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Tangent lines and normal lines

The derivative gives the slope of a curve at one point. How do you turn that single number into the line itself, and what can that line be used for?

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The power rule and basic differentiation rulesCalculus 1Parallel and perpendicular linesMath 8
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When you reach for it

The derivative at a point defined the tangent line to y = f(x) at x = a as the limiting position of secant lines, with slope f'(a), and the power rule now produces f'(a) for any polynomial or power of x in a line or two. Reach for the two together whenever a question asks for a line attached to a curve: the tangent at a point, the line perpendicular to it, a tangent with a required slope or through a point off the curve, a line tangent to two curves, or a quick estimate of a function value near a known point.

The procedure

The tangent line. A nonvertical line is fixed by one point and its slope. The tangent at x = a passes through (a, f(a)) and has slope f'(a), so point-slope form gives

y - f(a) = f'(a)(x - a).

Both f(a) and f'(a) are numbers, found by putting the fixed input a into f and f'. Solving for y gives the slope-intercept form y = f'(a)\,x + \big(f(a) - a f'(a)\big); the intercept f(a) - a f'(a) is what you compare when asking whether two tangent lines are the same line.

The normal line. The normal line to the curve at (a, f(a)) is the line through that point perpendicular to the tangent. Recall from Parallel and perpendicular lines that parallel nonvertical lines have equal slopes, and from Slopes of perpendicular lines that two nonvertical lines are perpendicular exactly when their slopes multiply to -1. So when f'(a) \ne 0 the normal has slope -\dfrac{1}{f'(a)}:

y - f(a) = -\frac{1}{f'(a)}(x - a).

When f'(a) = 0 the tangent is the horizontal line y = f(a), the formula above would divide by zero, and the normal is the vertical line x = a.

Running it backward. Often the point of tangency is the unknown. Call its input a and write down what the question demands of the tangent there.

  • A tangent of slope m: solve f'(a) = m. A horizontal tangent is the case m = 0, and a tangent parallel or perpendicular to a given line is the case where m is that line’s slope or its negative reciprocal.
  • A tangent through a point (p, q) not on the curve: the tangent at a must be satisfied by x = p, y = q, which gives one equation in a,
q - f(a) = f'(a)(p - a).
  • A line tangent to y = f(x) at x = a and to y = g(x) at x = b: the two tangent lines are the same line exactly when their slopes agree and their intercepts agree, f'(a) = g'(b) and f(a) - a f'(a) = g(b) - b g'(b). That is two equations in the two unknowns a and b.

The best local linear fit. Write the tangent as a function, L(x) = f(a) + f'(a)(x - a). It agrees with f at a, and the gap E(x) = f(x) - L(x) satisfies

\begin{aligned} \frac{E(x)}{x - a} &= \frac{f(x) - f(a)}{x - a} - f'(a) \\ &\to f'(a) - f'(a) = 0 \end{aligned}

as x \to a, because the difference quotient tends to f'(a). So the gap shrinks faster than the distance x - a itself. Any other line through (a, f(a)), say with slope m \ne f'(a), leaves a gap whose ratio to x - a tends to f'(a) - m \ne 0. The tangent is therefore the one line that hugs the curve best near a, and for x near a,

f(x) \approx f(a) + f'(a)(x - a).

This is the linear approximation (or local linearization) of f at a, previewed here and developed in Linear approximation and differentials.

Worked examples

Example 1

Tangent and normal at a point

Find the tangent line and the normal line to f(x) = 2x^3 - x^2 + 3 at x = 1.

Solution

The point: f(1) = 2 - 1 + 3 = 4, so the point is (1, 4). The slope: by the power rule term by term, f'(x) = 6x^2 - 2x, so f'(1) = 6 - 2 = 4.

Tangent: y - 4 = 4(x - 1), which simplifies to y = 4x.

Normal: its slope is the negative reciprocal of 4, namely \displaystyle -\frac{1}{4}, so \displaystyle y - 4 = -\frac{1}{4}(x - 1), that is, \displaystyle y = -\frac{1}{4}x + \frac{17}{4}. Check: the slopes multiply to \displaystyle 4 \cdot \left(-\frac{1}{4}\right) = -1, and both lines give y = 4 at x = 1.

