IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 ~10 min read

Gradients, Tangents & Normals

Now that you can find a derivative, let’s actually use it. Plug an x-value into f′(x) and you instantly know the curve’s gradient at that point. From there it’s a short step to writing down the equation of the tangent (the line that grazes the curve) or the normal (the line that cuts straight across at right angles). These are bread-and-butter Paper 1 questions.

📘 What you need to know

The gradient at a point is just f′(x) evaluated at that point’s x-coordinate.
The tangent at a point is a straight line that touches the curve there with the same gradient.
The normal is a straight line through the same point but perpendicular to the tangent.
Tangent gradient = f′(x₁) · Normal gradient = −1f′(x₁)
For the equation of either line, use y − y₁ = m(x − x₁) with the right gradient.
Always find the y-coordinate of the point first by substituting into the original f(x) — not the derivative.

Finding the gradient at a point

The derivative f′(x) is a function. Plug in a value of x and out comes the gradient of the curve at that point. That’s it.

For example, if f(x) = x² + 3x − 4, then f′(x) = 2x + 3.

Gradient at x = 1: f′(1) = 2(1) + 3 = 5
Gradient at x = −2: f′(−2) = 2(−2) + 3 = −1

📍

Two functions, two different uses

Use f(x) when you need a y-coordinate (height of the curve). Use f′(x) when you need a gradient. Don’t confuse them — putting a number into the wrong function is the most common error in this whole topic.

Your GDC can also evaluate a derivative at a point — look for the ddx(…) tool, usually under the calculus menu. It’ll save time on Paper 2 when the function is messy. But you’ll still need to be able to do it by hand for Paper 1.

The tangent and the normal

At any point on a curve, there are two special straight lines worth knowing:

Tangent and normal at a point on a curve

🔴 Tangent

A straight line that just touches the curve at the point — it has the same gradient as the curve there.

m_tangent = f′(x₁)

🟢 Normal

A straight line through the same point but at right angles to the tangent.

m_normal = −1f′(x₁)

🤔 Why is the normal’s gradient −1/m?

For two perpendicular lines, the product of their gradients is always −1. So if the tangent has gradient m, the normal has gradient that satisfies m × m_n = −1, meaning m_n = −1/m. Flip the fraction, then change the sign. That’s it.

🧠

“Flip and negate”

To go from a tangent gradient to a normal gradient: flip the fraction, then change the sign. So gradient 2 becomes −½. Gradient −¾ becomes +⁴⁄₃. Two quick moves — that’s the whole trick.

Writing the equation of either line

Both tangents and normals are straight lines. So once you know the gradient and a point, the equation is just an application of the standard line formula:

Equation of a straight line through (x₁, y₁)

y − y₁ = m(x − x₁)

✓ in formula booklet (under “geometry”)

The only difference between a tangent and a normal is the gradient you use:

Tangent at the point (x₁, y₁)

y − y₁ = f′(x₁)(x − x₁)

Normal at the point (x₁, y₁)

y − y₁ = −1f′(x₁)(x − x₁)

The 4-step recipe

Whether the question asks for a tangent or a normal, the process is the same. Just pick the right gradient at Step 3.

How to find the equation of a tangent or normal

Find the y-coordinate of the point. Substitute the given x-value into f(x) (the original function — NOT the derivative).
Differentiate to find f′(x).
Find the gradient at the point. Substitute the x-value into f′(x). For a normal, take −1/that.
Substitute everything into y − y₁ = m(x − x₁) and rearrange to whatever form the question asks for.

The form the question asks for

Exam questions often want your line equation in a specific format. The two most common:

Common form conversions

y − y₁ = m(x − x₁)

→

y = mx + c

→

ax + by + d = 0

To go from y = mx + c to ax + by + d = 0, just rearrange so everything’s on one side. Multiply through to clear any fractions if needed — exam questions often specify “where a, b, d are integers”.

Read the question carefully — if it says “in the form y = mx + c“, give a tidy slope-intercept answer. If it says “ax + by + d = 0 where a, b, d are integers”, clear all the fractions and put zero on one side. Marks are lost over format every year.

Worked examples

WE 1

Find the gradient at a point

A function is defined by f(x) = x³ + 6x² + 5x − 12.

(a) Find f′(x). (b) Find the gradient of y = f(x) when x = 1. (c) Find the gradient when x = −2.

Differentiate, then substitute the x-value into f′(x).part (a) — differentiate Apply the power rule term by term: f′(x) = 3x² + 12x + 5 f′(x) = 3x² + 12x + 5part (b) — gradient at x = 1 f′(1) = 3(1)² + 12(1) + 5 = 3 + 12 + 5 gradient = 20part (c) — gradient at x = −2 f′(−2) = 3(−2)² + 12(−2) + 5 = 12 − 24 + 5 gradient = −7 always sub into f′(x), never into f(x), when the question asks for a gradient!

