Question 203134
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Given: {{{system(y = x^2 - 5x + 6)}}}, where{{{system(a=1,b=-5,c=6)}}}}

Finding for X-intercepts via Quadratic:
{{{x^2-5x+6=0}}}, a perfect square
{{{(x-3)(x-2)=0}}}
{{{red(x=3)}}} & {{{red(x=2)}}}
X-intercepts --> (3,0) & (2,0) Answer


For the vertex:
{{{x=-b/2a=-(-5)/(2*1)=red(5/2)}}} ---> substitute to the eqn,


{{{y=(5/2)^2-5(5/2)+6=25/4-25/2+6}}}
{{{y=(25-50+24)/4=red(-1/4)}}}


The vertex: (5/2,-1/4) or (2.5,-.25) ---> Lies below the X - axis being the Y-Intercept is negative (-).

To clearly demonstrate this, we see the graph

{{{drawing(400,400,-4,8,-4,8,grid(1),graph(400,400,-4,8,-4,8,x^2-5x+6),blue(circle(2.5,-.25,.08)),green(circle(2,0,.08)),green(circle(3,0,.08)))}}}


Thank you,
Jojo</font>