Example 2

Horizontal tangents, vertical normals, and a given slope

Let f(x) = x^3 - 3x^2 - 9x + 4. (a) Find every point where the tangent is horizontal, with the tangent and normal lines there. (b) Find every point where the tangent has slope 15.

Solution

The derivative is f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1).

(a) A horizontal tangent has slope 0, and 3(x - 3)(x + 1) = 0 exactly when x = -1 or x = 3. Then f(-1) = -1 - 3 + 9 + 4 = 9 and f(3) = 27 - 27 - 27 + 4 = -23. The tangent lines are y = 9 at (-1, 9) and y = -23 at (3, -23). A line perpendicular to a horizontal line is vertical, so the normals are x = -1 and x = 3; neither can be written as y = mx + b.

(b) Solve 3x^2 - 6x - 9 = 15: dividing by 3 gives x^2 - 2x - 8 = 0, so (x - 4)(x + 2) = 0 and x = 4 or x = -2. The points are (4, f(4)) = (4, -16) and (-2, f(-2)) = (-2, 2).

Example 3

A tangent parallel to a given line

Find the line tangent to \displaystyle y = \sqrt{x} that is parallel to the line x - 4y = 3.

Solution

Solving the given line for y gives \displaystyle y = \frac{1}{4}x - \frac{3}{4}, slope \displaystyle \frac{1}{4}. Parallel lines have equal slopes, so we need \displaystyle y' = \frac{1}{4}.

Writing \displaystyle \sqrt{x} = x^{1/2}, the power rule gives \displaystyle y' = \frac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}} for x > 0. Setting \displaystyle \dfrac{1}{2\sqrt{x}} = \dfrac{1}{4} gives \displaystyle \sqrt{x} = 2, so x = 4 and the point is (4, 2).

The tangent is \displaystyle y - 2 = \frac{1}{4}(x - 4), or \displaystyle y = \frac{1}{4}x + 1. It is the only such tangent, since \displaystyle \dfrac{1}{2\sqrt{x}} takes each positive value once.

Example 4

Tangent lines through a point off the curve

Find every line through (1, -2) that is tangent to y = x^2 + 1.

Solution

The point is not on the curve, since 1^2 + 1 = 2 \ne -2, so the point of tangency is unknown. Call its input a. There f(a) = a^2 + 1 and f'(a) = 2a, so the tangent is y - (a^2 + 1) = 2a(x - a).

This line must pass through (1, -2). Putting x = 1, y = -2:

\begin{aligned} -2 - (a^2 + 1) &= 2a(1 - a) \\ -a^2 - 3 &= 2a - 2a^2 \\ a^2 - 2a - 3 &= 0, \end{aligned}

so (a - 3)(a + 1) = 0 and a = 3 or a = -1.

At a = 3: point (3, 10), slope 6, line y - 10 = 6(x - 3), or y = 6x - 8. At a = -1: point (-1, 2), slope -2, line y - 2 = -2(x + 1), or y = -2x. Check: at x = 1 these give 6 - 8 = -2 and -2.

−2 2 4 5 10 15 x y (1, −2) (3, 10) (−1, 2)
y = x^2 + 1
 y = 6x − 8 y = −2x
Figure 1From the point (1, -2) below the parabola y = x^2 + 1 exactly two tangent lines can be drawn, touching at x = -1 and x = 3 (Example 4).
Example 5

One line tangent to two curves

Find every line that is tangent to both y = x^2 and y = -x^2 + 6x - 5.

Solution

Let the line touch f(x) = x^2 at x = a and g(x) = -x^2 + 6x - 5 at x = b. With f'(x) = 2x and g'(x) = -2x + 6, the two tangent lines have

  • slopes 2a and -2b + 6,
  • intercepts f(a) - a f'(a) = -a^2 and g(b) - b g'(b) = b^2 - 5.

They are the same line when 2a = -2b + 6 and -a^2 = b^2 - 5. The first gives a = 3 - b. Substituting, -(3 - b)^2 = b^2 - 5, so -9 + 6b - b^2 = b^2 - 5, which rearranges to 2b^2 - 6b + 4 = 0, or (b - 1)(b - 2) = 0.

