Calculate Area Under y=-(x-3)²: Negative Quadratic Function

Q: Why can't this function ever be positive?

Because $ (x-3)^2 $ is always non-negative (≥ 0), and putting a negative sign in front makes it non-positive (≤ 0). The maximum value is 0 when x = 3.

Q: What does 'positive area' mean for a function?

Positive area refers to regions where the function values are above the x-axis (y > 0). Since this parabola opens downward and touches the x-axis at its highest point, there's no such region.

Q: How is this different from y = (x-3)²?

Without the negative sign, $ y = (x-3)^2 $ opens upward and is always ≥ 0. The negative sign flips the parabola downward, making it always ≤ 0.

Q: What if I need to find the area between the curve and x-axis?

You would calculate the absolute value of the area. Since the function is negative, the 'area' would be the integral of $ |-(x-3)^2| = (x-3)^2 $ over your interval.

Q: Can I graph this to check my answer?

Yes! Graph $ y = -(x-3)^2 $ and you'll see it's an upside-down parabola with vertex at (3,0) that never goes above the x-axis. This confirms there's no positive area.

Parabola Analysis with Negative Quadratic Functions

Find the positive area of the function

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the positive domain of the function

00:03 Use the abbreviated multiplication formulas and expand the parentheses

00:12 Negative times positive is always negative

00:15 Negative times negative is always positive

00:22 Let's look at the coefficient of X squared, it's negative

00:25 When the coefficient is negative, the function is concave down

00:30 Now we want to find the intersection points with the X-axis

00:34 At the intersection point with X-axis, Y=0, let's substitute and solve

00:40 Change from negative to positive

00:43 Take the square root to eliminate the exponent

00:47 Isolate X

00:50 This is the X value at the intersection point with the X-axis

00:56 Let's draw the function according to the intersection points and function type

01:02 The function is positive while it's above the X-axis

01:07 The function is always below the axis, therefore it's never positive

01:11 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the positive area of the function

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

Step-by-step solution

To determine the positive area of the function $y = -(x-3)^2$ , we start by examining the properties of the given function:

The function $y = -(x-3)^2$ is a quadratic in the form $y = -a(x-p)^2$ where $a = 1$ and $p = 3$ .
This represents a downward opening parabola with a vertex at $(3, 0)$ .

The function is defined as the negative of a square, so $y$ will always be less than or equal to zero. In equation form, $-(x-3)^2 \leq 0$ . We solve for the conditions under which this function could be positive:

For the expression $-(x-3)^2$ to be greater than zero, $-(x-3)^2 > 0$ , is impossible because the squared term $(x-3)^2$ is always non-negative and its negative is thus always non-positive.
At the vertex $x = 3$ , $-(x-3)^2 = 0$ , the function is exactly zero.

This analysis reveals that the function does not achieve positive values in any part of its domain.

Therefore, there is no positive area for the function $y = -(x-3)^2$ .

Final Answer

There is no positive area.

Key Points to Remember

Essential concepts to master this topic

Function Form: y = -(x-3)² opens downward with vertex at (3,0)
Analysis: Since -(x-3)² ≤ 0 always, function never positive
Verification: Test any x-value: -(2-3)² = -1 < 0 ✓

Common Mistakes

Avoid these frequent errors

Assuming negative quadratic functions have positive regions
Don't think that y = -(x-3)² can be positive somewhere = impossible result! The negative sign means the parabola opens downward and stays at or below the x-axis. Always remember that -(anything squared) ≤ 0.

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 can't this function ever be positive?

Because $(x-3)^2$ is always non-negative (≥ 0), and putting a negative sign in front makes it non-positive (≤ 0). The maximum value is 0 when x = 3.

What does 'positive area' mean for a function?

Positive area refers to regions where the function values are above the x-axis (y > 0). Since this parabola opens downward and touches the x-axis at its highest point, there's no such region.

How is this different from y = (x-3)²?

Without the negative sign, $y = (x-3)^2$ opens upward and is always ≥ 0. The negative sign flips the parabola downward, making it always ≤ 0.

What if I need to find the area between the curve and x-axis?

You would calculate the absolute value of the area. Since the function is negative, the 'area' would be the integral of $|-(x-3)^2| = (x-3)^2$ over your interval.

Can I graph this to check my answer?

Yes! Graph $y = -(x-3)^2$ and you'll see it's an upside-down parabola with vertex at (3,0) that never goes above the x-axis. This confirms there's no positive area.

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