← Differentiation methods

5 Tangents, normals and curve features

Tangent and normal lines

  • Tangent at x = a: gradient m = f′(a); point (a, f(a)); equation y − f(a) = m(x − a).
  • Normal at the same point: perpendicular to the tangent, gradient = −1/m (when m ≠ 0).
  • If f′(a) = 0, the tangent is horizontal (y = f(a)) and the normal is vertical (x = a).
  • Two lines are perpendicular exactly when their gradients multiply to −1 — a quick check on any tangent/normal pair.
Worked example Equation of a tangent, and its x-intercept Find the equation of the tangent to y = x³ − 4x + 1 at x = 2, and the point where it crosses the x-axis. y′ = 3x² − 4. At x = 2: y′ = 3(4) − 4 = 8 (the gradient). y(2) = 8 − 8 + 1 = 1, so the point is (2, 1). Tangent: y − 1 = 8(x − 2) ⟹ y = 8x − 15. x-intercept: 0 = 8x − 15 ⟹ x = 15/8. ✓ Tangent: y = 8x − 15; crosses the x-axis at x = 15/8 (= 1.875).

Finding maxima / minima

  • Turning points where dy/dx = 0.
  • Second-derivative test: d²y/dx² < 0 ⟹ maximum; > 0 ⟹ minimum.
  • Increasing where dy/dx > 0; decreasing where dy/dx < 0.
  • Points of inflection where d²y/dx² = 0 and concavity changes.

How many stationary points? Ask the discriminant

  • When dy/dx = 0 comes down to a quadratic, ax² + bx + c = 0, the number of stationary points is the number of real roots.
  • b² − 4ac > 0: two. = 0: exactly one. < 0: none.
  • So “the curve has a single stationary point” with an unknown constant k is an equation: set b² − 4ac = 0 and solve for k, then put k back to find x.
  • Factors like e3x or a squared denominator are never zero, so only the quadratic in the numerator decides.
✕

f″(x) = 0 does not by itself prove an inflection

If f″ = 0 at a stationary point, the second-derivative test is inconclusive — you must check the sign of f′ either side, or check that f″ actually CHANGES SIGN. y = x⁴ at x = 0 has f′(0) = 0 and f″(0) = 0, yet it is a genuine minimum, because f″ never changes sign there. "f″ = 0 therefore inflection" is the single most common wrong conclusion in this section — fall back to a sign table when the test fails; it always works. Worked example When the second-derivative test fails Find and classify the stationary point(s) of y = x⁴ − 4x³, given that the second-derivative test is inconclusive at one of them. y′ = 4x³ − 12x² = 4x²(x − 3). Setting y′ = 0 gives x = 0 (double root) and x = 3. y″ = 12x² − 24x. At x = 3: y″ = 108 − 72 = 36 > 0, so x = 3 is a minimum. At x = 0: y″ = 0 — the second-derivative test is inconclusive, so fall back to a sign table for y′. y′ just left of 0 (x = −1): 4(1)(−4) = −16, negative. y′ just right of 0 (x = 1): 4(1)(−2) = −8, negative. Same sign both sides. ✓ x = 3 is a minimum (y″ > 0). x = 0 is a stationary point of inflection — y′ does not change sign, so it is neither a maximum nor a minimum.

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