To find the line of reflection for the function \( f(x) = x^2 + 5x - 6 \) that results in \( f'(x) = -x^2 - 5x + 6 \), we first determine the vertex of the original quadratic function.
The vertex form of a quadratic equation is given by:
\[ x = -\frac{b}{2a} \]
where \( a = 1 \) and \( b = 5 \) in this case.
Calculating the x-coordinate of the vertex:
\[ x = -\frac{5}{2 \cdot 1} = -\frac{5}{2} \]
Next, we calculate the y-coordinate of the vertex by substituting \( x = -\frac{5}{2} \) back into \( f(x) \):
\[ f\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 6 \] \[ = \frac{25}{4} - \frac{25}{2} - 6 \] \[ = \frac{25}{4} - \frac{50}{4} - \frac{24}{4} = \frac{25 - 50 - 24}{4} = \frac{-49}{4} \]
So, the vertex of \( f(x) \) is \( \left(-\frac{5}{2}, -\frac{49}{4}\right) \).
Now, we identify the vertex of \( f'(x) = -x^2 - 5x + 6 \). The coefficient of \( x^2 \) is negative, and it is also a quadratic equation:
Using the same formula for the vertex:
\[ x' = -\frac{-5}{2 \cdot -1} = -\frac{5}{-2} = \frac{5}{2} \]
Now substituting into \( f'(x) \):
\[ f'\left(\frac{5}{2}\right) = -\left(\frac{5}{2}\right)^2 - 5 \left(\frac{5}{2}\right) + 6 \] \[ = -\frac{25}{4} - \frac{25}{2} + 6 \] \[ = -\frac{25}{4} - \frac{50}{4} + \frac{24}{4} = -\frac{25 + 50 - 24}{4} = -\frac{49}{4} \]
The vertex of \( f'(x) \) is \( \left(\frac{5}{2}, -\frac{49}{4}\right) \).
The reflection occurs across a horizontal line since both vertices have the same y-coordinate \( -\frac{49}{4} \). The line we are looking for is the line that lies exactly halfway between the two x-coordinates \( -\frac{5}{2} \) and \( \frac{5}{2} \).
The line of reflection can be found using the average of the two x-coordinates, but since both vertices share the same y-coordinate, the line of reflection is simply:
\[ y = -\frac{49}{4} \]
Thus, the equation for the line of reflection is:
\[ \boxed{-\frac{49}{4}} \]
1 answer