Nitayah's assertion can indeed be evaluated by looking at the characteristics of the given equation [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex].
### Step-by-Step Solution:
1. Understand the Equation:
The given equation is [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex]. This indicates two possible expressions for [tex]\( y \)[/tex]:
- [tex]\( y_1 = 3 + \sqrt{x + 2} \)[/tex]
- [tex]\( y_2 = 3 - \sqrt{x + 2} \)[/tex]
Here, [tex]\( \sqrt{x + 2} \)[/tex] denotes the square root of [tex]\( x + 2 \)[/tex].
2. Vertical Line Test:
A relation is considered a function if for each value of [tex]\( x \)[/tex], there is exactly one value of [tex]\( y \)[/tex]. To check if [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex] is a function, we can use the Vertical Line Test.
- The Vertical Line Test states that if a vertical line intersects the graph of a relation at more than one point, then the relation is not a function.
3. Evaluate for Non-unicity:
For any given value of [tex]\( x \)[/tex] in [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex]:
- Consider a specific value of [tex]\( x \)[/tex]. For example, let [tex]\( x = 2 \)[/tex].
- [tex]\( y_1 = 3 + \sqrt{2 + 2} = 3 + 2 = 5 \)[/tex]
- [tex]\( y_2 = 3 - \sqrt{2 + 2} = 3 - 2 = 1 \)[/tex]
- This approach shows that for [tex]\( x = 2 \)[/tex], we get both [tex]\( y = 5 \)[/tex] and [tex]\( y = 1 \)[/tex].
- This is true for all [tex]\( x \geq -2 \)[/tex] where the square root is defined (because for [tex]\( x < -2 \)[/tex], the expression inside the square root becomes negative and hence complex, which we generally avoid in real-valued functions).
4. Conclusion from the Vertical Line Test:
Since there are always two possible [tex]\( y \)[/tex] values (one positive and one negative deviation from [tex]\( 3 \)[/tex] due to [tex]\( \sqrt{x + 2} \)[/tex]), the graph fails the Vertical Line Test. Therefore, [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex] is indeed not a function because it does not produce one unique [tex]\( y \)[/tex] for a given [tex]\( x \)[/tex].
5. Inverting Non-function:
When trying to find the inverse of an equation, if the original equation is not a function, the inverse will also not be a function, due to the inherent ambiguity in mapping multiple values of one variable to a single value of another.
### Final Statement:
Nitayah is right because [tex]\( y = 3 \pm \sqrt{x + 2} \)[/tex] is not a function. As a consequence, its inverse cannot form a valid function either. Each [tex]\( x \)[/tex] in the given relation maps to two different [tex]\( y \)[/tex] values, thereby invalidating it as a function and thus confirming Nitayah's statement.
