Tangent and Normal Lines from a Derivative: Slope First, Equation Second

At a differentiable point (a, f(a)), the tangent slope is f′(a). A normal is perpendicular to the tangent: when the tangent slope m is finite and nonzero, the normal slope is −1/m.

Find the point as well as the slope

A slope does not determine one unique line without a point. First calculate the curve point from the stated input, then calculate the derivative there. Use the point-slope equation y − y₀ = m(x − x₀). This avoids confusing the derivative value with the y-coordinate.

The normal line is defined geometrically as perpendicular to the tangent. The negative-reciprocal slope rule is convenient only when both slopes are finite and the tangent slope is nonzero. Horizontal and vertical cases need their own line equations.

Worked example

For y = x² + 2x, find tangent and normal lines at x = 1. The point is (1, 3), since 1² + 2 = 3. The derivative is 2x + 2, so the tangent slope at 1 is 4.

The tangent is y − 3 = 4(x − 1), equivalently y = 4x − 1. The normal slope is −1/4, giving y − 3 = −(1/4)(x − 1). Both lines pass through the same curve point. Their slopes multiply to −1, checking perpendicularity.

Check by a local change

Near x = 1, the tangent predicts an output change of about 4h for a small input change h. The actual curve change is f(1 + h) − f(1) = 4h + h². The extra h² becomes small relative to h, which explains why the tangent provides the local linear approximation.

A tangent can cross a curve. It is defined by the local limiting slope, not by touching at exactly one point over the whole graph. For example, the tangent to a cubic at a particular point may intersect that cubic elsewhere.

Horizontal tangent case

At x = 0 on y = x², the tangent slope is zero. The tangent is the horizontal line y = 0, and its normal is the vertical line x = 0. Do not try to express the normal slope as the ordinary number −1/0.

Common mistake

Differentiate before substituting x = a. Substituting too early turns the function into one constant and loses its rate of change. Also check that you used the negative reciprocal, not only a negative sign, for the normal.

Finish by substituting the curve point into both proposed line equations. This simple check catches errors in rearranging the point-slope form and does not require expanding every expression.

Point aur slope alag nikalo; normal mein negative reciprocal lo.

Three practice questions with worked answers

1. Find the tangent to y = x³ at x = 1.

   The point is (1, 1) and derivative is 3x², giving slope 3. The tangent is y − 1 = 3(x − 1), or y = 3x − 2.

2. Find the normal to y = x² at x = 0.

   The point is (0, 0) and tangent slope is 0. Its normal is vertical, so the equation is x = 0.

3. Find the normal to y = 2x² at x = 2.

   The point is (2, 8). The tangent slope is 4x = 8, so the normal is y − 8 = −(1/8)(x − 2).

Related study guides

• Differentiation Formulas Class 12: Complete List with Examples
• Straight Lines Class 11 Notes: All Forms of a Line and Distance Formulas

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Defining the Derivative. No textbook or exam-board affiliation is claimed.