Equation of a Tangent Line

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.

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.