Question

Consider the function f(x)=x³+x²-6x+1. Find all the points on the graph of f where the tangent line has slope of -1?

Answer 1

Well, if we want to find when the slope of the tangent is -1, let's take a derivative!

`f'(x) = 3x^2 + 2x - 6`

Now `f'(x)` is the slope of the tangent line, so we set it equal to -1:

`-1 = 3x^2 + 2x - 6 implies 3x^2 + 2x -5 = 0`
This is a quadratic, which we can solve. I did it by manually factoring, but you can also use quadratic formula if you'd like.
`3x^2 + 2x - 5 = (3x+5)(x-1)`
So this is solved when `x in {1, -5/3}`

Plugging this into f(x) gives us the positions where the slope is -1:
`f(1) = 1 + 1 - 6 + 1 = -3`
`f(-5/3) = -125/27 + 25/9 + 10 + 1 = 247/27`

So our final answer is:
`(1, -3) and (-5/3, 247/27)`

Answer 2

`(1, -3), (-5/3, -158/27)`

Explanation:

The points at which the tangent line to `f(x)` has a slope of `-1` are the points at which the derivative of our function `f'(x)=-1`.

`f'(x)=3x^2+2x-6` (As per the Power Rule)

Now, let's solve this for `-1:`

`3x^2+2x-6=-1`

`3x^2+2x-5=0`

Using the quadratic formula:

`x=(-2+-sqrt(64))/6=(-2+-8)/6=1, -5/3`

So, at `x=1, -5/3,` the tangent line has a slope of `-1.` To find the coordinates, let's evaluate `f(1), f(-5/3):`

`f(1)=1^3+1^2-6(1)+1=-3`

`(1, -3)`

`f(-5/3)=-125/27+25/9-15/3+1=-125/27+75/27-135/27+27/27=-158/27`