Question

How do you find the vertex of `y= x^2 -16`?

Answer 1

show the answer below

Explanation:

The general parabola equation

`ax^2+bx+c=0`

in this equation

`y=x^2-16` you have

a=1 b=0 and c=-16

`x_ v=-b/(2a)=-0/2=0`

`y_v=(0)^2-16=-16`

the vertex is (0,-16)

graph{y=x^2-16 [-26.2, 25.1, -19.67, 6]}