# 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

### 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 1

### Solution

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) = x3 - 4x2 + 7

Then f'(x) = 3x2 - 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 2

### Solution

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) = ex

Then f'(x) = ex.

Derivative of ex

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.