Posted by **DANIELLE** on Tuesday, April 24, 2007 at 1:34pm.

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!!!

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