Question 136530: I am trying to help my son with a problem. He is doing parabolas. This is the one the teacher did in class and I don't understand what she did. (((y=x^2+8x+6)))....y-6+16 = x^2+8x+16....y+10=(x+4)^2....vertex (-4,-10) Domain is all real numbers, range is y is greater than or equal to -10, minimum is -10. Could you pleas explain her procedure here. Thank you!
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! a) subtracted the constant term (+6)
b) completed the square by adding the square of half of the coefficient of the x term
__ (8/2)^2=16
c) the equation is now in the "vertex" form __ y-k=(x-h)^2
__ where the vertex is (h,k)
the domain is what values x can be
__ in this case, any real number value for x gives a value for y
the range is what values y can be
__ since the x-h is squared, it can never be less than zero
__ so y+10 can never be less than zero __ so y can't be less than -10
the y value of the vertex is the minimum value for the function (and the lower bound of the range)
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