Zeros of a Fuction: Calculating geometric shapes

To find the area of triangle COB on the graph of the function $f(x) = -x^2 + 5x + 6$, follow these steps:

• Step 1: Find the x-intercepts
  To find the x-intercepts (points O and B), solve $-x^2 + 5x + 6 = 0$.
• Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = 5$, $c = 6$:
  $x = \frac{-5 \pm \sqrt{25 - 4 \cdot (-1) \cdot 6}}{2 \cdot -1} = \frac{-5 \pm \sqrt{49}}{-2}$
• Simplifying gives $x = \frac{-5 \pm 7}{-2}$.
  This results in the roots $x = 6$ and $x = -1$.
• The x-intercepts are points $O(0, 0)$ by definition at origin, and $B(6, 0)$.
• Step 2: Find the vertex
  The x-coordinate of the vertex $C$ is found using $x = -\frac{b}{2a}$:
  $x = -\frac{5}{2 \cdot (-1)} = \frac{5}{2}$.
  Substitute $x = \frac{5}{2}$ back into the function to find the y-coordinate:
  $f\left(\frac{5}{2}\right) = -\left(\frac{5}{2}\right)^2 + 5\left(\frac{5}{2}\right) + 6$.
• This simplifies to: $f\left(\frac{5}{2}\right) = -\frac{25}{4} + \frac{25}{2} + 6 = \frac{49}{4}$.
• Therefore, the vertex is $C\left(\frac{5}{2}, \frac{49}{4}\right)$.
• Step 3: Calculate the area of triangle COB
  Using the formula for the area of a triangle $\frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$, where $O(0, 0)$, $C\left(\frac{5}{2}, \frac{49}{4}\right)$, $B(6, 0)$:
• $\text{Area} = \frac{1}{2} \left| 0\left(0-\frac{49}{4}\right) + \frac{5}{2}(0 - 0) + 6\left(\frac{49}{4} - 0\right) \right|$
• This simplifies to: $\text{Area} = \frac{1}{2} \left| 6 \cdot \frac{49}{4} \right| = \frac{1}{2} \cdot \frac{294}{4} = \frac{147}{4} = 36\frac{3}{4}$.

Therefore, the area of triangle COB is $36\frac{3}{4}$.