Tangent to both pieces of a piecewise defined function

[train449]

Homework Statement

Find the line tangent to the curve f(x) = { (x+3)^2 | x < 0} , {-x^ + 8x -4 | x≥ 0} at two distinct points

Homework Equations

slope = Δy/Δx

The Attempt at a Solution

I only got as far as finding the slope of the line to be: a= -50p^2 + 50p. where p is a point on y = (x+3)^2.

my reasoning:

a = (s - q)/(r - p),
where (p,q) belongs to the curve (x + 3)^2 and (r,s) belongs to -x^ + 8x -4
then,

a = [-r^2 + 8r -4 -(p + 3)^2]/r - p = [-r^2 + 8r - p^2 -6p]/r-p

also,

d/dx (x + 3)^2 = 2x + 6
d/dx -x^2 + 8x -4 = -2x +8

these slopes should both be equal, and also to a
thus

2p + 6 = [-r^2 + 8r - p^2 -6p]/r-p
2pr - 2r -p^2 +12p = 0

-2r + 8 = [-r^2 + 8r - p^2 -6p]/r-p
-r^2 + 2pr -2p = 0

2pr - 2r -p^2 +12p = -r^2 + 2pr -2p
r = 7p

then a = [-(7p)^2 + 8(7p) - p^2 -6p]/7p-6p
a = -50p^2 +50p

As nice as the result looks, I'm now stumped, beating my head against the wall and feeling like I've gone down the complete wrong path.

Any help is appreciated

 

[train449]

my last line isn't completely simplified i realized it should be -50p + 50