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Which function has a range of [tex]$\{y \mid y \leq 5\}$[/tex]?

A. [tex]$f(x) = (x-4)^2 + 5$[/tex]

B. [tex][tex]$f(x) = -(x-4)^2 + 5$[/tex][/tex]

C. [tex]$f(x) = (x-5)^2 + 4$[/tex]

D. [tex]$f(x) = -(x-5)^2 + 4$[/tex]

Sagot :

GinnyAnsweridn
Let's analyze the given functions one by one to determine their ranges and identify which one has the range [tex]\(\{y \mid y \leq 5\}\)[/tex].

1. Function [tex]\( f(x) = (x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [5, \infty) \)[/tex].

2. Function [tex]\( f(x) = -(x-4)^2 + 5 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-4)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (4, 5) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 5 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 4 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 5] \)[/tex].

3. Function [tex]\( f(x) = (x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens upwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is positive).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens upwards, the minimum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] increases.
- Therefore, the range of this function is [tex]\( [4, \infty) \)[/tex].

4. Function [tex]\( f(x) = -(x-5)^2 + 4 \)[/tex]
- This is a quadratic function that opens downwards (since the coefficient of [tex]\( (x-5)^2 \)[/tex] is negative).
- The vertex of this parabola is at [tex]\( (5, 4) \)[/tex]. Since it opens downwards, the maximum value of [tex]\( f(x) \)[/tex] is [tex]\( 4 \)[/tex], and as [tex]\( x \)[/tex] moves away from [tex]\( 5 \)[/tex], the value of [tex]\( f(x) \)[/tex] decreases.
- Therefore, the range of this function is [tex]\( (-\infty, 4] \)[/tex].

Given these analyses, the function which has the range [tex]\(\{y \mid y \leq 5\}\)[/tex] is:

[tex]\[ f(x) = -(x-4)^2 + 5 \][/tex]

So, the function is:

[tex]\[ \boxed{f(x) = -(x-4)^2 + 5} \][/tex]
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