Determine the Symmetry Point of f(x) = 5x - x^2

To find the symmetry point of the quadratic function $f(x) = 5x - x^2$, we will determine the vertex of the parabola.

• Step 1: Express the quadratic function in standard form:
  The given function is already in standard form: $f(x) = -x^2 + 5x$, where $a = -1$ and $b = 5$.

• Step 2: Apply the vertex formula to find the x-coordinate of the vertex:
  For the quadratic function $ax^2 + bx + c$, the x-coordinate of the vertex is found using $x = -\frac{b}{2a}$.

• Step 3: Calculate the x-coordinate:
  $\begin{aligned} x &= -\frac{5}{2 \times (-1)} \\ &= -\frac{5}{-2} \\ &= \frac{5}{2} \end{aligned}$

• Step 4: Substitute $x = \frac{5}{2}$ back into the function to find the y-coordinate:
  $\begin{aligned} f\left(\frac{5}{2}\right) &= -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) \\ &= -\frac{25}{4} + \frac{25}{2} \\ &= -\frac{25}{4} + \frac{50}{4} \\ &= \frac{25}{4} \end{aligned}$

• Step 5: Determine the symmetry point:
  The symmetry point, and thus the vertex of the function, is $\left(\frac{5}{2}, \frac{25}{4}\right)$, or $(2\frac{1}{2}, 6\frac{1}{4})$.

Therefore, the symmetry point of the function is $(2\frac{1}{2}, 6\frac{1}{4})$.