Question 180347
set y = to zero when trying to find the x intercepts
:
{{{x^2-5x+6=0}}}
:
{{{(x-2)(x-3)=0}}}
:
x=2 and 3
:
since this has two intercepts we know that the vertex is NOT on the x axis
:
we also know that this parabola opens up(coefficent is positive in x squared term) therefore we know the vertex is below the x axis, but not by much because the x intercepts are very close to each other
:
we can find the exact vertex by completing the square
:
{{{y=x^2-5x+(25/4))6-25/4}}}
:
{{{y=(x-(5/2))^2-(1/4)}}}
:
so the vertex is at (5/2,-1/4)
:
{{{graph(300,300,-10,10,-10,10,x^2-5x+6)}}}