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The function [tex][tex]$f$[/tex][/tex] is defined by [tex][tex]$f(x) = a \sqrt{x+b}$[/tex][/tex], where [tex][tex]$a$[/tex][/tex] and [tex][tex]$b$[/tex][/tex] are constants. In the [tex][tex]$xy$[/tex][/tex]-plane, the graph of [tex][tex]$y=f(x)$[/tex][/tex] passes through the point [tex][tex]$(-24,0)$[/tex][/tex], and [tex][tex]$f(24)\ \textless \ 0$[/tex][/tex]. Which of the following must be true?

A. [tex]f(0) = 24[/tex]
B. [tex]f(0) = -24[/tex]
C. [tex]a \ \textgreater \ b[/tex]
D. [tex]a \ \textless \ b[/tex]

Sagot :

Given the function [tex]\( f(x) = a \sqrt{x + b} \)[/tex] and the points through which the graph passes, we can determine the nature of the constants [tex]\( a \)[/tex] and [tex]\( b \)[/tex].

First, let's use the information that the graph passes through the point [tex]\((-24, 0)\)[/tex]. This means that when [tex]\( x = -24 \)[/tex], [tex]\( f(x) = 0 \)[/tex].

[tex]\[ f(-24) = a \sqrt{-24 + b} = 0 \][/tex]

Since [tex]\( \sqrt{-24 + b} = 0 \)[/tex], we have:

[tex]\[ -24 + b = 0 \implies b = 24 \][/tex]

So, we have determined that [tex]\( b = 24 \)[/tex].

Next, we were given that [tex]\( f(24) < 0 \)[/tex]. Let's use this condition to understand the nature of [tex]\( a \)[/tex].

[tex]\[ f(24) = a \sqrt{24 + b} = a \sqrt{24 + 24} = a \sqrt{48} \][/tex]

Since [tex]\( \sqrt{48} \)[/tex] is positive, for [tex]\( f(24) < 0 \)[/tex], it must be that [tex]\( a \)[/tex] is negative. Thus,

[tex]\[ a < 0 \][/tex]

With [tex]\( b = 24 \)[/tex] and [tex]\( a < 0 \)[/tex], let's now evaluate the options provided:

- Option (A) [tex]\( f(0) = 24 \)[/tex]:

[tex]\[ f(0) = a \sqrt{0 + b} = a \sqrt{24} \][/tex]

Since [tex]\( a < 0 \)[/tex], [tex]\( a \sqrt{24} \)[/tex] cannot be 24 because it would be a positive number if [tex]\( a \)[/tex] were positive, which is not the case here. So, this option is incorrect.

- Option (B) [tex]\( f(0) = -24 \)[/tex]:

[tex]\[ f(0) = a \sqrt{0 + b} = a \sqrt{24} \][/tex]

We need to check if this equals -24:

[tex]\[ a \sqrt{24} = -24 \implies a = -\frac{24}{\sqrt{24}} = -\sqrt{24} \][/tex]

Given [tex]\( \sqrt{24} \)[/tex] is positive and [tex]\( a = -\sqrt{24} \)[/tex] is indeed negative, so [tex]\( f(0) = -24 \)[/tex] is a valid option.

- Option (C) [tex]\( a > b \)[/tex]:

We know [tex]\( a < 0 \)[/tex] and [tex]\( b = 24 \)[/tex]. Clearly [tex]\( a \)[/tex] is not greater than [tex]\( b \)[/tex] because [tex]\( a < 0 \)[/tex] and [tex]\( b = 24 \)[/tex]. So this option is incorrect.

- Option (D) [tex]\( a < b \)[/tex]:

As [tex]\( a < 0 \)[/tex] and [tex]\( b = 24 \)[/tex], it is true that [tex]\( a < b \)[/tex].

Thus, the options that are true based on the given conditions are:
- [tex]\( f(0) = -24 \)[/tex]
- [tex]\( a < b \)[/tex]

Thus the correct answers must include:

(B) [tex]\( f(0) = -24 \)[/tex]
(D) [tex]\( a < b \)[/tex]