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

pramodpandey | Student

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.

oldnick | Student



`f(y)=0`      `2y^2-48y+69=0` `y_1=1,5375`


`f'(x)= -8y+48`



`f''(y)=-8 max`


`y=6 ....simmetry``..axs`