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How do you write the vertex form equation of the parabola #y=x^2-2x-5#?

Answer 1

Isaiah Anthony

#y = (x - 1)^2 - 6#

#y = x^2 - 2 x -5# #y = (x -1)^2 -(-1)^2 -5# #y = (x - 1)^2 - 6#

since a coeficient of #(x-1)^2# is positive value, it has minimum at #-6# and it axis of simetry at #x = 1#.

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Answer 2

Eva Abramson

To write the vertex form equation of the parabola y=x^2-2x-5, complete the square on the expression. This involves rearranging terms to express the quadratic function in the form y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.

Starting with y = x^2 - 2x - 5:

y = (x^2 - 2x) - 5

y = (x^2 - 2x + 1) - 5 - 1 (adding and subtracting (2/2)^2 = 1 inside the parentheses)

y = (x - 1)^2 - 6

Thus, the vertex form equation of the parabola y=x^2-2x-5 is y = (x - 1)^2 - 6.

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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.