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

Question

.(1 point)
Responses

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

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 is greater than or equal to negative 9

domain R
; range y≤−9

Which quadratic function has a range that can be described as y≤10.5
?(1 point)
Responses

f(x)=−6(x−10.5)2+4
f left parenthesis x right parenthesis equals negative 6 left parenthesis x minus 10.5 right parenthesis squared plus 4

f(x)=10.5(x−4)2−6
f left parenthesis x right parenthesis equals 10.5 left parenthesis x minus 4 right parenthesis squared minus 6

f(x)=−6(x+4)2+10.5
f left parenthesis x right parenthesis equals negative 6 left parenthesis x plus 4 right parenthesis squared plus 10.5

f(x)=6(x+4)2+10.5

Graph the equation y=−5(x)2+1
. Which of the following statements about the graph is correct?(1 point)
Responses

The parabola will have a vertex at (1,0)
.
The parabola will have a vertex at left parenthesis 1 comma 0 right parenthesis .

The parabola will have a vertex at (0,0)
.
The parabola will have a vertex at left parenthesis 0 comma 0 right parenthesis .

The parabola will open downwards.
The parabola will open downwards.

The parabola will have a vertex at (−1,0)
.

Answer 1

Let's analyze each part of your query step by step.

Part 1: Determine the domain and range of the function \( f(x) = -6(x+1)^2 - 9 \)

Domain: Since this is a quadratic function, its domain is all real numbers.

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

Range: The function is in vertex form \( f(x) = a(x-h)^2 + k \). The vertex of this function is at the point \((-1, -9)\), where \( h = -1 \) and \( k = -9 \). Since the coefficient \( a = -6 \) is negative, the parabola opens downwards. The maximum value of the function is at the vertex, which is \( -9 \).

• Range: \( y \leq -9 \)

Part 2: Which quadratic function has a range that can be described as \( y \leq 10.5 \)?

To determine the function whose range is \( y \leq 10.5 \), we need to look for a function in the form of \( f(x) = a(x-h)^2 + k \) with \( a < 0 \) and \( k = 10.5 \) (the maximum point).

1. \( f(x) = -6(x - 10.5)^2 + 4 \) → This has a maximum at \( 4\), thus \( y \leq 4\).
2. \( f(x) = 10.5(x - 4)^2 - 6 \) → This opens upwards since \( a = 10.5\), so it does not fit the requirement.
3. \( f(x) = -6(x + 4)^2 + 10.5 \) → This has a maximum at \( 10.5\), thus \( y \leq 10.5\).
4. \( f(x) = 6(x + 4)^2 + 10.5 \) → This opens upwards since \( a = 6\), so it does not fit the requirement.

Correct answer:

• \( f(x) = -6(x + 4)^2 + 10.5 \)

Part 3: Analyze the graph of the equation \( y = -5(x)^2 + 1 \)

The vertex form for a parabola is defined as \( f(x) = a(x-h)^2 + k \).

1. In this case, we have \( a = -5 \), which indicates that the parabola opens downward.
2. The standard form here shows the vertex is at \( (0, 1) \) since the equation can be viewed as \( y = -5(x - 0)^2 + 1 \).
3. Based on these observations:

• The correct statement:
  • The parabola will open downwards.

Summary of Answers

1. Domain: \( \mathbb{R} \) or "the real numbers"; Range: \( y \leq -9 \)
2. Correct function: \( f(x) = -6(x + 4)^2 + 10.5 \)
3. Correct statement about the parabola: "The parabola will open downwards."