The big idea: A tangent line touches a curve at exactly one point without crossing it.
Its gradient equals the gradient of the curve at that point — found using f′(x).
To find the equation of a straight line you need two things: a gradient and a point on the line.
For a tangent, both of these come from the curve itself.
| What you need | Where it comes from |
|---|---|
| Gradient of the tangent at x = a | Evaluate f′(a) — substitute into the derivative |
| A point on the tangent | The point of tangency: (a, f(a)) — substitute x = a into the original function |
Point-slope form: Once you have the gradient m and the point (x₁, y₁), the tangent equation is:
y − y₁ = m(x − x₁)
You may also be asked to write it in the form y = mx + c — just rearrange.
Interactive graph: tangent line at a point
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The three steps: Step 1 — Differentiate: Find f′(x). Step 2 — Find the gradient: Substitute the given x-value into f′(x) to get a number m. Step 3 — Write the equation: Use y − y₁ = m(x − x₁) with the gradient m and point (x₁, y₁).
Worked example 1
Find the equation of the tangent to y = x² + 3x at the point where x = 1.
Give the answer in the form y = mx + c.
Step by step
- Step 1: Differentiate.
- Step 2: Find the gradient at x = 1.
- Find the y-coordinate of the point of tangency.
- Step 3: Apply point-slope form.
- Rearrange to y = mx + c.
Final answer
y = 5x − 1
Finding y₁: If the question gives you x but not y, always substitute x into the original function f(x) (not the derivative) to find y₁.
The derivative gives gradient, the original gives y-values.
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The big idea: IB sometimes gives you the full coordinates of the point of tangency.
You already have x₁ and y₁ — you still need to differentiate and find m.
Worked example 2
The curve f(x) = 2x³ − 5x passes through the point (2, 6).
Find the equation of the tangent at (2, 6).
Step by step
- Step 1: Differentiate f(x).
- Step 2: Find the gradient at x = 2.
- Step 3: Point is (2, 6), gradient is 19.
- Expand and simplify.
Final answer
y = 19x − 32
Checking the point lies on the curve: If the question says 'the curve passes through (2, 6)', you can verify: f(2) = 2(8) − 5(2) = 16 − 10 = 6 ✓.
This is a quick sanity check and worth doing when you have time.
Worked example 3 — find the gradient, leave in exact form
Find the equation of the tangent to f(x) = x³ − 2x² + 1 at x = −1.
Step by step
- Differentiate.
- Gradient at x = −1.
- y-value at x = −1.
- Point (−1, −2), gradient 7.
- Simplify.
Final answer
y = 7x + 5
The big idea: A harder variation: you are given the gradient of the tangent and asked to find where on the curve the tangent touches.
Set f′(x) = the given gradient and solve for x.
Worked example
The tangent to f(x) = x³ − 3x at a point has gradient 9.
Find the equation of the tangent.
Step by step
- Differentiate.
- Set f′(x) = 9 and solve.
- Two possible tangent points. Take x = 2: find y.
- Tangent at (2, 2) with m = 9.
- For x = −2: f(−2) = −8 + 6 = −2 → point (−2, −2).
Final answer
y = 9x − 16 (at x = 2) and y = 9x + 16 (at x = −2)
IB often wants only one of the tangents: When you get two x-values, check whether the question specifies a positive/negative x or gives some other condition.
If not, present both solutions.