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To determine which quadratic function represents a parabola that touches but does not cross the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex], we need to analyze the behavior of each given quadratic function at [tex]\(x = -6\)[/tex].
Here is the step-by-step approach:
1. Evaluate Each Function at [tex]\(x = -6\)[/tex]: The function must be zero at [tex]\(x = -6\)[/tex] since it touches the [tex]\(x\)[/tex]-axis at this point.
2. First Derivative Should be Zero: For the function to touch but not cross the [tex]\(x\)[/tex]-axis, there should be a critical point (a local extremum, minimum or maximum) at [tex]\(x = -6\)[/tex]. This means the first derivative of the function at [tex]\(x = -6\)[/tex] must be zero.
3. Second Derivative Non-zero: To ensure the parabola is not crossing the [tex]\(x\)[/tex]-axis and only touching it, we need the second derivative at [tex]\(x = -6\)[/tex] to be non-zero. If the second derivative is non-zero, it indicates that the critical point is indeed a minimum or maximum which supports the idea that the function touches but does not cross the [tex]\(x\)[/tex]-axis.
Let's analyze each of the given quadratic functions one by one:
### Function 1: [tex]\( f(x) = x^2 + 36x + 12 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = (-6)^2 + 36(-6) + 12 = 36 - 216 + 12 = -168 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 2: [tex]\( f(x) = x^2 - 36x - 12 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = (-6)^2 - 36(-6) - 12 = 36 + 216 - 12 = 240 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 3: [tex]\( f(x) = -x^2 + 12x + 36 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = -(-6)^2 + 12(-6) + 36 = -36 - 72 + 36 = -72 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 4: [tex]\( f(x) = -x^2 - 12x - 36 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = -(-6)^2 - 12(-6) - 36 = -36 + 72 - 36 = 0 \][/tex]
Since [tex]\( f(-6) = 0 \)[/tex], this function has the potential to touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
Next, Let's verify the derivatives for this function:
- First Derivative:
[tex]\[ f'(x) = \frac{d}{dx} (-x^2 - 12x - 36) = -2x - 12 \][/tex]
Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f'(-6) = -2(-6) - 12 = 12 - 12 = 0 \][/tex]
The first derivative at [tex]\(x = -6\)[/tex] is zero, indicating a critical point.
- Second Derivative:
[tex]\[ f''(x) = \frac{d}{dx}(-2x - 12) = -2 \][/tex]
Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f''(-6) = -2 \][/tex]
The second derivative at [tex]\(x = -6\)[/tex] is [tex]\(-2\)[/tex], which is non-zero and negative, indicating that [tex]\(x = -6\)[/tex] is a point of local maximum.
Since all the conditions are satisfied, the function [tex]\(f(x) = -x^2 - 12x - 36 \)[/tex] is the one Heather could be writing.
Therefore, the correct answer is:
[tex]\[ f(x) = -x^2 - 12x - 36 \][/tex]
Here is the step-by-step approach:
1. Evaluate Each Function at [tex]\(x = -6\)[/tex]: The function must be zero at [tex]\(x = -6\)[/tex] since it touches the [tex]\(x\)[/tex]-axis at this point.
2. First Derivative Should be Zero: For the function to touch but not cross the [tex]\(x\)[/tex]-axis, there should be a critical point (a local extremum, minimum or maximum) at [tex]\(x = -6\)[/tex]. This means the first derivative of the function at [tex]\(x = -6\)[/tex] must be zero.
3. Second Derivative Non-zero: To ensure the parabola is not crossing the [tex]\(x\)[/tex]-axis and only touching it, we need the second derivative at [tex]\(x = -6\)[/tex] to be non-zero. If the second derivative is non-zero, it indicates that the critical point is indeed a minimum or maximum which supports the idea that the function touches but does not cross the [tex]\(x\)[/tex]-axis.
Let's analyze each of the given quadratic functions one by one:
### Function 1: [tex]\( f(x) = x^2 + 36x + 12 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = (-6)^2 + 36(-6) + 12 = 36 - 216 + 12 = -168 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 2: [tex]\( f(x) = x^2 - 36x - 12 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = (-6)^2 - 36(-6) - 12 = 36 + 216 - 12 = 240 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 3: [tex]\( f(x) = -x^2 + 12x + 36 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = -(-6)^2 + 12(-6) + 36 = -36 - 72 + 36 = -72 \][/tex]
Since [tex]\( f(-6) \neq 0 \)[/tex], this function does not touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
### Function 4: [tex]\( f(x) = -x^2 - 12x - 36 \)[/tex]
1. Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f(-6) = -(-6)^2 - 12(-6) - 36 = -36 + 72 - 36 = 0 \][/tex]
Since [tex]\( f(-6) = 0 \)[/tex], this function has the potential to touch the [tex]\(x\)[/tex]-axis at [tex]\(x = -6\)[/tex].
Next, Let's verify the derivatives for this function:
- First Derivative:
[tex]\[ f'(x) = \frac{d}{dx} (-x^2 - 12x - 36) = -2x - 12 \][/tex]
Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f'(-6) = -2(-6) - 12 = 12 - 12 = 0 \][/tex]
The first derivative at [tex]\(x = -6\)[/tex] is zero, indicating a critical point.
- Second Derivative:
[tex]\[ f''(x) = \frac{d}{dx}(-2x - 12) = -2 \][/tex]
Evaluate at [tex]\(x = -6\)[/tex]:
[tex]\[ f''(-6) = -2 \][/tex]
The second derivative at [tex]\(x = -6\)[/tex] is [tex]\(-2\)[/tex], which is non-zero and negative, indicating that [tex]\(x = -6\)[/tex] is a point of local maximum.
Since all the conditions are satisfied, the function [tex]\(f(x) = -x^2 - 12x - 36 \)[/tex] is the one Heather could be writing.
Therefore, the correct answer is:
[tex]\[ f(x) = -x^2 - 12x - 36 \][/tex]
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