What are the coordinates of the vertex of y = x^2-2x-7?

Answer 1

vertex: #(1,-8)#

Converting #y=x^2-2x-7# into vertex form: #y=m(x-a)^2+b# (with vertex at #(a,b)#)
Complete the square #y=x^2-2xcolor(red)(+1) - 7 color(red)(-1)#
#y=(x-1)^2+(-8)# with the vertex at #(1,-8)#
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Answer 2

The coordinates of the vertex of the parabola ( y = x^2 - 2x - 7 ) can be found using the formula ( x = \frac{-b}{2a} ) where ( a = 1 ) (coefficient of ( x^2 )) and ( b = -2 ) (coefficient of ( x )). Substituting the values of ( a ) and ( b ) into the formula, we get ( x = \frac{-(-2)}{2(1)} = 1 ). To find the corresponding ( y )-coordinate, substitute ( x = 1 ) into the equation ( y = x^2 - 2x - 7 ). So, ( y = (1)^2 - 2(1) - 7 = -8 ). Therefore, the coordinates of the vertex are ( (1, -8) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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