# How do you write the quadratic in vertex form given #y=x^2-5#?

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To write (y = x^2 - 5) in vertex form, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex of the parabola, follow these steps:

Given (y = x^2 - 5), notice that the equation is already in a form close to the vertex form, but without the complete square. For this particular equation, (a = 1), and the equation lacks a linear (x) term, which simplifies the process since there's no need to complete the square.

The given equation directly translates to a vertex form of (y = 1(x - 0)^2 - 5), because:

- The coefficient (a = 1).
- There is no (x) term, which means the (h) value that would normally shift the vertex left or right is (0).
- The constant term, (-5), represents the (k) value, shifting the vertex down.

Thus, the vertex form of the equation (y = x^2 - 5) is (y = (x - 0)^2 - 5), or more simply, (y = x^2 - 5), with the vertex at ((0, -5)).

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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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