Question

Let f(x)=(x-6)(4x+2) find an equation for the tangent line to the graph of x=12?

Answer 1

`y = 74x - 660`

Explanation:

The equation of a tangent line at a specific point with respect to a given function is called the derivative.

We start with:
`f(x) = (x-6)(4x+2)`
`f(x) = 4x^2-24x+2x-12`
`f(x) = 4x^2 -22x-12`

Taking the derivative:
`f'(x) = 8x - 22`

Assessing at `x=12`:
`f'(12) = 8(12)-22 = 74`

This means the slope of the tangent line is 74.

Now we plug in `x=12` into the original `f(x)`:
`f(12) = 4(12)^2 - 22(12)-12 = 300`

This means the tangent line touches `f(x)` at the point `(12,300)`.

The equation of a line is:
`(y - y_0) = m (x-x_0)`
`(y - 300) = 74 (x - 12)`
`y - 300 = 74x - 960`
`y = 74x - 660`

Hence, the tangent line equation is:
`y = 74x - 660`

Answer 2

`74x-588=y`

Explanation:

Remember the product rule:

If `f(x)=g(x)*h(x)`, then `f'(x)=g'(x)*h(x)+g(x)*h'(x)`

Therefore,

`d/dx[(x-6)(4x+2)]=d/dx(x-6)*(4x+2)+(x-6)*d/dx(4x+2)`

`=>d/dx(x-6)*(4x+2)+(x-6)*d/dx(4x+2)`

We use the power rule:

`x^n=nx^(n-1)` where `n` is a constant.

When you are adding/subtracting a constant from a variable, then you could just eliminate it.

`=>1*x^(1-1)*(4x+2)+(x-6)*4*1x^(1-1)`

`=>(4x+2)+(x-6)*4`

`=>(4x+2)+(4x-24)`

`=>8x-22`

The slope of the tangent line is:

`f'(12)=8*12-22`

`f'(12)=74`

We also find `f(x)`

`f(12)=(12-6)(12*4+2)`

`f(12)=(6)(50)`

`f(12)=300`

We now use the slope-intercept form: `m(x-x_1)=y-y_1`

`74(x-12)=y-300`

`74x-888=y-300`

`74x-588=y`