Find two quadratic equations f(x) and g(x) such that the product h(x)=f(x)×g(x) is tangent to each.
Lets try the following two quadratic equations:
f(x) = x2-1 and g(x)=x2-3 and h(x) = (x2-1)( x2-3)
We can see that f(x) and g(x) must be in opposite directions in order to have a product graph that is tangent to both parabolas. Let us change the sign on g(x).
f(x) = x2-1 and g(x)=-x2+3 and h(x) = (x2-1)( -x2+3)
We see that our quadratic functions are symmetric about the y-axis and both curves share points of intersection. We can also see that the product curve coincides with the roots of both f(x) and g(x) and their points of intersection but it is not tangent to both curves as desired.
Playing with the constant values (c) on both curves I came upon the conclusion that the equations must take the following form in order to have h(x) tangent to both:
f(x) = ax2-n
g(x) = -(ax2-(n+1))
So now let us change the above equations to:
f(x) = x2-1 and g(x)=-x2+2 and h(x) = (x2-1)( -x2+2)
f(x) = 3x2-4 and g(x)=-3x2+5 and h(x) = (3x2-4)( -3x2+5)
Hence we can see that when the functions follow the generalized form f(x)=ax2-n and g(x)=-(ax2-(n+1)), then their product h(x)=f(x)g(x) is tangent to the curves at exactly two points on each curve.
We can also look at the equations in the form ax2+bx+c.
For such quadratic equations f(x) and g(x) must follow the form f(x)=ax2-bx-c and g(x)=-(ax2-bx-(c+1)).
For example lets look at f(x) = x2-2x-1 and g(x)= - x2+2x+2 and h(x) = (x2-2x-1)( - x2+2x+2)