# What is the equation of the tangent line of f(x) = (x^3 - 3x + 1)(x + 2) at the given point of (1, -3) ?

Oct 5, 2016

$y = - x - 2$

#### Explanation:

Find the derivative of the function.

Begin by using the Product Rule.

$f ' \left(x\right) = u v ' + u ' v$

$u = {x}^{3} - 3 x + 1$
$u ' = 3 {x}^{2} - 3$

$v = x + 2$
$v ' = 1$

$f ' \left(x\right) = \left({x}^{3} - 3 x + 1\right) \left(1\right) + \left(3 {x}^{2} - 3\right) \left(x + 2\right)$

$f ' \left(x\right) = \left({x}^{3} - 3 x + 1\right) + \left(3 {x}^{2} - 3\right) \left(x + 2\right)$

Substitute in the value of $x$ to get a numerical value for the slope, $m$.

$f ' \left(1\right) = \left({\left(1\right)}^{3} - 3 \left(1\right) + 1\right) + \left(3 {\left(1\right)}^{2} - 3\right) \left(\left(1\right) + 2\right)$

Simplify

$f ' \left(1\right) = \left(1 - 3 + 1\right) + \left(3 - 3\right) \left(1 + 2\right)$

$f ' \left(1\right) = - 1 + \left(0\right) \left(3\right) = - 1 + 0 = - 1$

The slope, $m$, is $- 1$.

Substitute in the point $\left(1 , - 3\right)$ and the slope, $m$, $- 1$, into the slope intercept formula, $y = m x + b$.

$- 3 = - 1 \left(1\right) + b$

Simplify

$- 3 = - 1 + b$

Add $1$ to both sides of the equation to isolate $b$.

$- 2 = b$

Write the equation of the tangent line using the above information.

$y = - 1 x - 2$

$\mathmr{and}$

$y = - x - 2$

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