Let's begin by identifying key information given to us:
[tex]\begin{gathered} y=4-x^2 \\ y=-x^2+4 \\ a=-1,b=0,c=4 \\ x_v=-\frac{b}{2a}=-\frac{0}{2(-1)}=0 \\ y_v=-\frac{b^2-4ac}{4a}=-\frac{0^2-4(-1)(4)}{4(-1)} \\ y_v=-\frac{0+16}{-4}=\frac{-16}{-4}=4 \\ y_v=4 \\ \\ \therefore The\text{ vertex of the equation is }(0,4) \end{gathered}[/tex]To know if the vertex is the maximum or minimum point, we will follow this below:
[tex]\begin{gathered} y_v=4 \\ \Rightarrow This\text{ is a minimum point} \end{gathered}[/tex]Hence, the answer is B.(0,4); minimum
