Question

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

Answer 1

`y=10x-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(x)=6x^2 + 10x -4`

now let't take the derivative using the power rule:
`f'(x)=12x + 10`

now we'll plug in `x=0` since that's the point we're interested in.
`f'(0) = 12(0) + 10`
`f'(0) = 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(0) = (-1)(4) = -4`
Our point is `(0,-4)`

Now let's use point-slope form to make an equation of the line.
`y-y_1=m(x-x_1)`
`y-(-4)=10(x-0)`
`y+4=10x`

`y=10x-4`