Answer:
Yes, the following equation is a quadratic function in vertex form
Step-by-step explanation:
we know that
The quadratic function of the vertical parabola into vertex form is equal to
[tex]y=a(x-h)^{2} +k[/tex]
where
(h,k) is the vertex of the parabola
If the coefficient a is > 0 ----> the parabola open upward (vertex is a minimum)
If the coefficient a is < 0 ----> the parabola open downward (vertex is a maximum)
in this problem we have
[tex]y=4(x-2)^{2} +6[/tex]
The squared term contain the x-coordinate of the vertex
[tex]h=2[/tex]
The constant term is the y-coordinate of the vertex
[tex]k=6[/tex]
The vertex is the point (2,6)
The coefficient is equal to
[tex]a=4[/tex] ----> open upward (vertex is a minimum)
