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To evaluate which statements about the graph of the function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] are true, let's analyze the function step by step.
### Step 1: Domain
The function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] is a polynomial, which means its domain is all real numbers, [tex]\( \mathbb{R} \)[/tex]. Therefore, the statement "The domain is [tex]\( \{x \mid x \leq -2\} \)[/tex]" is false.
### Step 2: Range
A quadratic function in the form of [tex]\( f(x) = ax^2 + bx + c \)[/tex] opens upwards if [tex]\( a > 0 \)[/tex] and opens downwards if [tex]\( a < 0 \)[/tex]. Since [tex]\( a = -1 \)[/tex] here, the function opens downwards. The vertex form of a quadratic function indicates the maximum or minimum point. For our function:
[tex]\[ x_{\text{vertex}} = \frac{-b}{2a} = \frac{4}{2(-1)} = 2 \][/tex]
[tex]\[ y_{\text{vertex}} = f(2) = -2^2 - 4(2) + 2 = -4 - 8 + 2 = -10 \][/tex]
Thus, the maximum value (since the parabola opens downwards) is [tex]\( -10 \)[/tex], and the range of the function is [tex]\( \{y \mid y \leq -10\} \)[/tex]. Therefore, the statement "The range is [tex]\( \{y \mid y \leq 6\} \)[/tex]" is false.
### Step 3: Increasing and Decreasing Intervals
To determine where the function is increasing or decreasing, we look at the first derivative:
[tex]\[ f'(x) = -2x - 4 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -2x - 4 = 0 \implies x = -2 \][/tex]
The function changes from increasing to decreasing at [tex]\( x = -2 \)[/tex].
- For [tex]\( x < -2 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] (the function is increasing).
- For [tex]\( x > -2 \)[/tex], [tex]\( f'(x) < 0 \)[/tex] (the function is decreasing).
Thus, the statement "The function is increasing over the interval [tex]\((- \infty, -2)\)[/tex]" is true, and the statement "The function is decreasing over the interval [tex]\((-4, \infty)\)[/tex]" is false.
### Step 4: Y-intercept
The y-intercept is found by evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 - 4(0) + 2 = 2 \][/tex]
Since [tex]\( f(0) = 2 \)[/tex], the y-intercept is positive. Therefore, the statement "The function has a positive y-intercept" is true.
### Conclusion
The three true statements are:
- The function is increasing over the interval [tex]\((- \infty, -2)\)[/tex].
- The function has a positive [tex]\( y \)[/tex]-intercept.
### Step 1: Domain
The function [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] is a polynomial, which means its domain is all real numbers, [tex]\( \mathbb{R} \)[/tex]. Therefore, the statement "The domain is [tex]\( \{x \mid x \leq -2\} \)[/tex]" is false.
### Step 2: Range
A quadratic function in the form of [tex]\( f(x) = ax^2 + bx + c \)[/tex] opens upwards if [tex]\( a > 0 \)[/tex] and opens downwards if [tex]\( a < 0 \)[/tex]. Since [tex]\( a = -1 \)[/tex] here, the function opens downwards. The vertex form of a quadratic function indicates the maximum or minimum point. For our function:
[tex]\[ x_{\text{vertex}} = \frac{-b}{2a} = \frac{4}{2(-1)} = 2 \][/tex]
[tex]\[ y_{\text{vertex}} = f(2) = -2^2 - 4(2) + 2 = -4 - 8 + 2 = -10 \][/tex]
Thus, the maximum value (since the parabola opens downwards) is [tex]\( -10 \)[/tex], and the range of the function is [tex]\( \{y \mid y \leq -10\} \)[/tex]. Therefore, the statement "The range is [tex]\( \{y \mid y \leq 6\} \)[/tex]" is false.
### Step 3: Increasing and Decreasing Intervals
To determine where the function is increasing or decreasing, we look at the first derivative:
[tex]\[ f'(x) = -2x - 4 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -2x - 4 = 0 \implies x = -2 \][/tex]
The function changes from increasing to decreasing at [tex]\( x = -2 \)[/tex].
- For [tex]\( x < -2 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] (the function is increasing).
- For [tex]\( x > -2 \)[/tex], [tex]\( f'(x) < 0 \)[/tex] (the function is decreasing).
Thus, the statement "The function is increasing over the interval [tex]\((- \infty, -2)\)[/tex]" is true, and the statement "The function is decreasing over the interval [tex]\((-4, \infty)\)[/tex]" is false.
### Step 4: Y-intercept
The y-intercept is found by evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -0^2 - 4(0) + 2 = 2 \][/tex]
Since [tex]\( f(0) = 2 \)[/tex], the y-intercept is positive. Therefore, the statement "The function has a positive y-intercept" is true.
### Conclusion
The three true statements are:
- The function is increasing over the interval [tex]\((- \infty, -2)\)[/tex].
- The function has a positive [tex]\( y \)[/tex]-intercept.
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