Tangents and normals

y − y₁ = m(x − x₁), where m is the gradient and (x₁, y₁) is the point of tangency.

1. Differentiate f(x) to get f′(x).\n2. Substitute x₁ into f′(x) to get the gradient m.\n3. Write y − y₁ = m(x − x₁) and simplify.

dy/dx = 2x. At x = 3: m = 6.

dy/dx = 2x → m = 4. y₁ = 5. Tangent: y − 5 = 4(x − 2) → y = 4x − 3.

Because f(x) gives y-values (heights). f′(x) gives gradients. You need the y-coordinate of the point of tangency — that comes from the original function.

Set f′(x) = given gradient and solve for x. There may be one or two solutions. Find y at each solution using f(x).

f′(x) = 3x². m = 3. f(−1) = −1 → point (−1, −1). Tangent: y + 1 = 3(x + 1) → y = 3x + 2.

The tangent is the best linear approximation to the curve at that point. It has exactly the same gradient as the curve at that point — but the curve will curve away from the tangent for x-values further away.

m_tangent × m_normal = −1, so m_normal = −1/m_tangent. The normal is perpendicular to the tangent.

m_n = −1/5.

m_n = −1/(−3) = 1/3. Two negatives cancel.

dy/dx = 2x − 2. m_t = 4. m_n = −1/4.

dy/dx = 2x → m_t = 6 → m_n = −1/6. Normal: y − 9 = −(1/6)(x − 3) → y = −(1/6)x + 19/2.

The normal is vertical: a line of the form x = x₁. You cannot divide −1 by zero.

True. Both lines pass through the point of tangency (x₁, y₁). They differ only in their gradients.

Using the tangent gradient (from f′) directly as the normal gradient, without applying m_n = −1/m_t. Always take the negative reciprocal.