# Thread: Dumb tangent line question

1. ## Dumb tangent line question

f(x) = x^3 - 3x^2 - x +3

I need to find the slope of the tangent line at x=2 using lim as x->2
(f(x) - f(2)) / (x-2)

I cannot for the life of me factor the equation in order to cancel out the x - 2. I feel pretty dumb, but I'm tired, and I need to do this for a write-up.

2. $f(x) - f(2) =$

$(x^3 - 3x^2 - x +3) - (-3) =$

$x^3 - 3x^2 - x + 6$

divide the cubic polynomial by $(x - 2)$ using long division or synthetic division, you will find that it factors as ...

$(x-2)(x^2-x-3)$

$\lim_{x \to 2} (x^2 - x - 3) = ?$

3. Originally Posted by alexmcentee
f(x) = x^3 - 3x^2 - x +3

I need to find the slope of the tangent line at x=2 using lim as x->2
(f(x) - f(2)) / (x-2)

I cannot for the life of me factor the equation in order to cancel out the x - 2. I feel pretty dumb, but I'm tired, and I need to do this for a write-up.
Using this definition, we see that $\lim_{x\to 2}\frac{x^3-3x^2-x+3-(-3)}{x-2}=\lim_{x\to 2}\frac{x^3-3x^2-x+6}{x-2}$

But $x^3-3x^2-x+6=(x-2)(x^2-x-3)$. You can use synthetic division with the root being 2, since x-2=0 => x=2 to get this factorized form.

Thus, $\lim_{x\to 2}\frac{x^3-3x^2-x+6}{x-2}=\lim_{x\to 2}x^2-x-3=\dots$

Does this make sense?

--Chris