How do you find the vertex for $y = x^{2} - 4$ $y=x^2-4$ ?

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How do you find the vertex for $y = x^{2} - 4$ $y=x^2-4$ ?

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Answer 1 · 2018-06-27T05:53:02

$(0, - 4)$ $(0, -4)$

Explanation:

To find the vertex of a standard quadratic equation $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ , first use the formula $\frac{- b}{2 a}$ $(-b)/(2a)$ to find the $x$ $x$ -coordinate of the vertex:
$x = \frac{- b}{2 a} = \frac{- 0}{2 (1)} = 0$ $x = (-b)/(2a) = (-0)/(2(1)) = 0$

Now plug the $x$ $x$ value of the vertex back into the equation to find the $y$ $y$ value:
$y = {(0)}^{2} - 4 = - 4$ $y = (0)^2 - 4 = -4$

Therefore, the vertex is at $(0, - 4)$ $(0, -4)$ .

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How do you find the vertex for $y = x^{2} - 4$ $y=x^2-4$ ?

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