Let's solve for the domain and range of the function [tex]\( f(x) = -3(x - 5)^2 + 4 \)[/tex].
### Finding the Domain
The domain of a function consists of all possible values of [tex]\( x \)[/tex] for which the function is defined.
1. The function [tex]\( f(x) = -3(x - 5)^2 + 4 \)[/tex] is a quadratic function.
2. Quadratic functions are defined for all real numbers because there are no restrictions on [tex]\( x \)[/tex].
Thus, the domain of [tex]\( f(x) \)[/tex] is all real numbers, which can be represented as:
[tex]\[ \text{Domain: } (-\infty, \infty) \][/tex]
### Finding the Range
The range of a function consists of all possible values of [tex]\( f(x) \)[/tex].
1. The standard form of the quadratic function we have is [tex]\( -3(x - 5)^2 + 4 \)[/tex], which can be written in the vertex form [tex]\( a(x - h)^2 + k \)[/tex] where [tex]\( (h, k) \)[/tex] is the vertex.
2. Here, [tex]\( a = -3 \)[/tex], [tex]\( h = 5 \)[/tex], and [tex]\( k = 4 \)[/tex]. The vertex of the parabola is at [tex]\( (5, 4) \)[/tex].
3. Since the coefficient of the [tex]\( (x - 5)^2 \)[/tex] term is negative ([tex]\( -3 \)[/tex]), the parabola opens downwards, meaning the vertex is the maximum point.
Since the maximum value of the function is 4 (at [tex]\( x = 5 \)[/tex]), the function will take on all values less than or equal to 4 as [tex]\( x \)[/tex] moves away from the vertex.
Thus, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ \text{Range: } (-\infty, 4] \][/tex]
### Conclusion
Combining the findings, we have:
[tex]\[ \text{Domain: } (-\infty, \infty) \][/tex]
[tex]\[ \text{Range: } (-\infty, 4] \][/tex]
