In general, the vertex of a parabola in the form y=ax^2y=ax2 where aa is any number is (0,0),(0,0), as the parabola has not been shifted up, down, left, or right at all.
We can prove this by comparing y=3/4x^2y=34x2 to the standard form of a quadratic, y=ax^2+bx+cy=ax2+bx+c
In this case, a=3/4, b=0, c=0a=34,b=0,c=0
The x-x−coordinate of the vertex is given by -b/(2a)−b2a. In this case, it would be (-0/(2*3/4))=0(−02⋅34)=0.
The y-y−coordinate of the vertex is given by plugging in the result of -b/(2a)−b2a into the equation. In this case, -b/(2a)=0, y=3/4(0^2)=0−b2a=0,y=34(02)=0
So, the vertex is (0,0)(0,0)
