Question

What is the equation of the tangent line of `y=(x^2-x)^2` at `x=-4`?

Answer 1

`232x+y=-528`

Explanation:

`y=(x^2-x)^2`
`rarr y=x^4-2x^3+x^2`

`y_(x=-4)=((-4)^2-(-4))^2 = (16+4)^2=20^2=400`
giving the point `(-4,400)`

The general slope is given by the derivative:
`(d y)/(d x) = 4x^3-6x+2x`

At `x=-4`, the slope becomes
`(d y_(x=-4))/(d x) = 4xx(-4)^3-6xx(-4)`
`color(white)("XXX")=4xx(-64)+4xx(6)`
`color(white)("XXX")=4xx(-64+6)`
`color(white)("XXX")=4xx(-58)`
`color(white)("XXX")=-232`

Using the slope formula for the tangent
`(y-400)/(x-(-4))=-232`

`y-400 = -232(x+4)`

`y-400 = -232x-928`

`232x+y=-528`