Find an equation of the tangent line to the curve $y = x\cos(x)$ at the point $(\pi, -\pi)$ I concluded that the equation is
$$(y + \pi) = (\cos(x) + x-\sin(x)) (x - \pi)$$
1) Is this correct so far?  Wolfram doesn't seem to process this correctly.
2) How would I expand this to get it in $y$-intercept form? I know I can plug $\pi$ into my derivative, but i'm not sure what to make of plugging $x$ into it. I feel like it wouldn't be correct, or it would be correct but it would be so verbose the data would be unusable. 

Your derivative, which we need for slope, is close, but $y'$ should be  $$y' =\underbrace{(1)}_{\frac d{dx}(x)}\cdot(\cos x) + (x)\underbrace{( -\sin x)}_{\frac d{dx}( \cos x)}= \cos x -x\sin x$$  Now, for slope itself, we evaluate $y'(\pi) = \cos (\pi) - \pi\sin(\pi) = -1 - 0 = -1$. 
That gives you the equation of the line: $$y+\pi = -(x -\pi)$$
To get the slope-intercept form, simply distribute the negative on the right, and subtract $\pi$ from each side to isolate $y$: $$y + \pi = -x+ \pi \iff y = -x$$
