Question 62265
graph Y=3x^2-8x+2 and give the vertex, domain and rage

This quadratic equation is in standard form: {{{y=ax^2+bx+c}}}
Your a=3, b=-8, c=2
We can find the x coordinate for the vertex with the formual {{{highlight(x=-b/2a)}}}
{{{x=-(-8)/(2(3))}}}
{{{x=8/6}}}
{{{x=4/3}}}
Plug x=4/3 into your equation to find the y-coordinate of the vertex.
{{{y=3(4/3)^2-8(4/3)+2}}}
{{{y=3(16/9)-32/3+2}}}
{{{y=16/3-32/3+2}}}
{{{y=16/3-32/3+6/3}}}
{{{y=(16-32+6)/3}}}
{{{y=-10/3}}}
The vertex is (4/3,-10/3)
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The domain (possible x values)  for all quadratic equations like this is ALL REAL NUMBERS.  (x can be anything.)  In set builder notation {x|x=R}, in interval notation (-infinity,infinity).
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The range (possible y values) is found by first looking at a.  a=3--it is positive.  That means that the parabola opens upward and the y coordinate of the vertex is it's lowest y.  Therefore the range is {y|y>=-10/3}.
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Happy Calculating!!!