To find the line of reflection for the function \( f(x) = x^2 + 5x - 6 \) and its reflection \( f'(x) = -x^2 - 5x + 6 \), we will first need to analyze the properties of these functions.
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Find the vertex of the original function: The function \( f(x) = x^2 + 5x - 6 \) is a quadratic function, and its vertex can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 5 \).
\[ x = -\frac{5}{2(1)} = -\frac{5}{2} \]
Next, substitute this \( x \) value back into \( f(x) \) to find the corresponding \( y \)-coordinate:
\[ f\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 6 \] \[ = \frac{25}{4} - \frac{25}{2} - 6 \] \[ = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} = \frac{25 - 50 - 24}{4} = \frac{-49}{4} \]
Thus, the vertex of \( f(x) \) is \( \left(-\frac{5}{2}, -\frac{49}{4}\right) \).
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Find the vertex of the reflected function: The reflected function \( f'(x) = -x^2 - 5x + 6 \) is also a quadratic function. Its vertex can be found using the same method, but we first compute \( x = -\frac{b}{2a} \) with \( a = -1 \) and \( b = -5 \):
\[ x = -\frac{-5}{2(-1)} = \frac{5}{2} \]
Now, substitute for \( f' \):
\[ f'\left(\frac{5}{2}\right) = -\left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 6 \] \[ = -\frac{25}{4} - \frac{25}{2} + 6 \] \[ = -\frac{25}{4} - \frac{50}{4} + \frac{24}{4} = \frac{-25 - 50 + 24}{4} = \frac{-51}{4} \]
Thus, the vertex of \( f'(x) \) is \( \left(\frac{5}{2}, -\frac{51}{4}\right) \).
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Finding the line of reflection: The line of reflection must be the perpendicular bisector between the two vertices \( \left(-\frac{5}{2}, -\frac{49}{4}\right) \) and \( \left(\frac{5}{2}, -\frac{51}{4}\right) \). The midpoint \( M \) of these points can be computed as follows:
\[ M_x = \frac{-\frac{5}{2} + \frac{5}{2}}{2} = 0 \] \[ M_y = \frac{-\frac{49}{4} - \frac{51}{4}}{2} = \frac{-\frac{100}{4}}{2} = -\frac{50}{4} = -\frac{25}{2} \]
The midpoint \( M \) is \( (0, -\frac{25}{2}) \).
The slope of the line segment connecting the two vertices is given by:
\[ \text{slope} = \frac{-\frac{51}{4} + \frac{49}{4}}{\frac{5}{2}+ \frac{5}{2}} = \frac{-\frac{2}{4}}{5} = \frac{-1}{10} \]
The slope of the perpendicular bisector is the negative reciprocal of \(-\frac{1}{10}\), which is \( 10 \).
Using the point-slope form of a line at the midpoint:
\[ y + \frac{25}{2} = 10(x - 0) \] \[ y = 10x - \frac{25}{2} \]
Thus, the line of reflection is:
\[ \boxed{y = 10x - \frac{25}{2}} \]