The vertex for a quadratic equation of the form
[tex]y=ax^2+bx+c[/tex]is given by
[tex]x=-\frac{b}{2a}[/tex]In our case, b = -4 and a = -1; therefore,
[tex]x=-\frac{-4}{2(-1)}=-2[/tex][tex]x=-2[/tex]this is the x-coordinate of the vertex, the y-coordinate is
[tex]\begin{gathered} y=-(-2)^2-4(-2)-1 \\ y=3 \end{gathered}[/tex]Hence, the coordinates of the vertex are
[tex](-2,3)[/tex]And since we have a negative sign on x^2, the parabola is concave down; therefore, the vertex is the maximum value.
The correct choice, therefore, is (-2, 3) maximum value
