# For f(x)=(3x-1)(2x+4)  what is the equation of the tangent line at x=0?

Nov 8, 2015

$y = 10 x - 4$

#### Explanation:

The slope of the tangent is found by taking the derivative.
lets rewrite the original equation by FOIL-ing it out
f(x)=(3x−1)(2x+4),
$f \left(x\right) = 6 {x}^{2} + 10 x - 4$

now let't take the derivative using the power rule:
$f ' \left(x\right) = 12 x + 10$

now we'll plug in $x = 0$ since that's the point we're interested in.
$f ' \left(0\right) = 12 \left(0\right) + 10$
$f ' \left(0\right) = 10$

So, the slope of the original function at $x = 0$ is $10$.

To find the tangent line we will need the slope and a point on the line.
The point we will find is the point at $x = 0$ on the original function so...
f(0)=(3(0)−1)(2(0)+4),
$f \left(0\right) = \left(- 1\right) \left(4\right) = - 4$
Our point is $\left(0 , - 4\right)$

Now let's use point-slope form to make an equation of the line.
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 4\right) = 10 \left(x - 0\right)$
$y + 4 = 10 x$

$y = 10 x - 4$