How do you identify the vertex, axis of symmetry, direction of opening, and min/max value of this problem. f(y)= -4(y-6)^2+6
Let us write `x=f(y)`
so equation become
`(y-6)^2>=0` always so `6/4-x/4>=0`
This suggest ,graph is defined only for `x<=6` ,It is parabola and open down.
Vertex put `6/4-x/4=0` i.e x=6 then y=6
vertex is (6,6)
For one value of x<6 ,we have two values of y-6.
This means y-6=0 is axis of symmetry.
Graph is opening down it suggest maximum , It occurs at x=6.
`f(y)=0` `2y^2-48y+69=0` `y_1=1,5375`