Gwhat is the minimum vertical distance between the parabolas y=x^2+1 and y=x-x^2

As you can notice, the that the graphs are opening in different directions. The positive sign in the first equation means that the graph is opening upwards and the second graph is opening downwards. 

Find the vertices of both the parabolas since these points are the lowest/highest points of the entire equation. 

1) y=x²+1

x=-b/2a 
x=-0/2(1) 
x=0 

y=(0)+1 
y=1 

The vertex is (0, 1) 

2) y=-x²+x 

x=-b/2a (this formula helps find the x coordinate of a quadratic function) 
x=-1/2(-1) 
x=1/2 

y=-(1/2)²+(1/2) 
y=1/4 

Since you only need to find the vertical distance, all you need to be concerned with is the y coordinates. 

Use distance formula just for the y coordinates. 

√(y₁-y₂)²
√(1-1/4)²
√3/4 
=0.8660254038 ≈0.9 units 

Hope I helped :)