Calculate Area Under y-4=-(x-3)²: Shifted Parabola Problem

Video Solution

Solution Steps

00:00 Find the positive domain of the function

00:03 The positive domain means above the X-axis

00:07 We'll use the abbreviated multiplication formulas and expand the brackets

00:13 Let's arrange the equation as a function

00:30 Collect like terms

00:35 Notice that the coefficient of X squared is negative

00:39 When the coefficient is negative, the function is concave down

00:44 Now we want to find the intersection points with the X-axis

00:51 At the intersection points with the X-axis, Y=0, substitute and solve

00:58 Change from negative to positive

01:05 Take the square root

01:08 Remember when taking a square root there are 2 solutions (positive and negative)

01:16 Solve each solution, isolate X

01:31 These are the intersection points with the X-axis

01:37 Let's graph the function according to the intersection points and function type:

01:44 The function is positive as long as it's above the X-axis

01:50 And this is the solution to the problem

Step-by-Step Solution

To solve this problem, we'll follow these steps:

Step 1: Identify the critical points where the parabola is zero.
Step 2: Solve the inequality to determine where the parabola is above the x-axis.
Step 3: Use these results to specify the correct domain of x-values.

Step 1: Finding the critical points where the function is zero, solve:
$- (x-3)^2 + 4 = 0$ .

Re-organize to find:
$-(x-3)^2 = -4$ simplifies to $(x-3)^2 = 4$ .

Step 2: Take the square root of both sides:
$x-3 = \pm2$ .

This gives critical points $x = 3 + 2 = 5$ and $x = 3 - 2 = 1$ .

Step 3: Since the parabola opens downwards, the function is positive between these roots. Therefore, the interval where the function is positive is:

$1 < x < 5$ .

Therefore, the solution to the problem is $1 < x < 5$ .