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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

User SquareCog
by
8.2k points

1 Answer

4 votes

Answer:


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

User Hzrari
by
8.9k points

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