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y=3x^2+6x-10 find the value of x where y has the smallest value. What is the smallest value of y?
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look for vertex
notice the parabola is opening up and so its smallest y value is the vertex. We now it opens up because the # in front of x^2 is positive
|dw:1350400171492:dw|
the x-value for the vertex is x=-b/2a where we have an equation of the form \[ax^2+bx+c=f(x) \]
identify b and a
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but how do I get rid of the 3 to make x^2 by itself?
you don't need to do any separating. to find the vertex it is all plug and play.
|dw:1350400335193:dw|
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