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!