To analyze the function [tex]\( f(x) = -(x + 1)^2 + 4 \)[/tex], we need to identify the vertex, domain, and range step-by-step.
1. Identifying the Vertex:
The given function [tex]\( f(x) = -(x + 1)^2 + 4 \)[/tex] is in the vertex form of a quadratic function:
[tex]\[ f(x) = a(x - h)^2 + k, \][/tex]
where [tex]\( (h, k) \)[/tex] is the vertex.
Comparing [tex]\( f(x) = -(x + 1)^2 + 4 \)[/tex] with the vertex form:
- The expression inside the parentheses is [tex]\( (x + 1) \)[/tex], which can be rewritten as [tex]\( (x - (-1)) \)[/tex].
- Therefore, [tex]\( h = -1 \)[/tex].
- The constant outside the parentheses is [tex]\( +4 \)[/tex], so [tex]\( k = 4 \)[/tex].
Thus, the vertex of the function is at:
[tex]\[
(-1, 4).
\][/tex]
2. Identifying the Domain:
Quardratic functions, such as this one, are defined for all real values of [tex]\( x \)[/tex]. Therefore, the domain is:
[tex]\[
\text{all real numbers} \quad (or \; \mathbb{R}).
\][/tex]
3. Identifying the Range:
Since the coefficient of the quadratic term [tex]\((x + 1)^2\)[/tex] is negative ([tex]\(-1\)[/tex]), the parabola opens downwards.
The vertex [tex]\((h, k)\)[/tex] represents the maximum point of the function, given that a downward parabola has its maximum at its vertex.
Therefore, the maximum value of [tex]\( f(x) \)[/tex] is 4 (the y-value of the vertex). This means that the values of [tex]\( f(x) \)[/tex] can be anything less than or equal to 4.
Thus, the range is:
[tex]\[
y \leq 4.
\][/tex]
In summary:
The vertex is [tex]\((-1, 4)\)[/tex], the domain is all real numbers, and the range is [tex]\( y \leq 4 \)[/tex].
So the correct choices are:
- The vertex is [tex]\((-1, 4)\)[/tex].
- The domain is all real numbers.
- The range is [tex]\( y \leq 4 \)[/tex].
