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

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Answer 1 · 2017-12-12T23:03:08

I know this question is one year old....

But for the others who may want to know how to do this, here is the solution:

To find vertexes, there are two methods.

Method one: Using the formula

You can easily find the $x$ -coordinate of the vertex by using the formula:
$-b/(2a)$ for a quadratic equation $ax^2 + bx + c$

Therefore, for vertex $(x, y)$ ,
$x = \frac{-0}{2 * 1} = 0$

Then, you calculate the $y$ -coordinate with the given equation:
$y = x^2 - 4 = 0^2 - 4 = -4$

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

This method is more formal and some tests and exams require you to use this method.

This method will find the vertex form $y = a(x - p)^2 + q$ where the vertex is $(p, q)$ .

However, the equation $y = x^2 - 4$ is already in vertex form (This is equal to $y = 1(x - 0)^2 + (-4)$ .

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

Hope that makes sense!

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