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

Q: Why does the parabola open downward?

The negative coefficient in front of $-(x-3)^2$ makes the parabola open downward. When the coefficient of the squared term is negative, the parabola always opens down!

Q: How do I find where the function crosses the x-axis?

Set the entire function equal to zero: $y = -(x-3)^2 + 4 = 0$. Then solve for x by isolating the squared term and taking the square root of both sides.

Q: What does 'positive area' mean in this problem?

Positive area refers to the interval where the function values are above the x-axis (where y > 0). Since this parabola opens downward, it's only positive between its two zeros.

Q: Why is the vertex at (3, 4)?

The vertex form $y - 4 = -(x - 3)^2$ directly shows the vertex coordinates: (h, k) = (3, 4). The values inside the parentheses give the x-coordinate, and the constant gives the y-coordinate.

Q: How can I verify my interval is correct?

Pick a test point between your zeros, like x = 3: $y = -(3-3)^2 + 4 = 4 > 0$ ✓. Pick a point outside, like x = 0: \(y = -(0-3)^2 + 4 = -5 ✓

Q: What if I can't factor the equation easily?

Use the square root method! Isolate the squared term: $(x-3)^2 = 4$, then take the square root of both sides: $x-3 = ±2$. This gives you both solutions directly.

Parabola Area with Vertex Form

Find the positive area of the function

$y-4=-(x-3)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:07 Let's find the positive domain of the function.

00:10 This means, where the graph is above the X-axis.

00:14 We'll use shortcut multiplication formulas and expand the brackets.

00:20 First, we'll arrange the equation as a function.

00:37 Next, collect like terms.

00:42 Notice the coefficient of X squared is negative.

00:46 A negative coefficient means the curve faces down.

00:51 Now, find where it crosses the X-axis.

00:58 At these points, Y equals zero. Substitute and solve.

01:05 Change from negative to positive.

01:12 Take the square root.

01:15 Remember, the square root gives two answers: positive and negative.

01:23 Solve each separately to find X.

01:38 These are where the graph meets the X-axis.

01:44 Now, let's draw the graph using these points and the curve's direction.

01:51 The graph is positive where it's above the X-axis.

01:57 And that's the solution to our problem! Great job!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive area of the function

$y-4=-(x-3)^2$

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 $1 < x < 5$ .

Final Answer

$1 < x < 5$

Key Points to Remember

Essential concepts to master this topic

Rule: Find zeros by setting transformed parabola equal to zero
Technique: Solve $-(x-3)^2 + 4 = 0$ gives critical points x = 1, 5
Check: Downward parabola is positive between roots: 1 < x < 5 ✓

Common Mistakes

Avoid these frequent errors

Forgetting the parabola opens downward
Don't assume the parabola is positive outside the roots = wrong interval! The negative coefficient means it opens downward, so it's only positive between the zeros. Always check the direction the parabola opens before determining the positive region.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x-2)^2 $$

With the X

FAQ

Everything you need to know about this question

Why does the parabola open downward?

The negative coefficient in front of $-(x-3)^2$ makes the parabola open downward. When the coefficient of the squared term is negative, the parabola always opens down!

How do I find where the function crosses the x-axis?

Set the entire function equal to zero: $y = -(x-3)^2 + 4 = 0$ . Then solve for x by isolating the squared term and taking the square root of both sides.

What does 'positive area' mean in this problem?

Positive area refers to the interval where the function values are above the x-axis (where y > 0). Since this parabola opens downward, it's only positive between its two zeros.

Why is the vertex at (3, 4)?

The vertex form $y - 4 = -(x - 3)^2$ directly shows the vertex coordinates: (h, k) = (3, 4). The values inside the parentheses give the x-coordinate, and the constant gives the y-coordinate.

How can I verify my interval is correct?

Pick a test point between your zeros, like x = 3: $y = -(3-3)^2 + 4 = 4 > 0$ ✓. Pick a point outside, like x = 0: $y = -(0-3)^2 + 4 = -5 < 0$ ✓

What if I can't factor the equation easily?

Use the square root method! Isolate the squared term: $(x-3)^2 = 4$ , then take the square root of both sides: $x-3 = ±2$ . This gives you both solutions directly.

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