You have the following function:
[tex]y=-x^2+6x[/tex]the general form of a quadratic function is given by:
[tex]y=ax^2+bx+c[/tex]By comparing the given function with the general form of a quadratic function, you have that a=-1 anf b=6.
The x coordinate of the vertex of a quadractic function can be obained by using the following formula:
[tex]x=-\frac{b}{2a}[/tex]By replacing you obtain:
[tex]x=-\frac{6}{2(-1)}=3[/tex]Thus, the y cooridante of the vertex is:
[tex]y=-(3)^2+6(3)=-9+18=9[/tex]Then, the vertex is (3,9)
The previous point is a maximum of the function. It can be appreciated in the fact that the dominant term is negative in the function. This makes that for larger values of x the function tends to - infinity and so, the vertex is a a maximum.
