Question 759456
Generally you would have y=ax^2+bx+c, and you are given the two zeros of the parabola.  Try to narrow some of the equation using them.


Point (5,0):
{{{0=a*5*5+b*5+c}}}
{{{25a+5b+c=0}}}


Point (6,0):
{{{a*6*6+b*6+c=0}}}
{{{36a+6b+c=0}}}


Also you were given point B(4,6):
{{{a*4*4+b*4+c=6}}}
{{{16a+4b+c=6}}}


NOW you have a system of three equations in three unknowns (yes, they should be constants but currently you do not know their values) of a, b, and c.
---------------------
SYSTEM:
{{{16a+4b+c=6}}}
{{{36a+6b+c=0}}}
{{{25a+5b+c=0}}}
---------------------


Can you take the rest of the solution process from there?



Omitting the steps in solving that system, I found:
{{{a=3}}}
{{{b=-33}}}
{{{c=90}}}
EQUATION: {{{highlight(y=3x^2-33x+90)}}}
{solution checks as correct}.