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Sagot :
To determine which statements about the function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] are true, let's analyze its key characteristics.
### 1. Finding the y-intercept
The y-intercept of a function is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 - 4(0) + 2 = 2 \][/tex]
So, the y-intercept is [tex]\( 2 \)[/tex], which is indeed positive.
True Statement: The function has a positive [tex]\( y \)[/tex]-intercept.
### 2. Finding the vertex of the function
The function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] is a quadratic function in standard form [tex]\( ax^2 + bx + c \)[/tex]. The vertex form for the x-coordinate of a parabola is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here [tex]\( a = -1 \)[/tex] and [tex]\( b = -4 \)[/tex], so:
[tex]\[ x = -\frac{-4}{2(-1)} = -\frac{4}{-2} = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
Plug this value back into the function to find the y-coordinate:
[tex]\[ f(-2) = -(-2)^2 - 4(-2) + 2 = -4 + 8 + 2 = 6 \][/tex]
The vertex is at [tex]\( (-2, 6) \)[/tex], which means the function reaches its maximum value, 6, at [tex]\( x = -2 \)[/tex].
### 3. Determining the range of the function
Since the parabola opens downwards (as [tex]\( a = -1 \)[/tex] is negative), the maximum value is 6 (at the vertex). Hence, the range of the function is:
[tex]\[ \{ y \mid y \leq 6 \} \][/tex]
True Statement: The range is [tex]\( \{ y \mid y \leq 6 \} \)[/tex].
### 4. Finding intervals of increase and decrease
To determine where the function is increasing or decreasing, consider the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = -2x - 4 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ -2x - 4 = 0 \][/tex]
[tex]\[ -2x = 4 \][/tex]
[tex]\[ x = -2 \][/tex]
Since the coefficient of [tex]\( x^2 \)[/tex] in the original function is negative, the parabola opens downwards. Therefore:
- The function is increasing for [tex]\( x < -2 \)[/tex].
- The function is decreasing for [tex]\( x > -2 \)[/tex].
True Statement: The function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
### 5. Evaluating other statements
- The domain is [tex]\( \{ x \mid x \leq -2 \)[/tex] is false: The correct domain for [tex]\( f(x) \)[/tex] is all real numbers, [tex]\( \mathbb{R} \)[/tex].
Therefore, from the analysis, three true statements are:
- The range is [tex]\( \{ y \mid y \leq 6 \} \)[/tex].
- The function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
- The function has a positive y-intercept.
### 1. Finding the y-intercept
The y-intercept of a function is found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 - 4(0) + 2 = 2 \][/tex]
So, the y-intercept is [tex]\( 2 \)[/tex], which is indeed positive.
True Statement: The function has a positive [tex]\( y \)[/tex]-intercept.
### 2. Finding the vertex of the function
The function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] is a quadratic function in standard form [tex]\( ax^2 + bx + c \)[/tex]. The vertex form for the x-coordinate of a parabola is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here [tex]\( a = -1 \)[/tex] and [tex]\( b = -4 \)[/tex], so:
[tex]\[ x = -\frac{-4}{2(-1)} = -\frac{4}{-2} = 2 \][/tex]
[tex]\[ x = -2 \][/tex]
Plug this value back into the function to find the y-coordinate:
[tex]\[ f(-2) = -(-2)^2 - 4(-2) + 2 = -4 + 8 + 2 = 6 \][/tex]
The vertex is at [tex]\( (-2, 6) \)[/tex], which means the function reaches its maximum value, 6, at [tex]\( x = -2 \)[/tex].
### 3. Determining the range of the function
Since the parabola opens downwards (as [tex]\( a = -1 \)[/tex] is negative), the maximum value is 6 (at the vertex). Hence, the range of the function is:
[tex]\[ \{ y \mid y \leq 6 \} \][/tex]
True Statement: The range is [tex]\( \{ y \mid y \leq 6 \} \)[/tex].
### 4. Finding intervals of increase and decrease
To determine where the function is increasing or decreasing, consider the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = -2x - 4 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ -2x - 4 = 0 \][/tex]
[tex]\[ -2x = 4 \][/tex]
[tex]\[ x = -2 \][/tex]
Since the coefficient of [tex]\( x^2 \)[/tex] in the original function is negative, the parabola opens downwards. Therefore:
- The function is increasing for [tex]\( x < -2 \)[/tex].
- The function is decreasing for [tex]\( x > -2 \)[/tex].
True Statement: The function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
### 5. Evaluating other statements
- The domain is [tex]\( \{ x \mid x \leq -2 \)[/tex] is false: The correct domain for [tex]\( f(x) \)[/tex] is all real numbers, [tex]\( \mathbb{R} \)[/tex].
Therefore, from the analysis, three true statements are:
- The range is [tex]\( \{ y \mid y \leq 6 \} \)[/tex].
- The function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
- The function has a positive y-intercept.
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