Question 81554
Find the vertex of y= x^2 - 5x + 3:
x-value of the vertex: -b/2a
x-value=-b/2a=-(-5)/2(1)=5/2=2.5
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Plug-in (x=2.5) and solve for the y-value of the vertex:
y= x^2 - 5x + 3
y= (2.5)^2 - 5(2.5) + 3 
y=-3.25
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So, the vertex is (2.5, -3.25). 
Axis of symetry = x-value of the vertex=2.5
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Table of values:  pick some values for the x-term and solve for the y-term:
Plot the vertex at point (2.5, -3.25)
Let x=0, than y = (0)^2 - 5(0) + 3=3.  Plot point (0, 3)
Let x=1, than y=-1.  Plot points (1, -1)
Let x=-1, than y=9.  Plot points (-1, 9)
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Let y=0 
0=x^2-5x+3  [solve for x using the quadratic formula]
x=.69 and x=4.31 [these are the x-intercepts of the graph]
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{{{ graph( 300, 200, -6, 5, -10, 10,  x^2-5x+3) }}}