Calculate Area Under y=-(x-4)²+1: Quadratic Function Problem

Q: Why does the parabola open downward?

The negative coefficient in front of $ (x-4)^2 $ makes it open downward! When you see $ y = -(x-4)^2 + 1 $, the negative sign flips the parabola upside down.

Q: How do I know the parabola is positive between x = 3 and x = 5?

Since the parabola opens downward and crosses the x-axis at x = 3 and x = 5, it must be above the x-axis (positive) between these two points. The vertex at x = 4 is the highest point!

Q: What does 'positive area' actually mean?

Positive area refers to the region where the function has positive y-values (above the x-axis). We need to find the x-interval where $ y > 0 $.

Q: Can I just look at the graph instead of calculating?

While graphing helps visualize, you should always calculate the x-intercepts algebraically to get exact values. This ensures you have the precise interval boundaries!

Q: What if I made an error finding the roots?

Double-check by substituting your roots back: $ -(3-4)^2 + 1 = -1 + 1 = 0 $ and $ -(5-4)^2 + 1 = -1 + 1 = 0 $. Both should equal zero!

Quadratic Functions with Positive Domain Intervals

Find the positive area of the function

$y=-(x-4)^2+1$

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Understand the problem

Find the positive area of the function

$y=-(x-4)^2+1$

Step-by-step solution

To solve this problem, follow these steps:

Step 1: Identify the vertex form of the parabola, which is given as $y = -(x-4)^2 + 1$ .
Step 2: Convert the equation to find the x-intercepts (roots), set $y = 0$ :

We set the equation equal to zero:

$-(x-4)^2 + 1 = 0$

$-(x-4)^2 = -1$

$(x-4)^2 = 1$

Step 3: Solve for $x$ by taking the square root of both sides:

$x-4 = \pm 1$

This gives two solutions for $x$ :

$x = 4 + 1 = 5$

$x = 4 - 1 = 3$

Step 4: Determine the interval where $y$ is positive by analyzing intervals around the roots 3 and 5.
Check intervals: $x < 3$ , $3 < x < 5$ , and $x > 5$ .
Since the vertex (highest point) occurs at $x = 4$ , we know the parabola is positive between its roots from 3 to 5.

Thus, the segment of the domain where the parabola lies above the x-axis is:

$3 < x < 5$

The correct choice is: $3 < x < 5$ .

Final Answer

$3 < x < 5$

Key Points to Remember

Essential concepts to master this topic

Rule: Find x-intercepts by setting the quadratic function equal to zero
Technique: Solve $-(x-4)^2 + 1 = 0$ to get x = 3 and x = 5
Check: Test x = 4 in original function: $-(4-4)^2 + 1 = 1 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing positive area with positive y-values
Don't think 'positive area' means the entire parabola = includes negative regions too! This leads to wrong domain answers. Always find where y > 0 by solving for x-intercepts first, then identify the interval between roots where the parabola is above the x-axis.

Practice Quiz

Test your knowledge with interactive questions

Find the corresponding algebraic representation of the drawing:

FAQ

Everything you need to know about this question

Why does the parabola open downward?

The negative coefficient in front of $(x-4)^2$ makes it open downward! When you see $y = -(x-4)^2 + 1$ , the negative sign flips the parabola upside down.

How do I know the parabola is positive between x = 3 and x = 5?

Since the parabola opens downward and crosses the x-axis at x = 3 and x = 5, it must be above the x-axis (positive) between these two points. The vertex at x = 4 is the highest point!

What does 'positive area' actually mean?

Positive area refers to the region where the function has positive y-values (above the x-axis). We need to find the x-interval where $y > 0$ .

Can I just look at the graph instead of calculating?

While graphing helps visualize, you should always calculate the x-intercepts algebraically to get exact values. This ensures you have the precise interval boundaries!

What if I made an error finding the roots?

Double-check by substituting your roots back: $-(3-4)^2 + 1 = -1 + 1 = 0$ and $-(5-4)^2 + 1 = -1 + 1 = 0$ . Both should equal zero!

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