If b = 1, then a = 2: slope 4, intercept -4, the line y = 4x - 4, touching y = x^2 at (2, 4) and the other parabola at (1, 0). If b = 2, then a = 1: slope 2, intercept -1, the line y = 2x - 1, touching at (1, 1) and (2, 3). There are exactly two such lines.

2 3 4 2 4 6 x y
y = x^2
y = -x^2 + 6x - 5
 y = 4x − 4 y = 2x − 1
Figure 2The two lines tangent to both y = x^2 and y = -x^2 + 6x - 5 (Example 5). Each touches the two curves at different points.
Example 6

Hema’s cube: the tangent line as an estimate

Hema designs a cube with edges of 10 centimeters, but the printed cube comes out with edges of 10.2 centimeters. Use the tangent line to V(s) = s^3 at s = 10 to estimate the printed volume, and compare with the true volume.

Solution

V(10) = 1000 cubic centimeters and V'(s) = 3s^2, so V'(10) = 300 cubic centimeters per centimeter of edge. The tangent line is

L(s) = 1000 + 300(s - 10),

so V(10.2) \approx L(10.2) = 1060 cubic centimeters. The true volume is 10.2^3 = 1061.208 cubic centimeters, so the estimate is short by 1.208.

The gap has an exact form. Since s^3 - 1000 = (s - 10)(s^2 + 10s + 100),

\begin{aligned} V(s) - L(s) &= (s - 10)(s^2 + 10s - 200) \\ &= (s - 10)^2(s + 20). \end{aligned}

At s = 10.2 this is (0.2)^2(30.2) = 1.208, as found. Because of the factor (s - 10)^2, the gap divided by s - 10 tends to 0, and halving the change in edge roughly quarters the error. For s > -20, s \ne 10, the gap is positive, so the estimate is low.

Example 7

Where the normal line meets the curve again

Find the tangent and the normal to y = \dfrac{x^2}{4} at (2, 1), and find the other point where the normal crosses the curve.

Solution

y' = \dfrac{x}{2}, so the slope at x = 2 is 1. Tangent: y - 1 = 1(x - 2), or y = x - 1. Normal: slope -1, so y - 1 = -(x - 2), or y = -x + 3.

The normal meets the curve where \dfrac{x^2}{4} = -x + 3, that is, x^2 + 4x - 12 = 0, or (x + 6)(x - 2) = 0. The root x = 2 is the starting point; the other is x = -6, where \displaystyle y = \frac{36}{4} = 9. The normal crosses the parabola again at (-6, 9).

−6 −4 −2 2 2 6 8 10 x y (2, 1) (−6, 9) tangent normal
y = \dfrac{x^2}{4}
Figure 3The tangent and the normal to \displaystyle y = \frac{x^2}{4} at (2, 1). The normal crosses the curve again at (-6, 9) (Example 7).

Boundaries

  • The derivative must exist at a. A graph with a corner or a cusp, such as y = x^{2/3} at x = 0, has no tangent line there. A graph can also have a vertical tangent, as \displaystyle y = \sqrt[3]{x} does at the origin: the secant slopes grow without bound, f'(0) does not exist, and the tangent is the line x = 0, which no equation y - f(a) = m(x - a) can express. Its normal is the horizontal line y = 0.
  • The normal-slope formula needs f'(a) \ne 0. At a horizontal tangent the normal is vertical, x = a.
  • “Touches once” is not the definition. A tangent line may cross the curve at the point of tangency (y = 0 is tangent to y = x^3 at the origin and crosses it there) and may meet the curve again far away. Tangency is a statement about slope at one point.
  • The linear fit is only local. The error in Example 6 is (s - 10)^2(s + 20), small near s = 10 and large far from it.

Where this goes next

Linear approximation and differentials uses the linear approximation to turn a small error in a measured input into an estimated error in a computed output. Newton's method uses tangent lines over and over: the point where a tangent crosses the x-axis is usually closer to a root than the point it started from. Curves that are not graphs of functions, such as circles, get tangent lines through Implicit differentiation, and curves traced by a moving point through Derivatives of parametric curves.

