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Sagot :
Let's analyze the function [tex]\( f(x) = (x + 4)(x - 6) \)[/tex] step-by-step to determine the correctness of each statement.
First, let's rewrite the function in standard quadratic form:
[tex]\[ f(x) = (x + 4)(x - 6) = x^2 - 6x + 4x - 24 = x^2 - 2x - 24 \][/tex]
### Finding the Vertex
The vertex formula for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -24 \)[/tex]. Let's calculate the x-coordinate of the vertex:
[tex]\[ x = -\frac{-2}{2 \times 1} = \frac{2}{2} = 1 \][/tex]
Now, substitute [tex]\( x = 1 \)[/tex] back into the function to find the y-coordinate:
[tex]\[ f(1) = (1 + 4)(1 - 6) = 5 \cdot (-5) = -25 \][/tex]
So, the vertex of the function is at [tex]\( (1, -25) \)[/tex].
### Analyzing the Statements
1. The vertex of the function is at [tex]\((1, -25)\)[/tex].
- This statement is true, as we calculated the vertex to be at [tex]\( (1, -25) \)[/tex].
2. The vertex of the function is at [tex]\((1, -24)\)[/tex].
- This statement is false, as the correct vertex is [tex]\( (1, -25) \)[/tex].
### Investigating the Intervals
Let's determine where the function is increasing or decreasing. Since [tex]\( f(x) = x^2 - 2x - 24 \)[/tex] is a parabola that opens upwards (a > 0), it decreases to the vertex and then increases.
3. The graph is increasing only on the interval [tex]\( -4 < x < 6 \)[/tex].
- To determine where the graph is increasing or decreasing, note that the function decreases until the vertex at [tex]\( x = 1 \)[/tex] and increases after that. Therefore, this statement is false. It increases on the interval [tex]\( 1 < x < \infty \)[/tex], not [tex]\( -4 < x < 6 \)[/tex].
4. The graph is positive only on one interval, where [tex]\( x < -4 \)[/tex].
- To see where the function is positive:
Factor the polynomial into roots:
[tex]\[ f(x) = (x + 4)(x - 6) \][/tex]
Setting [tex]\( f(x) > 0 \)[/tex], we test:
- When [tex]\( x + 4 > 0 \)[/tex] and [tex]\( x - 6 > 0 \)[/tex] (i.e., [tex]\( x > 6 \)[/tex])
- When [tex]\( x + 4 < 0 \)[/tex] and [tex]\( x - 6 < 0 \)[/tex] (i.e., [tex]\( x < -4 \)[/tex])
So, the function is positive on the intervals [tex]\( x < -4 \)[/tex] and [tex]\( x > 6 \)[/tex]. This statement is false because there is another positive interval, [tex]\( x > 6 \)[/tex].
5. The graph is negative on the entire interval [tex]\( -4 < x < 6 \)[/tex].
- Between the roots [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex], the function takes negative values, as the quadratic is below the x-axis. This statement is true.
Conclusion:
The true statements are:
- The vertex of the function is at [tex]\((1, -25)\)[/tex].
- The graph is negative on the entire interval [tex]\( -4 < x < 6 \)[/tex].
First, let's rewrite the function in standard quadratic form:
[tex]\[ f(x) = (x + 4)(x - 6) = x^2 - 6x + 4x - 24 = x^2 - 2x - 24 \][/tex]
### Finding the Vertex
The vertex formula for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -24 \)[/tex]. Let's calculate the x-coordinate of the vertex:
[tex]\[ x = -\frac{-2}{2 \times 1} = \frac{2}{2} = 1 \][/tex]
Now, substitute [tex]\( x = 1 \)[/tex] back into the function to find the y-coordinate:
[tex]\[ f(1) = (1 + 4)(1 - 6) = 5 \cdot (-5) = -25 \][/tex]
So, the vertex of the function is at [tex]\( (1, -25) \)[/tex].
### Analyzing the Statements
1. The vertex of the function is at [tex]\((1, -25)\)[/tex].
- This statement is true, as we calculated the vertex to be at [tex]\( (1, -25) \)[/tex].
2. The vertex of the function is at [tex]\((1, -24)\)[/tex].
- This statement is false, as the correct vertex is [tex]\( (1, -25) \)[/tex].
### Investigating the Intervals
Let's determine where the function is increasing or decreasing. Since [tex]\( f(x) = x^2 - 2x - 24 \)[/tex] is a parabola that opens upwards (a > 0), it decreases to the vertex and then increases.
3. The graph is increasing only on the interval [tex]\( -4 < x < 6 \)[/tex].
- To determine where the graph is increasing or decreasing, note that the function decreases until the vertex at [tex]\( x = 1 \)[/tex] and increases after that. Therefore, this statement is false. It increases on the interval [tex]\( 1 < x < \infty \)[/tex], not [tex]\( -4 < x < 6 \)[/tex].
4. The graph is positive only on one interval, where [tex]\( x < -4 \)[/tex].
- To see where the function is positive:
Factor the polynomial into roots:
[tex]\[ f(x) = (x + 4)(x - 6) \][/tex]
Setting [tex]\( f(x) > 0 \)[/tex], we test:
- When [tex]\( x + 4 > 0 \)[/tex] and [tex]\( x - 6 > 0 \)[/tex] (i.e., [tex]\( x > 6 \)[/tex])
- When [tex]\( x + 4 < 0 \)[/tex] and [tex]\( x - 6 < 0 \)[/tex] (i.e., [tex]\( x < -4 \)[/tex])
So, the function is positive on the intervals [tex]\( x < -4 \)[/tex] and [tex]\( x > 6 \)[/tex]. This statement is false because there is another positive interval, [tex]\( x > 6 \)[/tex].
5. The graph is negative on the entire interval [tex]\( -4 < x < 6 \)[/tex].
- Between the roots [tex]\( x = -4 \)[/tex] and [tex]\( x = 6 \)[/tex], the function takes negative values, as the quadratic is below the x-axis. This statement is true.
Conclusion:
The true statements are:
- The vertex of the function is at [tex]\((1, -25)\)[/tex].
- The graph is negative on the entire interval [tex]\( -4 < x < 6 \)[/tex].
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