Answer:
Option (3) is correct.
The vertex of given equation [tex]y=-x^2[/tex] is (0,0)
Step-by-step explanation:
Given equation [tex]y=-x^2[/tex]
We have to find the vertex of the graph of given equation.
Consider the equation [tex]y=-x^2[/tex]
Since, given equation is a form of quadratic equation [tex]ax^2+bx+c[/tex] with a =-1 , b=c=0, then ,
If the quadratic is written in the form [tex]y = a(x-h)^2 + k[/tex] then the vertex is the point (h, k).
Writing the given equation in the form of [tex]y = a(x-h)^2 + k[/tex] , we get,
[tex]y=-x^2 \Rightarrow y=-(x-0)^2+0 [/tex]
On comparing, we get,
h = k= 0 and a = -1
Thus, the vertex is (0 , 0)
Option (3) is correct.
The vertex of given equation [tex]y=-x^2[/tex] is (0,0)
