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

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pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

f(y)= -4(y-6)^2+6

Let us write `x=f(y)`

so equation become

`x=-4(y-6)^2+6`

`x-6=-4(y-6)^2`

`(y-6)^2=(x-6)/(-4)`

`(y-6)^2=6/4-x/4`

`(y-6)^2>=0`  always so `6/4-x/4>=0`

`6>=x`

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's profile pic

oldnick | (Level 1) Valedictorian

Posted on

`f(x)=-4(y^2-12y+36)+6=-4y^2+48y-144+6=```

`=-4y^2+48y-138`

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

`y_2=-22,4625`

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

`f'(y)=0`

`y=6`

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

`f(6)=6`

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

`V=(6;6)`

``

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