Find Negative Area of y-4=-(x-4)²: Quadratic Function Analysis

Q: What does 'negative area' mean for a parabola?

The negative area refers to where the function values are negative (below the x-axis). For parabola $ y = -(x-4)^2 + 4 $, this happens when y

Q: How do I know which way the parabola opens?

Look at the coefficient of the squared term! If it's negative (like -1 here), the parabola opens downward. If positive, it opens upward.

Q: Why do I need to find the x-intercepts?

The x-intercepts are where the parabola crosses the x-axis (y = 0). These points divide the graph into regions where the function is positive or negative.

Q: How can I verify my answer is correct?

Test a point in each region! For example, try x = 0: $ y = -(0-4)^2 + 4 = -12 $. Since this is negative and 0

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

Use the square root method like we did here! When you have $ (x-4)^2 = 4 $, take the square root: $ x-4 = ±2 $, giving x = 6 and x = 2.

Quadratic Functions with Vertex Form Analysis

Find the negative area of the function

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the negative domain of the function

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

00:14 Arrange the equation so that it remains a function

00:19 Notice that the coefficient of X squared is negative

00:23 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 points with the x-axis, Y=0, substitute and solve

00:39 Change from negative to positive

00:44 Extract the root

00:50 When extracting a root there are always 2 solutions (positive and negative)

00:54 Solve each possibility to find the points, isolate X

01:09 These are the intersection points with the x-axis

01:14 Graph the function according to the intersection points and type of function

01:20 The function is negative while it's below the x-axis

01:30 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the negative area of the function

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

Step-by-step solution

The given equation is $y - 4 = -(x - 4)^2$ .

First, identify the vertex: the equation is in vertex form $y = -(x - 4)^2+4$ with vertex at $(4, 4)$ . The parabola opens downwards because the coefficient of $(x-4)^2$ is negative.

Next, find the x-intercepts by setting $y = 0$ :

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

$(x - 4)^2 = 4$

Taking the square root of both sides gives $x - 4 = \pm 2$ .

So, the solutions are $x = 6$ and $x = 2$ .

The parabola is negative between these x-intercepts. Since it opens downwards, the function is negative outside the interval $[2, 6]$ , i.e., for $x < 2$ or $x > 6$ .

Thus, the negative area of the parabola is for $x < 2$ or $x > 6$ , matching choice 2.

Final Answer

$x < 2$ o $x > 6$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: $y = -(x-4)^2 + 4$ opens downward with vertex at (4,4)
X-intercepts: Set y=0 to get $(x-4)^2 = 4$ , so x = 2 and x = 6
Check: Function is negative outside interval [2,6], meaning x < 2 or x > 6 ✓

Common Mistakes

Avoid these frequent errors

Confusing where function is positive vs negative
Don't assume the parabola is negative between the x-intercepts = completely wrong answer! Since this parabola opens downward, it's positive between the roots and negative outside them. Always check whether the parabola opens up or down first.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

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

With the Y

FAQ

Everything you need to know about this question

What does 'negative area' mean for a parabola?

The negative area refers to where the function values are negative (below the x-axis). For parabola $y = -(x-4)^2 + 4$ , this happens when y < 0.

How do I know which way the parabola opens?

Look at the coefficient of the squared term! If it's negative (like -1 here), the parabola opens downward. If positive, it opens upward.

Why do I need to find the x-intercepts?

The x-intercepts are where the parabola crosses the x-axis (y = 0). These points divide the graph into regions where the function is positive or negative.

How can I verify my answer is correct?

Test a point in each region! For example, try x = 0: $y = -(0-4)^2 + 4 = -12$ . Since this is negative and 0 < 2, our answer x < 2 is correct!

What if I can't factor the quadratic easily?

Use the square root method like we did here! When you have $(x-4)^2 = 4$ , take the square root: $x-4 = ±2$ , giving x = 6 and x = 2.

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