A line L is tangent to two curves $x^2 + 1$ and $-2x^2 - 3$, find L's equation Some things I'd like to clarify beforehand:
From my understanding, for this situation to occur these conditions need to be fulfilled:
1) $f(a) = g(a)$
2) $f'(a) = g'(a)$
However, both of these conditions are not fulfilled.
What I first did:
$a^2 + 1 = -2a^2 - 3$
I get $3a^2 + 4 = 0$.
Am I on the right track here? The thing is my class doesn't allow calculators, which is why I believe I'm wrong as the equation would get too complicated following this logic. 
Thanks.
 A: Condition 1 needs to be fulfilled if $L$ is tangent to the two curves at the same point.  This doesn't appear to be a requirement.  Actually, it can't be a requirement because these two curves don't have any points in common.
A modified version of condition 2 does need to be fulfilled whether $L$ is tangent to both curves at the same point or not.  This is because $L$ is just a line, and a line has the same slope at all points.  This modified version is $f'(a_1) = g'(a_2)$, where $a_1$ is the $x$-coordinate of the point of tangency of $L$ with $f$, and $a_2$ is the $x$-coordinate of the point of tangency of $L$ with $g$.
I recommend starting by writing the equation of the line $L$ as $y = mx + b$.  Then note that $m = f'(a_1) = g'(a_2)$, and $ma_1 + b = f(a_1)$ and $ma_2 + b = g(a_2)$.
Can you take it from here?
A: Note that the two curves do not intersect anywhere, so setting the equations equal to each other yields no real solutions. In this problem it is not possible for a line to be tangent to both curves at one point. But there are in fact TWO lines that are tangent to both curves. This can be done with straight forward math, i.e. no calculus would be needed. The curves are parabolas.
Let the tangents be of the form  $y=mx+b$. Setting equal to find points of intersections, we get $x^2+1=mx+b$ which translates into $x^2-mx+1-b=0$. In order to have tangency, we want the discriminant $b^2-4ac$ to be zero, so $m^2-4(1-b)=0$ which is $m^2=4(1-b)$. Follow exactly the same procedure for $y=mx+b$ to arrive at $m^2=8(3+b)$ (Please verify!). Now the simple trick: Setting equal to solve for $b$ to find $b=-\frac{5}{3}$. There is only one $b$ here, does that make graphically sense?. From here of course you find two $m$ values. Can you finish the problem from here?  
A: $f(x) = x^2 +1\\
g(x) = -2x^2 - 3$
both are tangent to the line $L: y = mx+b$
If the curve is tangent, when we subtract one from the other, we have a root of multiplicity.  The discriminant in the quadratic formula is zero.
$f(x) = x^2 -mx +1-b\\
m^2 - 4(1-b) = 0$
$g(x) = -2x^2 -mx -3-b\\
m^2 - (-2)(-3-b)(4) = 0$
$4-4b = 24 + 8b\\
12b = -20\\
b = -\frac {5}{3}\\
m = \pm \sqrt {\frac {32}3}$
There are two lines:
$y_1 = 4\sqrt {\frac 23} x - \frac {5}{3}\\
y_2 = -4\sqrt {\frac 23} x - \frac {5}{3}$
Now it is probably wroth checking that each line is tangent to each curve
$y_1$ is tangent to $f(x)$ at $2\sqrt{\frac 23}$ and $g(x)$ at $2\sqrt{\frac 23}$ 
At the points of tangency $f(x) = y(x)$ and $f'(x) = y'(x).$  
