Question

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

Answer 1

`10x-y=4`

Explanation:

Step 1: Find the general slope of `f(x)`

`color(white)("XXX")`Using the Chain Rule:

`color(white)("XXX")(d[(3x-1) * (2x+4)])/(dx)=(d(3x-1))/(dx) * (2x+4)+(3x-1) * (d(2x+4))/(dx)`
`color(white)("XXX")=3(2x+4)+(3x-1)2`
`color(white)("XXX")=6x+12+6x-2`
`color(white)("XXX")=12x+10`

You could alternately (with some effort) multiply the factors of `f(x)` to get
`f(x)=6x^2+10x-8`
and from there, using the Exponent Rule, deduce `(df)/(dx)=12x+10`

Step 2: Find the slope of the tangent at `x=0`
If `(df_x)/(dx)=12x+10`
then `(df_0)/(dx)=12 * 0 +10 =10`

Step 3: Find the `y` coordinate value at `f(x=0)`
`y_0=f(0)=(3 * 0-1) * (2 * 0 +4) =-4`

Step 4: Use the point-slope form to write the equation
With a slope of `10` and the point `(0,-4)`:
`color(white)("XXX")(y-(-4))/(x-0)=10 rarr y+4=10x`

Step 5: Rewrite the equation in standard form
`color(white)("XXX")y+4=10x rarr 10x-y=4`