What is the vertex of # y=x^2-4 #?

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Answer 1 · 2018-03-04T01:13:41

#Vertex(0,-4)#

.

#y=x^2-4#

If the equation of a parabola is in the form:

#y=ax^2+bx+c#

we can find the #x#-coordinate of its vertex using the following formula:

#x_(vertex)=-b/(2a)#

Comparing the problem equation with the form above, we see:

#a=1, b=0, c=-4#

#x_(vertex)=-0/(2(1))=0#

Now, we can plug this into the equation to find the #y#-coordinate:

#y_(vertex)=(0)^2-4=0-4=-4#

Therefore,

#Vertex(0,-4)#

You can see the graph of this parabola below:

graph{x^2-4 [-10, 10, -5, 5]}