Hich function has a range of y ≤ 5? f(x) = (x – 4)2 + 5 f(x) = –(x – 4)2 + 5 f(x) = (x – 5)2 + 4 f(x) = –(x – 5)2 + 4

Question

asked Mar 24, 2020 3.9k views

1 Answer

Hzrari · Answer 1 · 2020-03-31T02:41:41+0000

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]

$y\leq 4$