Exercises

Tangent and normal lines at a point

  1. Find the tangent line to f(x) = x^3 - 2x^2 + 5 at x = 2.
  2. Find the normal line to the same f(x) = x^3 - 2x^2 + 5 at x = 2.
  3. Find the tangent line and the normal line to g(x) = \dfrac{16}{x^2} at x = 2.
  4. Find the tangent line and the normal line to h(x) = x^4 - 2x^2 at x = 1.
  5. Find the tangent line and the normal line to \displaystyle y = 2\sqrt{x} at x = 9.

Where the slope is given

  1. Find every point where f(x) = 2x^3 + 3x^2 - 12x + 1 has a horizontal tangent.
  2. Find the tangent lines to y = x^3 - 9x + 1 that have slope 3.
  3. Find the line tangent to y = x^2 - 5x + 2 that is parallel to y = 3x + 1.
  4. Find the line tangent to \displaystyle y = 4\sqrt{x} that is perpendicular to the line 2x + y = 5.
  5. At what point of y = x^2 is the normal line parallel to \displaystyle y = \frac{1}{4}x? Give that normal line.
  6. Find the least slope of any tangent line to y = x^3 + 2x - 7, and the point where it occurs. Explain why the graph has no horizontal tangent.

Outside points, shared tangents, unknown coefficients

  1. Find every line through (2, 3) that is tangent to y = x^2.
  2. Show that no line through (1, 2) is tangent to y = x^2, and say where (1, 2) sits relative to the parabola.
  3. Find every line through (2, 0) that is tangent to y = x^3.
  4. Find the line tangent to both y = x^2 + 2 and y = x^2 - 4x.
  5. Find the number c for which the line y = 5x + c is tangent to y = x^2 + x.
  6. Find a and b so that the line y = 3x - 4 is tangent to f(x) = ax^2 + bx at x = 2.

The tangent line as a linear fit

  1. Ninad wants \displaystyle \sqrt{102} without a calculator. Use the tangent line to \displaystyle f(x) = \sqrt{x} at x = 100 to estimate it, and decide whether the estimate is too high or too low.
  2. Srinivas cuts a square tile meant to be 6 centimeters on a side, and it comes out 6.1 centimeters. Use the tangent line to A(s) = s^2 at s = 6 to estimate the tile’s area, and find the exact error.

