Question 7124
To find the x-intercepts (the zeros), set the quadratic = 0 and solve for x:
{{{x^2 + x - 6 = 0}}} Factor.
{{{(x + 3)(x - 2) = 0}}} Apply the zero product principle.
{{{x + 3 = 0}}}, {{{x = -3}}} 
{{{x - 2 = 0}}}, {{{x =  2}}}

The x-intercepts are: (-3, 0) and (2, 0)

The x-coordinate of the vertex is at: -b/2a = -1/2.  (ax^2 + bx + c = 0)Substitute this into the quadratic equation and solve for the y-coordinate.
{{{y = (-1/2)^2 + (-1/2) - 6}}}
{{{y = 1/4 - 1/2 - 6}}}
{{{y = -26/4}}} = - 6 1/4

The vertex is at: (-1/2, -6 1/4)

The graph of this quadratic equationis:
{{{graph(300,200,-5,5,-7,10,x^2+x-6)}}}

The domain (the set all possible values of the independent variable, that's x) is the set of all real numbers.
The range (the set of all possible values of the dependent variable, that's y) is also the set of all real numbers.