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?

2 Answers
Mar 22, 2018

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)

Mar 22, 2018

(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