Tangent Line Calculator

Q: How do you find the equation of a tangent line?

Three steps: (1) find the point (a, f(a)); (2) find the slope m = f'(a) by differentiating; (3) plug into point-slope form y - f(a) = m(x - a) and simplify to y = mx + b.

Q: What is the slope of the tangent line?

It is the derivative evaluated at the point, f'(a). It is also the instantaneous rate of change of f at a.

Q: What is the instantaneous rate of change?

The rate at which f changes at one exact instant, which is the derivative f'(a). Compare it with the average rate of change \frac{f(b)-f(a)}{b-a}, which is the slope of a secant line.

Enter a curve and a point. Get the slope, the tangent line equation and the normal line, each step shown, with the tangent laid on the graph like a ruler.

Curve y = f(x)

Reads as

x^{3} - 2 x

Try:

At x =

Tangent line at x = 1

y = x - 2

slope $m = 1$ · point $(1, -1)$

Drag on the graph to move the point. Dashed line = normal.

Step by step

Find the point
Evaluate f at the given x.
$f(1) = -1 \;\Rightarrow\; (1,\, -1)$
Differentiate
The derivative gives the slope at every x.
$f'(x) = 3 x^{2} - 2$
Slope at the point
$m = f'(1) = 1 = 1$
Point-slope form
y − y₁ = m(x − x₁), then solve for y.
$y - -1 = 1\left(x - 1\right) \;\Rightarrow\; y = x - 2$
Normal line
Perpendicular slope is −1/m.
$y = -x$

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The slope of a curve, made concrete

A curve doesn’t have one slope; it has a slope at every point. Lay a ruler against the curve so it just touches at one spot and runs in the same direction: that ruler is the tangent line, and its slope is the derivative $f'(a)$ . This is also called the instantaneous rate of change: if $f(t)$ is position, $f'(a)$ is the speed at time $a$ .

Worked example

Tangent to $y = x^2$ at $x = 3$ : the point is $(3, 9)$ , $f'(x) = 2x$ so $m = 6$ , and $y - 9 = 6(x-3)$ gives $y = 6x - 9$ . The normal line has slope $-\tfrac16$ : $y = -\tfrac16 x + \tfrac{19}{2}$ .

Tangent lines are also the best straight-line approximation of a curve near a point. Put that to work in the linear approximation calculator, or see where the idea comes from on what is a derivative?

Questions students ask

How do you find the equation of a tangent line?

Three steps: (1) find the point $(a, f(a))$ ; (2) find the slope $m = f'(a)$ by differentiating; (3) plug into point-slope form $y - f(a) = m(x - a)$ and simplify to $y = mx + b$ .

What is the slope of the tangent line?

It is the derivative evaluated at the point, $f'(a)$ . It is also the instantaneous rate of change of $f$ at $a$ .

What is the normal line?

The line through the same point perpendicular to the tangent. Its slope is $-\frac{1}{f'(a)}$ . If the tangent is horizontal, the normal is the vertical line $x = a$ .

What is the instantaneous rate of change?

The rate at which $f$ changes at one exact instant, which is the derivative $f'(a)$ . Compare it with the average rate of change $\frac{f(b)-f(a)}{b-a}$ , which is the slope of a secant line.

Can a tangent line cross the curve?

Yes. A tangent line touches the curve at the point of tangency but can cross it elsewhere, and at an inflection point it even crosses right at the point (try x^3 at a = 0).