Reasoning

  1. Show that for f(x) = px^2 + qx + r and its tangent line L at x = a, f(x) - L(x) = p(x - a)^2. What does this say about which side of its tangent lines a parabola lies on?
  2. Show that the tangent to y = x^3 at x = 1 meets the curve at a second point, and find it.
  3. Explain why the normal line at a point where f'(a) = 0 cannot be written in the form y = mx + b, and write its equation.
  4. Show that every tangent line to y = \dfrac{1}{x} at a point with x > 0 forms, with the two coordinate axes, a triangle of area 2.
  5. For y = x^3 and the line through (1, 1) with slope m, show that the gap between curve and line, divided by x - 1, tends to 3 - m as x \to 1. Explain how this singles out the tangent line.
Answers
  1. y = 4x - 3. f(2) = 8 - 8 + 5 = 5; f'(x) = 3x^2 - 4x, so f'(2) = 4, and y - 5 = 4(x - 2).
  2. \displaystyle y = -\frac{1}{4}x + \frac{11}{2}. The slope is \displaystyle -\frac{1}{4}, so \displaystyle y - 5 = -\frac{1}{4}(x - 2).
  3. Tangent y = -4x + 12; normal \displaystyle y = \frac{1}{4}x + \frac{7}{2}. g(2) = 4 and g'(x) = -32x^{-3}, so g'(2) = -4.
  4. Tangent y = -1; normal x = 1. h(1) = -1 and h'(x) = 4x^3 - 4x gives h'(1) = 0, a horizontal tangent, so the normal is vertical.
  5. Tangent \displaystyle y = \frac{1}{3}x + 3; normal y = -3x + 33. y(9) = 6 and y' = x^{-1/2}, so the slope is \displaystyle \frac{1}{3} and the normal slope is -3.
  6. (-2, 21) and (1, -6). f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1).
  7. y = 3x - 15 and y = 3x + 17. 3x^2 - 9 = 3 gives x = \pm 2; the points are (2, -9) and (-2, 11).
  8. y = 3x - 14. 2x - 5 = 3 gives x = 4 and the point (4, -2).
  9. \displaystyle y = \frac{1}{2}x + 8. The line has slope -2, so we need slope \displaystyle \frac{1}{2}: \displaystyle \frac{2}{\sqrt{x}} = \frac{1}{2} gives x = 16 and the point (16, 16).
  10. At (-2, 4); normal \displaystyle y = \frac{1}{4}x + \frac{9}{2}. The normal slope at a is \displaystyle -\frac{1}{2a}, and \displaystyle -\frac{1}{2a} = \frac{1}{4} gives a = -2.
  11. Least slope 2, at (0, -7). y' = 3x^2 + 2 \ge 2 with equality only at x = 0; since y' is never 0, no tangent is horizontal.
  12. y = 2x - 1 and y = 6x - 9. The tangent at a is y = 2ax - a^2; through (2, 3): a^2 - 4a + 3 = 0, so a = 1 or a = 3.
  13. Through (1, 2) the tangent y = 2ax - a^2 needs a^2 - 2a + 2 = 0, whose discriminant is 4 - 8 = -4 < 0: no real a. The point is above the parabola, inside its cup, since 2 > 1^2.
  14. y = 0 and y = 27x - 54. The tangent at a is y = 3a^2x - 2a^3; through (2, 0): 6a^2 - 2a^3 = 0, so a = 0 or a = 3.
  15. \displaystyle y = -3x - \frac{1}{4}. Tangent to the first at a: slope 2a, intercept 2 - a^2. Tangent to the second at b: slope 2b - 4, intercept -b^2. Then a = b - 2 and 2 - (b - 2)^2 = -b^2 give \displaystyle b = \frac{1}{2}, \displaystyle a = -\frac{3}{2}.
  16. c = -4. The slope 2x + 1 = 5 at x = 2, where y = 6; then 6 = 10 + c.
  17. a = 1, b = -1. The line gives f(2) = 2 and f'(2) = 3: 4a + 2b = 2 and 4a + b = 3.
  18. \displaystyle \sqrt{102} \approx 10.1, too high. \displaystyle L(x) = 10 + \frac{1}{20}(x - 100), so L(102) = 10.1, while \displaystyle \sqrt{102} \approx 10.0995; indeed 10.1^2 = 102.01 > 102.
  19. About 37.2 square centimeters; the exact area is 37.21, so the error is 0.01. L(s) = 36 + 12(s - 6), and the gap is (s - 6)^2 = 0.01.
  20. L(x) = pa^2 + qa + r + (2pa + q)(x - a). Subtracting and expanding, the constant and linear parts cancel and leave p(x - a)^2. If p > 0 the parabola lies above every tangent line, touching only at the point of tangency; if p < 0 it lies below.
  21. The tangent is y = 3x - 2. Then x^3 - 3x + 2 = (x - 1)^2(x + 2) = 0 also at x = -2, so the line meets the curve again at (-2, -8).
  22. The normal is perpendicular to the horizontal tangent y = f(a), so it is vertical, and a vertical line has no slope m. Its equation is x = a.
  23. At \displaystyle (a, \frac{1}{a}) the slope is \displaystyle -\frac{1}{a^2}, so the tangent is \displaystyle y = -\frac{1}{a^2}x + \frac{2}{a}. Its intercepts are x = 2a and \displaystyle y = \frac{2}{a}, and the area is \displaystyle \frac{1}{2}(2a)\left(\frac{2}{a}\right) = 2 for every a > 0.
  24. The gap is x^3 - 1 - m(x - 1) = (x - 1)(x^2 + x + 1 - m), so the ratio is x^2 + x + 1 - m \to 3 - m. It is 0 only for m = 3 = f'(1), the tangent’s slope: every other line through (1, 1) leaves a gap comparable to x - 1 itself.
kg1128 · updated 2026-10-21 · CC-BY-SA-4.0