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