ALGEBRA 1

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Please check for me.Without drawing the graph of the given equation determine
(a)how many x-intercepts the parabola has
(b)whether it vertex lies above, below or on the axis.

1. y=x^2-5x+6
I use the determinant
sqrt b^2 -4ac
(-5)^2-4(1)(6))=1
There are two real numbers
The parabola has two X- intercepets
The parabola opens downward and its vertex lies above the axis.

2.Y=-X^2+2X-1
(2)^2-4(-1)(-1)=0
There is one real number
one x-intercept
The parabola opens upward and its vertex lies on the axis.
Thanks Much!!!

well done


Thanks Reiny

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