# Equation of a Tangent Line

How to find the equation of the tangent line to y = f(x) at a point: formula, 2 examples, and their solutions.

## Formula

The equation of the tangent line to y = f(x)

at (a, f(a)) is

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

Recall that

the slope of the tangent line to y = f(x) at x = a is f'(a).

And the tangent point on y = f(x) at x = a is (a, f(a)).

Then the linear equation of the tangent line

in point-slope form is

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

## Example

Find f(1)

by putting x = 1 into f(x).

Then f(1) = 4.

So the tangent point is (1, 4).

To find f'(1),

find f'(x).

f(x) = x^{3} - 4x^{2} + 7

Then f'(x) = 3x^{2} - 8x.

Derivative of a Polynomial

Put x = 1 into f'(x).

Then f'(1) = -5.

This is the slope of the tangent line.

The tangent point is (1, 4).

And the slope of the tangent line is f'(1) = -5.

Then the linear equation of the tangent line

in point-slope form is

y = -5(x - 1) + 4.

Change y = -5(x - 1) + 4 to slope-intercept form.

Then y = -5x + 9.

So

y = -5x + 9

is the tangent line.

## Example

Find f(0)

by putting x = 0 into f(x).

Then f(0) = 1.

So the tangent point is (0, 1).

To find f'(0),

find f'(x).

f(x) = e^{x}

Then f'(x) = e^{x}.

Derivative of e^{x}

Put x = 0 into f'(x).

Then f'(0) = 1.

This is the slope of the tangent line.

The tangent point is (0, 1).

And the slope of the tangent line is f'(0) = 1.

Then the linear equation of the tangent line

in point-slope form is

y = 1(x - 0) + 1.

y = 1(x - 0) + 1 is

y = x + 1.

So

y = x + 1

is the tangent line.