Question

Hich function has a range of {y|y ≤ 5}?

Answer 1

Answer:

$f(x)=-(x-4)^{2}+5$

Step-by-step 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]

$y\leq 4$