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

Dec 24, 2015

$y = 4 x - 2$

#### Explanation:

Step 1: Find derivative of the equation

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + \frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 x\right) + \frac{d}{\mathrm{dx}} \left(2\right)$

$f ' \left(x\right) = 3 {x}^{2} + 4 x - 3$

Step 2: Find the slope of the tangent line at $x = 1$
$m = f ' \left(1\right) = 3 \left({1}^{2}\right) + 4 \left(1\right) - 3 = 4$

Step 3: Find the $y$ coordinate of the function when $x = 1$

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

$f \left(1\right) = 2$

So, the original point of the graph is $\left(1 , 2\right)$

Step 4: Find the equation of the tangent line using point slope formula

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

$m = 4 \text{ " " } \left(1 , 2\right)$

$y - 2 = 4 \left(x - 1\right)$
$\implies y - 2 = 4 x - 4$
$\implies y = 4 x - 4 + 2$
$\implies y = 4 x - 2$