WE 2

Equation of a tangent

Find the equation of the tangent to the curve y = x² − 4x + 7 at the point where x = 3, in the form y = mx + c.

Follow the 4-step recipe.step 1 — find y at x = 3 y = (3)² − 4(3) + 7 = 9 − 12 + 7 = 4 Point is (3, 4).step 2 — differentiate f′(x) = 2x − 4step 3 — gradient at x = 3 f′(3) = 2(3) − 4 = 2step 4 — equation of line y − 4 = 2(x − 3) y = 2x − 6 + 4 y = 2x − 2 always state the point (3, 4) before going to step 4 — earns method marks!

WE 3

Equation of a normal

Find the equation of the normal to the curve y = x² − 6x + 10 at the point where x = 4, in the form y = mx + c.

Same recipe — flip and negate the gradient at step 3.step 1 — find y at x = 4 y = (4)² − 6(4) + 10 = 16 − 24 + 10 = 2 Point is (4, 2).step 2 — differentiate f′(x) = 2x − 6step 3 — gradient of normal Tangent gradient: f′(4) = 2(4) − 6 = 2 Flip and negate: m_normal = −1/2step 4 — equation y − 2 = −½(x − 4) y = −½x + 2 + 2 y = −½x + 4 classic move — flip 2 to ½, then change sign to get −½!

WE 4

Tangent and normal at the same point

The function f(x) is defined by f(x) = 2x⁴ + 3x², x ≠ 0.

(a) Find the equation of the tangent to y = f(x) at the point where x = 1, in the form y = mx + c.
(b) Find the equation of the normal at the same point, in the form ax + by + d = 0, where a, b, d are integers.

Rewrite first (3/x² = 3x⁻²), then apply the recipe.setup — rewrite & differentiate Rewrite: f(x) = 2x⁴ + 3x⁻² Differentiate: f′(x) = 8x³ − 6x⁻³ Find y at x = 1: y = 2(1) + 3/1 = 5 → point (1, 5) Tangent gradient: f′(1) = 8 − 6 = 2part (a) — tangent y − 5 = 2(x − 1) y = 2x + 3part (b) — normal Flip & negate: m_normal = −1/2 y − 5 = −½(x − 1) y = −½x + ½ + 5 = −½x + ¹¹⁄₂ Multiply by 2 to clear fractions: 2y = −x + 11 x + 2y − 11 = 0 when the form requires integers, multiply through to clear fractions before rearranging!

WE 5

Working backwards: find x given the gradient

For the curve y = x³ − 3x + 2, find the x-coordinates where the tangent is parallel to the line y = 9x − 4.

Parallel lines have the same gradient → set f′(x) equal to the line’s gradient. Differentiate: f′(x) = 3x² − 3 Line y = 9x − 4 has gradient 9. So: 3x² − 3 = 9 3x² = 12 → x² = 4 x = ±2 x = 2 and x = −2 “parallel” → equal gradients · “perpendicular” → product of gradients = −1!

💡 Top tips

Always find the y-coordinate first. Sub the x-value into f(x), not f′(x). The derivative gives gradient, not height.
Show every step. Examiners reward method marks for stating the point, finding f′(x), and substituting clearly — even if you slip up on arithmetic.
“Flip and negate” is the move from tangent gradient to normal gradient. Two operations, in that order.
Use y − y₁ = m(x − x₁) rather than y = mx + c. It’s faster and more direct.
Match the requested form exactly: y = mx + c, or ax + by + d = 0 with integer coefficients. Both come up.
For “parallel” or “perpendicular” lines, set f′(x) equal to the right gradient (same for parallel, negative reciprocal for perpendicular) and solve.
If working with roots or fractions, rewrite as powers of x first before differentiating (just like in the previous note).
Sketch a quick diagram if you’re stuck — it’ll show whether your gradient is positive, negative, or zero.

⚠ Common mistakes

Using f(x) instead of f′(x) when the question wants a gradient. The original function gives heights; the derivative gives gradients.
Forgetting to find the y-coordinate of the point. You need both x₁ AND y₁ for the line equation.
Forgetting to flip OR forgetting to negate when finding the normal’s gradient. Both are required — neither alone is enough.
Using the y-coordinate from the line instead of the curve. Always sub into f(x) for the y-value of the point on the curve.
Not reading the required form. Writing y = mx + c when the question wants ax + by + d = 0 (or vice versa) costs marks.
Leaving fractional coefficients when the question asks for integers. Multiply through to clear them.
Sign errors in the substitution. (−2)² = 4, not −4. (−2)³ = −8, not 8. Take care.
Confusing tangent and normal. Tangent shares gradient with the curve; normal is perpendicular to it.

You’ve now turned the derivative into a working tool — finding gradients, tangents, and normals are the most common Paper 1 calculus questions you’ll see. The next note uses the same idea to figure out where a function is increasing, decreasing, or stationary — which is the gateway to finding maxima and minima.

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