Question 171217
{{{x^2-5x+6=0}}} given....... x intercepts are found when y is 0
:
{{{(x-2)(x-3)}}} so x intercepts are 2 and 3.
:
vertex (-B/2A , C - B^2/4A)  (5/2(1),6-(25/4)...(5/2,-1/4) 
so the vertex is below the x axis
:
{{{-x^2+2x-1=0}}}
vertex for x is  -2/2(-1)=1 plug that into equation to find y
:
{{{(-1)^2+2(-1)-1=y}}}=0....so vertex is at (1,0) which is on the x axis meaning there is no x intercepts

there is one x intercept at x=1
 :
*[invoke solve_quadratic_equation -1, 2, -1]