Common tangent(s) to two parabolas without calculus We have the parabolas: $y = 2x^{2} + 2x + 1$ and $y = 2x^{2} - 2x - 1$.
Finding the common tangents with calculus is as straight forward as just solving a system of equations with the derivatives of both parabolas and $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$; but how about in analytical geometry? 
 A: A line $y=kx+m$ is a tangent to the parabola $y=ax^2+bx+c$ iff the quadratic equation 
$$
kx+m=ax^2+bx+c
$$
has exactly one solution, which makes the discriminant to be zero. In our case
$$
kx+m=2x^2+2x+1\iff 2x^2+(2-k)x+1-m=0,\\
kx+m=2x^2-2x-1\iff 2x^2-(2+k)x-1-m=0
$$
should each have a unique solution $x$, that is the discriminants are zeros for both equations
$$
\begin{cases}
(2-k)^2-8(1-m)&=&0,\\
(2+k)^2+8(1+m)&=&0.
\end{cases}
$$
Solve the system: $k=-2$, $m=-1$.
A: This pair of parabolas are meant to make the problem easy. Complete the square to get 
 $$2y = 4x^2 + 4x + 2 = (2x+1)^2 + 1 = 4(x + \frac12)^2 - 1 + 2$$  for the first parabola and
 $$2y = 4x^2 - 4x -2 = (2x-1)^2 - 3 = 4(x - \frac12)^2 - 1 - 2$$  for the second.
The vertex of the first is at $\,(-\frac12,\frac12)\,$ and the second at $\,(\frac12,-\frac32).\,$
The line between them has slope $\,-2,\,$ which is the slope of the common tangent line. The parabolas intersect at the difference $\, 4x + 2 = 0 \,$ which is $\, x = -\frac12. \,$ Now the line with slope $\, -2 \,$ that passes through the intersection is $\, y = -2x + \frac12.\,$ That line intersects the second parabola at solutions to $\,  2x^2 - \frac32 = 0 \,$ but the average of the two solutions is $\, x = 0. \,$ The point on the  second parabola with this
$\,x\,$ value is $\, (0,-1) \,$ and is the point of tangency of the parabola with the line
$\, y = -2x - 1. \,$
To check, we find that the line $\, y = -2x + \frac12\,$ intersects the first parabola at solutions to $\,  2x^2 + 4x + \frac12 = 0. \,$ and the average of the two solutions is $\, x = -1.\,$ The point on the first parabola with this $\,x\,$
value is $\, (-1,1) \,$ which is the point of tangency with the line
$\, y = -2x - 1. \,$
To summarize, think of the special case of $\, y = (x-1)^2 \,$ and $\, y = (x+1)^2. \,$ The line $\, y = 1 \,$ passes through their intersection point
$\,(0,1)\,$ and also another point on each parabola. The line $\, y = 0 \,$
is the common tangent. The geometry of this special case is affinely equivalent to the general case of any two parabolas which are translates of each other.
A: Refer to this question here for the equation of a tangent to a parabola. 
First parabola: $$y-\tfrac 12=2(x-\tfrac 12)^2$$
Tangent at $(h, k= 2h^2+2h+1)$ : 
$$\tfrac 12 ((y-\tfrac 12)+(k-\tfrac 12))=2(x+\tfrac12)(h+\tfrac12)\\
y=2(1+2h)x+1-2h^2\qquad (1)$$
Second parabola: 
$$y+\tfrac 32=2(x-\tfrac 12)^2$$
Tangent at $(m, n=2m^2-2m-1)$:
$$\tfrac 12 ((y+\tfrac 32)+(m+\tfrac 32))=2(x-\tfrac12)(m-\tfrac12)\\y=2(2m-1)x-1-2m^2\qquad (2)$$
Equating coefficients in $(1), (2)$ gives $(m,n)=(0,-1)$ and $(h,k)=(-1,1)$.  
Hence, from either $(1)$ or $(2)$, equation of common tangent is 
$$\color{red}{y=-2x-1}$$
A: The common tangent is parallel to the line joining the two vertices, hence its equation is of the form $y=-2x+k$. 
To find $k$ we can use the fact that this tangent has only one point in common with any of the parabolas (the second one, for instance). That is, the system
$$
\cases{y=-2x+k\\
y=2x^2-2x-1
}
$$
must have only one solution. 
Eliminating $y$ we find $2x^2-1=k$, which has a single solution only if $k=-1$.
