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

Use the product rule to compute the first derivative:

`dy/dx = (d(3x-1))/dx(2x+4) + (3x-1)(d(2x+4))/dx`

`dy/dx = 3(2x+4) + 2(3x-1)`

`dy/dx = 6x+12+6x-2`

`dy/dx= 12x+10`

The slope of the tangent line, m, is the first derivative evaluated at `x = 0`:

`m = 12(0)+10`

`m = 10`

Because the given point, `(0,-4)`, is the y intercept, one can use the slope-intercept form to write the equation of the tangent line:

`y = mx + b`

`y = 10x -4`