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Sagot :
To determine 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.
1. Domain:
The domain of a quadratic function (parabola) is always all real numbers, because you can substitute any real number for [tex]\( x \)[/tex]. Thus, the domain is [tex]\( \{x \mid x \in \mathbb{R}\} \)[/tex].
2. Vertex:
The vertex of the parabola [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] can be found using the formula for the [tex]\( x \)[/tex]-coordinate of the vertex, [tex]\( x = -\frac{b}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex] and [tex]\( b = -4 \)[/tex]:
[tex]\[ x = -\frac{-4}{2(-1)} = \frac{4}{2} = 2 \][/tex]
The function given is [tex]\( a = -1 \)[/tex], so the correct calculation should be:
[tex]\[ x = -\frac{-4}{2(-1)} = -2 \][/tex]
Substituting [tex]\( x = -2 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f(-2) = -(-2)^2 - 4(-2) + 2 = -4 + 8 + 2 = 6 \][/tex]
The vertex is at [tex]\( (-2, 6) \)[/tex].
3. Range:
Since the parabola opens downwards (determined by the negative coefficient of [tex]\( x^2 \)[/tex]), the maximum value of [tex]\( y \)[/tex] is at the vertex, 6. Therefore, the range is [tex]\( \{y \mid y \leq 6\} \)[/tex].
4. Intervals of Increase and Decrease:
- The function is increasing on the interval to the left of the vertex. Thus, the function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
- The function is decreasing on the interval to the right of the vertex. Thus, the function is decreasing over the interval [tex]\( (-2, \infty) \)[/tex].
5. y-intercept:
The [tex]\( y \)[/tex]-intercept is calculated by evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -(0)^2 - 4(0) + 2 = 2 \][/tex]
So, the y-intercept is 2, which is positive.
Therefore, the three true statements from the options provided 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.
Thus, the correct options 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.
These correspond to the statements (2), (3), and (5).
1. Domain:
The domain of a quadratic function (parabola) is always all real numbers, because you can substitute any real number for [tex]\( x \)[/tex]. Thus, the domain is [tex]\( \{x \mid x \in \mathbb{R}\} \)[/tex].
2. Vertex:
The vertex of the parabola [tex]\( f(x) = -x^2 - 4x + 2 \)[/tex] can be found using the formula for the [tex]\( x \)[/tex]-coordinate of the vertex, [tex]\( x = -\frac{b}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex] and [tex]\( b = -4 \)[/tex]:
[tex]\[ x = -\frac{-4}{2(-1)} = \frac{4}{2} = 2 \][/tex]
The function given is [tex]\( a = -1 \)[/tex], so the correct calculation should be:
[tex]\[ x = -\frac{-4}{2(-1)} = -2 \][/tex]
Substituting [tex]\( x = -2 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate of the vertex:
[tex]\[ f(-2) = -(-2)^2 - 4(-2) + 2 = -4 + 8 + 2 = 6 \][/tex]
The vertex is at [tex]\( (-2, 6) \)[/tex].
3. Range:
Since the parabola opens downwards (determined by the negative coefficient of [tex]\( x^2 \)[/tex]), the maximum value of [tex]\( y \)[/tex] is at the vertex, 6. Therefore, the range is [tex]\( \{y \mid y \leq 6\} \)[/tex].
4. Intervals of Increase and Decrease:
- The function is increasing on the interval to the left of the vertex. Thus, the function is increasing over the interval [tex]\( (-\infty, -2) \)[/tex].
- The function is decreasing on the interval to the right of the vertex. Thus, the function is decreasing over the interval [tex]\( (-2, \infty) \)[/tex].
5. y-intercept:
The [tex]\( y \)[/tex]-intercept is calculated by evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -(0)^2 - 4(0) + 2 = 2 \][/tex]
So, the y-intercept is 2, which is positive.
Therefore, the three true statements from the options provided 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.
Thus, the correct options 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.
These correspond to the statements (2), (3), and (5).
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