Answer :
To determine which function Heather could be writing, we need to identify which of the given quadratic functions touches but does not cross the [tex]$x$[/tex]-axis at [tex]$x = -6$[/tex].
This condition means that the function value at [tex]$x = -6$[/tex] should be 0, and its derivative (indicating the slope of the function) at [tex]$x = -6$[/tex] should also be 0. Let's check each function one by one.
### Function 1: \( f(x) = x^2 + 36x + 12 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = (-6)^2 + 36(-6) + 12 = 36 - 216 + 12 = -168 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 2: \( f(x) = x^2 - 36x - 12 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = (-6)^2 - 36(-6) - 12 = 36 + 216 - 12 = 240 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 3: \( f(x) = -x^2 + 12x + 36 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = -(-6)^2 + 12(-6) + 36 = -36 - 72 + 36 = -72 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 4: \( f(x) = -x^2 - 12x - 36 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = -(-6)^2 - 12(-6) - 36 = -36 + 72 - 36 = 0 \][/tex]
Since \( f(-6) = 0 \), this function touches the [tex]$x$[/tex]-axis at \( x = -6 \).
Next, we check the derivative of each function at \( x = -6 \). The derivative of a quadratic function \( ax^2 + bx + c \) is \( 2ax + b \).
### Function 4: \( f(x) = -x^2 - 12x - 36 \)
1. The derivative of \( f(x) \) is:
[tex]\[ f'(x) = -2x - 12 \][/tex]
2. Calculate \( f'(-6) \):
[tex]\[ f'(-6) = -2(-6) - 12 = 12 - 12 = 0 \][/tex]
Since both \( f(-6) = 0 \) and \( f'(-6) = 0 \), the function \( f(x) = -x^2 - 12x - 36 \) touches but does not cross the [tex]$x$[/tex]-axis at \( x = -6 \).
Therefore, the function Heather could be writing is:
[tex]\[ f(x) = -x^2 - 12x - 36 \][/tex]
This condition means that the function value at [tex]$x = -6$[/tex] should be 0, and its derivative (indicating the slope of the function) at [tex]$x = -6$[/tex] should also be 0. Let's check each function one by one.
### Function 1: \( f(x) = x^2 + 36x + 12 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = (-6)^2 + 36(-6) + 12 = 36 - 216 + 12 = -168 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 2: \( f(x) = x^2 - 36x - 12 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = (-6)^2 - 36(-6) - 12 = 36 + 216 - 12 = 240 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 3: \( f(x) = -x^2 + 12x + 36 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = -(-6)^2 + 12(-6) + 36 = -36 - 72 + 36 = -72 \][/tex]
Since \( f(-6) \neq 0 \), this function does not touch the [tex]$x$[/tex]-axis at \( x = -6 \).
### Function 4: \( f(x) = -x^2 - 12x - 36 \)
1. Calculate \( f(-6) \):
[tex]\[ f(-6) = -(-6)^2 - 12(-6) - 36 = -36 + 72 - 36 = 0 \][/tex]
Since \( f(-6) = 0 \), this function touches the [tex]$x$[/tex]-axis at \( x = -6 \).
Next, we check the derivative of each function at \( x = -6 \). The derivative of a quadratic function \( ax^2 + bx + c \) is \( 2ax + b \).
### Function 4: \( f(x) = -x^2 - 12x - 36 \)
1. The derivative of \( f(x) \) is:
[tex]\[ f'(x) = -2x - 12 \][/tex]
2. Calculate \( f'(-6) \):
[tex]\[ f'(-6) = -2(-6) - 12 = 12 - 12 = 0 \][/tex]
Since both \( f(-6) = 0 \) and \( f'(-6) = 0 \), the function \( f(x) = -x^2 - 12x - 36 \) touches but does not cross the [tex]$x$[/tex]-axis at \( x = -6 \).
Therefore, the function Heather could be writing is:
[tex]\[ f(x) = -x^2 - 12x - 36 \][/tex]
