Tangent and Normal Lines to a Graph
The tangent line to the graph of f at the point (a, f(a)) is the line through that point with slope f′(a), and the normal line is the line through the same point perpendicular to the tangent. In brief: for f(x) = x2 at a = 3: f(3) = 9 and f′(3) = 6, so the tangent is y − 9 = 6(x − 3), i.e. y = 6x − 9, and the normal has the slope −1/6: y − 9 = −(1/6)(x − 3).
The tangent line is the best linear approximation of the function near the point of tangency. It is used to estimate values, to describe the direction of a curve and to solve geometric problems about slopes, angles between curves and distances from a point to a curve. This guide gives the equations, shows how to find tangents at a given point, from a given slope and from an external point, how to find the angle between curves and the linear approximation, with a figure and a table of examples.
What are the equations of the tangent and the normal?
If f is differentiable at a, the tangent line at (a, f(a)) is
y = f(a) + f′(a)(x − a),
which is the point-slope form with the slope m = f′(a). The normal line is perpendicular to the tangent, so its slope is −1/f′(a) (if f′(a) ≠ 0):
y = f(a) − (x − a)/f′(a).
If f′(a) = 0, the tangent is horizontal, y = f(a), and the normal is vertical, x = a. If the tangent is vertical (infinite slope), the normal is horizontal.
What does a tangent line look like?
Near the point of tangency, the curve and the tangent are very close: the tangent is the line that “best fits” the graph locally.
How do you solve typical problems?
Case 1: tangent at a given point
f(x) = x3 − 2x + 1 at x = 2. f(2) = 8 − 4 + 1 = 5 and f′(x) = 3x2 − 2, so f′(2) = 10. Tangent: y − 5 = 10(x − 2), i.e., y = 10x − 15. Normal: y − 5 = −(1/10)(x − 2), i.e. y = −0.1x + 5.2.
Case 2: tangent with a given slope
Where does y = x2 − 4x + 5 have a tangent parallel to y = 2x + 7? Solve f′(x) = 2x − 4 = 2: x = 3, f(3) = 2. The tangent: y − 2 = 2(x − 3), i.e., y = 2x − 4. A horizontal tangent: f′(x) = 0, x = 2, the vertex (2, 1).
Case 3: tangent from an external point
Find the tangent lines to y = x2 through the point P(1, −3), which is not on the parabola. Let the point of tangency be (t, t2). The tangent there: y = 2t(x − t) + t2 = 2tx − t2. It passes through P: −3 = 2t − t2, so t2 − 2t − 3 = 0, (t − 3)(t + 1) = 0, giving t = 3 or t = −1. The tangents: y = 6x − 9 and y = −2x − 1. Check that P lies on both: 6 − 9 = −3 ✓ and −2 − 1 = −3 ✓.
Case 4: implicit curves
The circle x2 + y2 = 25 at (3, 4). Implicit differentiation: y′ = −x/y = −3/4. The tangent: y − 4 = −(3/4)(x − 3), or 3x + 4y = 25. The normal has the slope 4/3 and passes through the center (0, 0): y = (4/3)x ✓.
Case 5: the common tangent point
When are y = x2 and y = −x2 + 4x − 2 tangent to each other? At a point where the values and the slopes agree: x2 = −x2 + 4x − 2 gives 2x2 − 4x + 2 = 0, so x = 1 (a double root), and the slopes are 2 and −2 + 4 = 2 ✓. They touch at (1, 1), and the common tangent is y = 2x − 1.
How is the tangent line used for approximation?
Near a, f(x) ≈ L(x) = f(a) + f′(a)(x − a), the linearization of f at a. The approximation is good when x is close to a.
Example 6
√(4.1): use f(x) = √x at a = 4. f(4) = 2, f′(4) = 1/(2 · 2) = 0.25. L(4.1) = 2 + 0.25 · 0.1 = 2.025. The exact value is 2.02485…, so the error is about 0.00015.
Example 7
(1.02)10 with f(x) = x10 at 1: f′(1) = 10, so ≈ 1 + 10 · 0.02 = 1.2. The exact value is 1.21899, so the linear estimate is rough because the step 0.02 times the exponent is not small enough.
Example 8: small angles
sin x ≈ x near 0 (tangent at 0: slope cos 0 = 1), and ex ≈ 1 + x, ln(1 + x) ≈ x. For x = 0.05: sin = 0.04998, e0.05 = 1.05127, ln(1.05) = 0.04879.
How do you find the angle between two curves?
The angle between two curves at an intersection point is the angle between their tangent lines. If the slopes are m1 and m2, then tan θ = |(m2 − m1)/(1 + m1m2)|. The curves are perpendicular if m1m2 = −1.
Example 9
y = x2 and y = x intersect at (1, 1). The slopes: 2 and 1. tan θ = |(1 − 2)/(1 + 2)| = 1/3, so θ ≈ 18.43°. And y = x2 and y = 2 − x2 intersect at (1, 1) with slopes 2 and −2: tan θ = |(−2 − 2)/(1 − 4)| = 4/3, θ ≈ 53.13°.
Table of tangent and normal lines
| Function and point | f′(a) | Tangent | Normal |
|---|---|---|---|
| x2 at x = 3 | 6 | y = 6x − 9 | y = −x/6 + 9.5 |
| x3 − 2x + 1 at x = 2 | 10 | y = 10x − 15 | y = −0.1x + 5.2 |
| √x at x = 4 | 1/4 | y = x/4 + 1 | y = −4x + 18 |
| 1/x at x = 2 | −1/4 | y = −x/4 + 1 | y = 4x − 7.5 |
| ex at x = 0 | 1 | y = x + 1 | y = −x + 1 |
| sin x at x = 0 | 1 | y = x | y = −x |
Where does the tangent problem come from?
The tangent to a circle was studied by Euclid. The tangent problem for general curves was one of the two central problems that led to calculus: Descartes and Fermat in the 1630s developed algebraic methods for finding tangents, Barrow found the connection between tangents and areas, and Newton and Leibniz generalised the methods into the derivative. The word tangent is from Latin tangere, “to touch”.
Where is it used?
Optics uses the normal line to a surface to describe the reflection and refraction of light: the angle of incidence equals the angle of reflection, measured from the normal. In engineering, the tangent to a road or a rail curve gives the direction of travel, and the normal gives the direction of the centripetal force. Newton’s method for solving f(x) = 0 repeatedly intersects the tangent line with the x-axis. In computer graphics, normal vectors determine the shading of surfaces.
Which mistakes are common?
Using f′(x) instead of f′(a) for the slope
The slope of the tangent at a specific point is a number, f′(a), not a function of x.
Forgetting the point on the curve
The tangent line passes through (a, f(a)). Compute f(a) as well.
Wrong slope of the normal
The normal slope is −1/f′(a), the negative reciprocal. For f′(a) = 6, it is −1/6, not −6 or 1/6.
Assuming that a tangent touches the curve only once
A tangent can cross the curve elsewhere: the tangent to y = x3 at x = 1 is y = 3x − 2, and it meets the curve again at x = −2, since x3 − 3x + 2 = (x − 1)2(x + 2).