Find Negative Area: Solving y+1=(x+3)² Parabola Problem

Q: Why is the negative area between the roots and not outside them?

Since the parabola opens upward (coefficient of $ x^2 $ is positive), it's shaped like a U. The lowest part is between the roots, so that's where \( y .

Q: How do I know the parabola opens upward?

Look at the coefficient of $ x^2 $ in $ y = (x+3)^2 - 1 $. Since it's positive (+1), the parabola opens upward like a smile!

Q: What if I chose a different test point?

Any point between -4 and -2 will work! Try $ x = -2.5 $: \( y = (-2.5+3)^2 - 1 = 0.25 - 1 = -0.75 . Same result!

Q: Why do we write the interval as -4 < x < -2?

We use strict inequalities ($ y = 0 $, not \( y .

Q: What does 'negative area' actually mean?

Negative area refers to the region where the function has negative y-values (below the x-axis). We're finding the x-interval where this happens, not calculating an actual area.

Parabola Intervals with Negative Values

Find the negative area of the function

$y+1=(x+3)^2$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the negative domain of the function

00:03 Use the shortened multiplication formulas and open the brackets

00:08 Arrange the equation so that it remains a function

00:15 Note the coefficient of X squared is positive

00:19 When the coefficient is positive, the function is smiling

00:23 Now we want to find the intersection points with the X-axis

00:27 At the intersection points with the X-axis, Y=0, we'll substitute and solve

00:37 Take the square root

00:43 When taking a square root there are always 2 solutions (positive and negative)

00:50 Solve each possibility to find the points, isolate X

01:04 These are the intersection points with the X-axis

01:09 Let's draw the function according to the intersection points and type of function:

01:22 The function is negative while it's below the X-axis

01:32 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 negative area of the function

$y+1=(x+3)^2$

Step-by-step solution

To find the negative area of the given parabola, we need to determine where the function $y = (x + 3)^2 - 1$ is below the x-axis. This corresponds to finding when the parabola is negative.

First, let's set the equation $y = (x + 3)^2 - 1$ equal to zero and solve for $x$ to find the roots:

Set $(x + 3)^2 - 1 = 0$ .
This simplifies to $(x + 3)^2 = 1$ .
Taking square roots gives $x + 3 = \pm 1$ .
Thus, $x + 3 = 1$ gives $x = -2$ , and $x + 3 = -1$ gives $x = -4$ .

The roots are $x = -4$ and $x = -2$ . The parabola opens upwards since the coefficient of $x^2$ is positive. Therefore, it is negative (below the x-axis) between these roots.

To verify, choose a test point between the roots, say $x = -3$ :

Plug into the equation: $y = ((-3) + 3)^2 - 1 = 0 - 1 = -1$ , which is negative.

Therefore, the function is negative on the interval $-4 < x < -2$ .

The correct answer is $-4 < x < -2$ .

Final Answer

$-4 < x < -2$

Key Points to Remember

Essential concepts to master this topic

Rule: Find where parabola is below x-axis by solving y = 0
Technique: Solve $(x + 3)^2 = 1$ to get roots x = -4, -2
Check: Test x = -3: $y = (-3+3)^2 - 1 = -1 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Finding only the roots without determining the correct interval
Don't just solve for x = -4 and x = -2 and assume that's the answer! This only gives the boundary points. Always check which interval between the roots makes the parabola negative by testing a point or using the upward opening direction.

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

Why is the negative area between the roots and not outside them?

Since the parabola opens upward (coefficient of $x^2$ is positive), it's shaped like a U. The lowest part is between the roots, so that's where $y < 0$ .

How do I know the parabola opens upward?

Look at the coefficient of $x^2$ in $y = (x+3)^2 - 1$ . Since it's positive (+1), the parabola opens upward like a smile!

What if I chose a different test point?

Any point between -4 and -2 will work! Try $x = -2.5$ : $y = (-2.5+3)^2 - 1 = 0.25 - 1 = -0.75 < 0$ . Same result!

Why do we write the interval as -4 < x < -2?

We use strict inequalities (< not ≤) because at x = -4 and x = -2, the parabola touches the x-axis where $y = 0$ , not $y < 0$ .

What does 'negative area' actually mean?

Negative area refers to the region where the function has negative y-values (below the x-axis). We're finding the x-interval where this happens, not calculating an actual area.

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