Question 118068
The x-intercepts occur where {{{g(x) = 0}}}
The y-intercepts occur where {{{x = 0}}}
First set {{{g(x) = 0}}}
{{{g(x) = x^2 + x - 6}}}
{{{x^2 + x - 6 = 0}}}
{{{(x + 3)(x - 2) = 0}}}
This is true if either
{{{x = -3}}} 1st x-intercept
or
{{{x = 2}}} 2nd x-intercept
These are called the roots of the equation. The {{{x}}} co-
ordinate of the vertex is exactly midway between them, so
it,s at V(-(1/2), ). Now find g(-(1/2))
(Notice that if I add {{{2.5}}} to {{{-(1/2)}}}, I get
{{{x = 2}}} and if I subtract {{{2.5}}} from it, I get
{{{x = -3}}} as I should)
{{{g(x) = x^2 + x - 6}}}
{{{g(-(1/2)) = (-(1/2))^2 + (-(1/2)) - 6}}}
{{{g(-(1/2)) = 1/4 - 1/2 - 6}}}
{{{g(-(1/2)) = -(25/4)}}}
So the vertex is at (-1/2 , -25/4) answer
The y-intercept is at {{{g(0)}}}
{{{g(0) = 0^2 + 0 - 6}}}
{{{g(0) = -6}}}
So the y-intercept is at (0 , -6) answer 
Here's the sketch
{{{ graph( 600, 600, -6, 6, -10, 10, x^2 + x - 6) }}}