from the derivative

WE 2

Find points where gradient takes a specific value

The curve y = 2x³ − 9x² + 12x. Find the coordinates of all points on the curve where the gradient is 0.

Differentiate and set equal to 0 dy/dx = 6x² − 18x + 12 = 0 Factor 6(x² − 3x + 2) = 0 6(x − 1)(x − 2) = 0 x = 1 or x = 2 Substitute into y to get coordinates y(1) = 2 − 9 + 12 = 5 → (1, 5) y(2) = 16 − 36 + 24 = 4 → (2, 4) Points are (1, 5) and (2, 4) don’t forget to find the y-coordinates — “the points” needs both x AND y

WE 3

Equation of tangent — basic, in y = mx + c form

Find the equation of the tangent to y = x² − 3x + 4 at the point where x = 2. Give your answer in the form y = mx + c.

Find y at x = 2 y = 4 − 6 + 4 = 2 → point is (2, 2) Differentiate; get gradient dy/dx = 2x − 3 m = 2(2) − 3 = 1 Apply point-slope form y − 2 = 1(x − 2) y = x − 2 + 2 = x Tangent: y = x when c = 0, the tangent passes through the origin — geometrically clean answer

WE 4

Tangent — rewrite required first

Find the equation of the tangent to y = 5√x − 6x at the point where x = 1. Give your answer in the form y = mx + c.

Rewrite y as powers of x y = 5x^(1/2) − 6x⁻¹ Find y at x = 1 y = 5 − 6 = −1 → point is (1, −1) Differentiate; get gradient at x = 1 dy/dx = (5/2)x^(−1/2) + 6x⁻² = 5/(2√x) + 6/x² m = 5/2 + 6 = 17/2 Point-slope form y − (−1) = (17/2)(x − 1) y + 1 = (17/2)x − 17/2 y = (17/2)x − 19/2 Tangent: y = (17/2)x − 19/2 careful with double-negatives: y − (−1) = y + 1, not y − 1

WE 5

Equation of normal — integer-coefficient form

Find the equation of the normal to y = x² + 3x at the point where x = −1. Give your answer in the form ax + by + c = 0, where a, b, c are integers.

Find y at x = −1 y = 1 − 3 = −2 → point is (−1, −2) Tangent gradient at x = −1 dy/dx = 2x + 3 → m = 2(−1) + 3 = 1 Normal gradient = −1/m = −1 Point-slope form for the normal y − (−2) = −1(x − (−1)) y + 2 = −(x + 1) y + 2 = −x − 1 y = −x − 3 Rearrange to ax + by + c = 0 x + y + 3 = 0 Normal: x + y + 3 = 0 verify: substitute (−1, −2) → −1 − 2 + 3 = 0 ✓

WE 6

Multi-part: tangent, normal, and where the normal meets the x-axis

The curve y = x³ − 4x + 5 has a tangent and a normal at the point P where x = 2.

(a) Find the y-coordinate of P. (b) Find the equation of the tangent at P. (c) Find the equation of the normal at P. (d) Find the coordinates of the point where the normal meets the x-axis.

(a) Substitute x = 2 y = 8 − 8 + 5 = 5 → P = (2, 5) (b) Differentiate; gradient at P dy/dx = 3x² − 4 → m = 3(4) − 4 = 8 Tangent: y − 5 = 8(x − 2) y = 8x − 11 (c) Normal gradient = −1/8 Normal: y − 5 = −(1/8)(x − 2) y = 5 − (x − 2)/8 = (42 − x)/8 y = −(1/8)x + 21/4 (d) Set y = 0 0 = (42 − x)/8 → 42 − x = 0 → x = 42 (a) y = 5; (b) y = 8x − 11; (c) y = −(1/8)x + 21/4; (d) (42, 0) tangent gradient × normal gradient = 8 × (−1/8) = −1 ✓ — perpendicular check

💡 Top tips

Use point-slope form y − y₁ = m(x − x₁) — it’s faster than substituting into y = mx + c.
For tangent + normal at the same point, compute m ONCE; use m for tangent, −1/m for normal.
Sanity check: tangent gradient × normal gradient = −1 always.
For ax + by + c = 0, multiply through by any denominator to clear fractions; integer coefficients are usually expected.
GDC d/dx tool gives m numerically — useful when the algebra is messy or you want to verify.

⚠ Common mistakes

Using f(x₁) where you mean f′(x₁) — gradient comes from the DERIVATIVE.
Forgetting to find y₁ — you need the y-coordinate of the point, not just x₁.
Using +1/m for normal gradient — it’s the NEGATIVE reciprocal, −1/m.
Sign errors when expanding y − (−2) or x − (−1) — double-negatives become additions.
Rearrangement slips — after writing y = mx + c, double-check by plugging the point back in.

Next: Increasing & Decreasing Functions. The sign of f′(x) tells you whether f is increasing (f′ > 0), decreasing (f′ < 0), or stationary (f′ = 0). Solving the inequalities f′(x) > 0 and f′(x) < 0 partitions the domain into increasing and decreasing intervals — a fast way to map the shape of a curve without sketching it.

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