Question

Question 1 (2 points)

Answer 1

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

$f(x) = -(x+1)^2 + 4$

Solving (a): The vertex;

Assume the general form of a function is

$f(x) = a(x-h)^2+k$

The vertex is determined by $(h,k)$

By comparison, we have:

$-h = 1$

Solve for h, we have:

$h = -1$

$k = 4$

So, the vertex is:

$(h,k) = (-1,4)$

The domain; x is all real numbers

From the function;

$f(x) = -(x+1)^2 + 4$

This can be rewritten as:

$f(x) = 4 - (x+1)^2$

This implies that:

Whatever the value of $(x + 1)^2$ is, it will be subtracted from 4 to give y or f(x);

Hence:

y is less than 4

Option B answers the question