Question

Determine the domain and range of the quadratic function f(x)=−6(x+1)2−9
.(1 point)
Responses

domain R
; range y≤9
domain the real numbers ; range y less-than-or-equal-to 9

domain R
; range y≤−9
domain the real numbers ; range y less-than-or-equal-to negative 9

domain R
; range y≥9
domain the real numbers ; range y is greater than or equal to 9

domain R
; range y≥−9

Answer 1

To determine the domain and range of the quadratic function \( f(x) = -6(x+1)^2 - 9 \), we can analyze the function step by step.

**Domain:**
The function is a polynomial (a quadratic), and polynomials are defined for all real numbers. Therefore, the domain of \( f(x) \) is:

**Domain: \( \mathbb{R} \)** (or "the real numbers").

**Range:**
Next, let's determine the range. The function is in vertex form:

\[ f(x) = a(x-h)^2 + k \]

where \( (h, k) \) is the vertex of the parabola. Here, \( a = -6 \), \( h = -1 \), and \( k = -9 \). Since \( a < 0 \), the parabola opens downwards.

The vertex \((-1, -9)\) is the maximum point of the function because the parabola opens downwards.

The maximum value of \( f(x) \) is \( -9 \), and since the parabola opens downward, \( f(x) \) can go to negative infinity. Therefore, the range is:

**Range: \( y \leq -9 \)**.

In conclusion, the correct response is:
**Domain: the real numbers; Range: \( y \leq -9 \)**.