Gradient from f', point from f: The tangent at x = a is the straight line touching the curve there.
Its gradient is f'(a) and it passes through (a, f(a)).
Build it with y − y₁ = m(x − x₁).
A tangent touching y = x²: its gradient is f'(a) and it passes through the point of contact.
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You need a gradient AND a point: Find f'(a) for the gradient and f(a) for the y-coordinate — a line needs both.
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Gradient, point, then the line: Method: (1) differentiate and find f'(a) for the gradient; (2) find f(a) for the point; (3) put them into y − y₁ = m(x − x₁) and simplify.
IB-style question — tangent equation
Find the equation of the tangent to y = x² − 3x + 2 at the point where x = 3.
Step by step
- Gradient: f'(x) = 2x − 3, so f'(3) = 3.
- Point: f(3) = 9 − 9 + 2 = 2. Use y − y₁ = m(x − x₁).
Final answer
y = 3x − 7.
Don't forget the y-coordinate: Substitute x = a into the original f(x) for y₁ — not into f'(x).
IB-style question — reading from a given tangent
The line y = 5x − 2 is the tangent to the curve y = f(x) at the point where x = 1.
Write down (a) f(1) and (b) f′(1).
Step by step
- (a) The tangent touches the curve at x = 1, so f(1) equals the line's value there.
- (b) f′(1) is the gradient of the curve at that point = the gradient of the tangent line.
Final answer
(a) f(1) = 3. (b) f′(1) = 5.
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Match the gradient to the condition: A horizontal tangent has gradient 0 → solve f'(x) = 0.
A tangent parallel to a line shares that line's gradient → solve f'(x) = (that gradient).
IB-style question — horizontal tangents
Find the equations of the horizontal tangents to y = x³ − 12x.
Step by step
- Horizontal → f'(x) = 0.
- y-values: f(2) = −16, f(−2) = 16.
Final answer
The horizontal tangents are y = −16 (at x = 2) and y = 16 (at x = −2).
Horizontal tangent → y = constant: A horizontal tangent has the form y = (a number) — just the y-coordinate of the point.