If an equation of a parabola in the xy-plane is `f(x)=-(x+2)^(2)-1` , what are the coordinates of the vertex of the parabola defined by `g(x)=f(x-2)`?

Options

A. `(0, -1)`
B. `(4, -1)`
C. `(-2, -3)`
D. `(-2, 1)`

Answer

To find the coordinates of the vertex of the parabola defined by \( g(x) = f(x-2) \), where \( f(x) = -(x+2)^2 - 1 \), we will follow these steps:

### Step 1: Understand the function \( f(x) \)
The given function is:
\[ f(x) = -(x + 2)^2 - 1 \]
This is a parabola that opens downwards, and its vertex can be found by identifying the values of \( h \) and \( k \) in the standard form \( f(x) = a(x - h)^2 + k \).

### Step 2: Identify the vertex of \( f(x) \)
In the function \( f(x) = -(x + 2)^2 - 1 \):
- The expression \( (x + 2) \) indicates a horizontal shift. The vertex \( h \) is at \( -2 \) (since \( x + 2 = 0 \) gives \( x = -2 \)).
- The constant term \( -1 \) indicates the vertical position of the vertex. Thus, \( k = -1 \).

So, the vertex of \( f(x) \) is:
\[ (-2, -1) \]

### Step 3: Find \( g(x) = f(x-2) \)
Now, we need to find \( g(x) \):
\[ g(x) = f(x-2) \]
Substituting \( x-2 \) into \( f(x) \):
\[ g(x) = -((x-2) + 2)^2 - 1 \]
This simplifies to:
\[ g(x) = -(x)^2 - 1 \]

### Step 4: Identify the vertex of \( g(x) \)
The function \( g(x) = -x^2 - 1 \) is also a parabola that opens downwards. In this case, it is in the standard form \( g(x) = a(x - h)^2 + k \), where:
- The vertex \( h \) is \( 0 \) (since there is no horizontal shift).
- The vertex \( k \) is \( -1 \).

Thus, the vertex of \( g(x) \) is:
\[ (0, -1) \]

### Final Answer
The coordinates of the vertex of the parabola defined by \( g(x) \) are:
\[ (0, -1) \]

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