State the coordinates of the vertex for each of the following

[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)

Step-by-step 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.

send it over

step-by-step explanation: