Question

How do you find the equation of the tangent line to the curve `y = (3x-1)(2x+4)` at the point of (0,-4) ?

Answer 1

`y=10x-4`

Explanation:

`y=6x^2+10x-4`-> FOIL or multiply it out first
`y'=12x+10`-> find the derivative
`y'(0)=10`->Put in 0 for x to find slope
`y+4=10(x-0)` ->Put the slope `m` and the point`(x_1,y_1)` into the point slope form of an equation of a line
`y=10x-4`-Write the answer in slope-intercept form

Note you can start by finding the derivative using the product rule whatever is easier for you go with it.