Question

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

Answer 1

`Vertex(0,-4)`

Explanation:

.

`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]}