SOLUTION: For the following function, find the vertex and the maximun or min value: h(x)=-2x^2+6x-3 so far I have done is -2(x^2-3x)-3 I'm not sure if I am suppose to find the per

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Question 22027: For the following function, find the vertex and the maximun or min value:
h(x)=-2x^2+6x-3

so far I have done is -2(x^2-3x)-3
I'm not sure if I am suppose to find the perfect square or what.
Thanks for your help.
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
h(x)=-2x^2+6x-3

so far I have done is -2(x^2-3x)-3....OK
I'm not sure if I am suppose to find the perfect square ....YES DO IT ...YOU HAVE TO USE THE GIVEN X^2 AND X TERMS TO MAKE A PERFECT SQUARE
=-2(X^2-2*1.5*X+1.5^2-1.5^2)-3
=-2{X-1.5)^2-1.5^2}-3
=-2(X-1.5)^2+2*1.5*1.5-3
=-2(X-1.5)^2+1.5=Y SAY...Y=1.5-2(X-1.5)^2
NOW ,WE REASON OUT THAT (X-1.5)^2 BEING PERFECT SQUARE IT IS ALWAYS POSITIVE. SO ITS MIIMUM VALUE IS ZERO WHEN X=1.5
NOW WHEN THIS IS ZERO,Y WILL BECOME MAXIMUM SINCE WE ARE SUBTRACTING THE MINIMUM POSSIBLE VALUE FROM 1.5....YO Y HAS A MAXIMUM VALUE OR VERTEX AT X=1.5