Answer:
[a] y=x^2+3, vertex, V(0,3)
[b] y=2x^2, vertex, V(0,0)
[c] y=-x^2 + 4, vertex, V(0,4)
[d] y= (1/2)x^2 - 5, vertex, V(0,-5)
Explanation:
The vertex, V, of a quadratic can be found as follows:
1. find the x-coordinate, x0, by completing the square
2. find the y-coordinate, y0, by substituting the x-value of the vertex.
[a] y=x^2+3, vertex, V(0,3)
y=(x-0)^2 + 3
x0=0, y0=0^2+3=3
vertex, V(0,3)
[b] y=2x^2, vertex, V(0,0)
y=2(x-0)^2+0
x0 = 0, y0=0^2 + 0 = 0
vertex, V(0,0)
[c] y=-x^2 + 4, vertex, V(0,4)
y=-(x^2-0)^2 + 4
x0 = 0, y0 = 0^2 + 4 = 4
vertex, V(0,4)
y = (1/2)(x-0)^2 -5
x0 = 0, y0=(1/2)0^2 -5 = -5
vertex, V(0,-5)
Conclusion:
When the linear term (term in x) is absent, the vertex is at (0,k)
where k is the constant term.