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Equation of the tangent line

Basic Calculus · Derivatives

The Equation of the Tangent Line

Three moves, every time: find the point on the curve, get the slope from the derivative, drop both into point-slope form.

The problem

What is the equation of the tangent to y = x² at the point (3, 9)?

Move one: the point.

x = 3 → y = 3² = 9

the point of tangency: (3, 9)

If the problem only gives x, compute y from the CURVE — the tangent touches where the curve actually is. A tangent "at x = 3" to y = x² has no choice but to pass through (3, 9).

Move two: the slope, from the derivative.

y′ = 2x

the slope machine

m = y′(3) = 6

EVALUATED at the point — m is a number

Writing "y = 2x·x + …" with the un-evaluated derivative as slope is the error to fear: the tangent's slope is 6, not the formula 2x.

Move three: point-slope, then tidy.

y − 9 = 6(x − 3)

point and slope, dropped in

y = 6x − 9

expanded to slope-intercept

When a graph shows the tangent passing through two marked points, read its slope from them instead: through (1, 4) and (3, 0), m = (0 − 4) ÷ (3 − 1) = −2. Then point-slope as usual: y − 4 = −2(x − 1), so y = −2x + 6.

Bonus: the normal line is one flip away.

normal slope = −16

perpendicular: flip and switch the sign

y − 9 = −16(x − 3)

same point, perpendicular direction

The slope is given: find the point first.

Find the tangent to y = x² − x that is parallel to y = 3x + 2. (Same work answers “y = 3x + k is tangent to y = x² − x; find k”.)

parallel → m = 3

perpendicular would be −13

y′ = 2x − 1 = 3 → x = 2

run the slope machine backwards

y = 4 − 2 = 2 → point (2, 2)

height from the curve

y − 2 = 3(x − 2) → y = 3x − 4

so k = −4

Do not reuse the given line's intercept

y = 3x + 2 has the right slope but the wrong place — it never touches the curve at (2, 2). The tangent shares only the slope; its intercept comes from the point.

Variations: a negative x, a slope of 0, a fraction slope.

A negative x on a cubic. Find the tangent to y = x³ − 4x at x = −1.

y = (−1)³ − 4(−1) = −1 + 4 = 3 → (−1, 3)

point from the curve, brackets on −1

y′ = 3x² − 4 → m = 3 − 4 = −1

(−1)² = 1

y − 3 = −1(x + 1) → y = −x + 2

x − (−1) = x + 1

What is different: three sign traps — in y, in the slope, and in x − (−1). Check: at x = −1 the line gives 1 + 2 = 3 ✓.

A slope of 0. Find the tangent and the normal to y = x² + 6x + 4 at x = −3.

y = 9 − 18 + 4 = −5 → (−3, −5)

the vertex of this parabola

y′ = 2x + 6 → m = 0

flat

tangent: y = −5

a horizontal line

normal: x = −3

vertical; −10 has no value

What is different: the flip rule −1m breaks at m = 0. The normal to a flat tangent is straight up and down.

The normal is not y = −3

A perpendicular to a horizontal line is vertical, so its equation is x = (a number). Writing y = −3 gives another horizontal line.

A fraction slope. Find the tangent to y = 6x at x = 3.

y = 63 = 2 → (3, 2)

the point

y = 6x⁻¹ → y′ = −6x² → m = −69 = −23

rewrite as a power first

y − 2 = −⅔(x − 3) → 3y − 6 = −2x + 6

× 3 to clear the fraction

2x + 3y = 12

check: 2(3) + 3(2) = 12 ✓

What is different: the slope is a fraction, so the tidy answer is often given in standard form, not y = mx + b.

Tangents through a point that is not on the curve.

Find the tangents to y = x² that pass through (1, −3).

touch at (a, a²), slope 2a

(1, −3) is below the curve, so it is not the touch point

y − a² = 2a(x − a) → y = 2ax − a²

the tangent at a general point

−3 = 2a(1) − a² → a² − 2a − 3 = 0

make it pass through (1, −3)

(a − 3)(a + 1) = 0 → a = 3 or −1

two touch points

y = 6x − 9 and y = −2x − 1

check at x = 1: −3 and −3 ✓

Using the slope at x = 1 (which is 2) would give a line through (1, −3) that touches nothing. The derivative must be taken where the line touches, not where it passes.

On the exam

The plain item gives x; the harder ones give the slope (parallel, perpendicular, y = mx + k) or a point off the curve. Every version is “point from the curve, slope from f′ at that point”. Check the final line two ways: it passes through the touch point, and its slope equals f′ there.

Check: touches at the point, sits below elsewhere.

at x = 3 line 9, curve 9 they meet ✓

at x = 4 line 15, curve 16 the parabola pulls above — a tangent, not a secant ✓

Remember the flow

Point from the curve · Slope from f′, evaluated · Point-slope form

m is a number. Always a number.

Now you try

Practice problem

The problem

Find the tangent to y = x³ at x = 1.

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