Question

Hich function has a range of y?

f(x) = (x – 4)2 + 5
f(x) = –(x – 4)2 + 5
f(x) = (x – 5)2 + 4
f(x) = –(x – 5)2 + 4

Answer 1

`f(x)=-(x-4)^(2)+5`

Explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex of the parabola

if a>0 -----> the parabola open upward (vertex is a minimum)

if a<0 -----> the parabola open downward (vertex is a maximum)

Verify each case

case A)
`f(x)=(x-4)^(2)+5`

The vertex is the point
`(4,5)`

`a=1`

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [5,∞)

`y\geq5`

case B)
`f(x)=-(x-4)^(2)+5`

The vertex is the point
`(4,5)`

`a=-1`

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,5]

`y\leq 5`

case C)
`f(x)=(x-5)^(2)+4`

The vertex is the point
`(5,4)`

`a=1`

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [4,∞)

`y\geq4`

case D)
`f(x)=-(x-5)^(2)+4`

The vertex is the point
`(5,4)`

`a=-1`

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,4]

`y\leq 4`