Answer:
The vertex is (-1, 4), the domain is all real numbers, and the range is y less than 4
Step-by-step explanation:
Given
[tex]f(x) = -(x+1)^2 + 4[/tex]
Solving (a): The vertex;
Assume the general form of a function is
[tex]f(x) = a(x-h)^2+k[/tex]
The vertex is determined by [tex](h,k)[/tex]
By comparison, we have:
[tex]-h = 1[/tex]
Solve for h, we have:
[tex]h = -1[/tex]
[tex]k = 4[/tex]
So, the vertex is:
[tex](h,k) = (-1,4)[/tex]
The domain; x is all real numbers
From the function;
[tex]f(x) = -(x+1)^2 + 4[/tex]
This can be rewritten as:
[tex]f(x) = 4 - (x+1)^2[/tex]
This implies that:
Whatever the value of [tex](x + 1)^2[/tex] is, it will be subtracted from 4 to give y or f(x);
Hence:
y is less than 4
Option B answers the question