Calculate Triangle Area from Quadratic Graph: f(x) = -x² + 5x + 6

Video Solution

Solution Steps

00:00 Find the area of triangle COB

00:03 Point B is on the X-axis, therefore the Y value = 0

00:09 Substitute Y = 0 in the function and solve to find X

00:17 Convert from negative to positive

00:23 Factor into the appropriate trinomial

00:26 Solve what makes each factor in the trinomial equal to zero

00:29 These are the intersection points with the X-axis of the function

00:32 Point B is at X=6 since we can see it's on the positive side

00:36 Point O is the origin, its coordinates are always (0,0)

00:41 Point C is the vertex point, so we need to find it

00:45 We'll use the formula to calculate the vertex point

00:49 Let's examine the coefficients of the function

00:52 Substitute appropriate values and solve to find X

00:57 This is the X value at point C

01:00 Now let's substitute this value in the function to find the Y value at point C

01:04 Calculate and solve

01:17 This is the Y value at point C

01:25 This is point C, now we have all the points of the triangle

01:29 We'll use the formula for calculating triangle area

01:33 (height multiplied by base) divided by 2

01:44 Side DB equals the distance from origin to B (6)

01:47 The height equals the Y value of point C

01:50 Substitute and solve to find the triangle area

01:58 And this is the solution to the problem

Step-by-Step Solution

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 $36\frac{3}{4}$ .