Parabola of the Form y=(x-p)²: Identify the positive and negative domain

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 .

In the first step, we place 0 in place of Y:

0 = (x-2)²

We perform a square root:

0=x-2

x=2

And thus we reveal the point

(2, 0)

This is the vertex of the parabola.

Then we decompose the equation into standard form:

y=(x-2)²

y=x²-4x+2

Since the coefficient of x² is positive, we learn that the parabola is a minimum parabola (smiling).

If we plot the parabola, it seems that it is actually positive except for its vertex.

Therefore the domain of positivity is all X, except X≠2.

The function $y = (x+2)^2$ describes a parabola that opens upwards and has its vertex at $(-2, 0)$. Since the equation involves a perfect square, it yields only non-negative values for all $x$ and always lies on or above the x-axis. Therefore, there is no part of this parabola that crosses below the x-axis, resulting in no "negative area" above or below the x-axis.

In conclusion, the correct answer among the choices is: There is no negative area.

The given function is $y = (x+6)^2$. This function is a parabola open upwards with a vertex at $x = -6$.

The expression $(x+6)^2$ signifies the square of a number, which is always non-negative for all real numbers $x$. This means $(x+6)^2 \geq 0$.

To find when the area under the curve is positive, solve for when $(x+6)^2 > 0$. The square of any non-zero number is positive. Therefore, we require:

$(x+6) \neq 0$.

Simplifying this equation, we find:

• $(x+6) = 0$ when $x = -6$.
• Hence, $(x+6)^2$ is positive wherever $x \neq -6$.

Conclusively, the positive area of this parabola exists at all points except precisely at $x = -6$, where the function equals zero.

Looking at the multiple-choice options, the correct answer that aligns with our solution is:

$x \neq -6$.

To solve this problem, we'll follow these steps:

• Step 1: Identify where the quadratic expression $(x+5)^2$ equals zero, as this determines when $y=0$.
• Step 2: Solve the equation $(x+5)^2 = 0$ to find values of $x$.
• Step 3: Determine the values of $x$ where $(x+5)^2 > 0$.

Now, let's work through each step:

Step 1: We need to analyze the expression $(x+5)^2$.

Step 2: Solve $(x+5)^2 = 0$.
The equation simplifies to:

$(x+5) = 0$
Solve for $x$:
$x = -5$

Step 3: Determine $x$ values where $(x+5)^2$ is positive.
The expression $(x+5)^2$ is positive for any $x \neq -5$, because the square of a non-zero real number is always positive.

Therefore, the quadratic $(x+5)^2$ is positive for $x \neq -5$, meaning the positive area applies for all $x$ except $x = -5$.

The correct choice is: For each $x \neq -5$.

Therefore, the solution to the problem is For each $x \neq 5$.

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.

To find the negative area of the function $y + 4 = (x + 6)^2$, we need to determine where the function is below the x-axis, i.e., where $y < 0$.

The equation can be rewritten as:

$y = (x + 6)^2 - 4$.

We need to solve the inequality:

$(x + 6)^2 - 4 < 0$.

Adding $4$ to both sides gives:

$(x + 6)^2 < 4$.

Taking the square root gives:

$-2 < x + 6 < 2$.

Subtracting $6$ from all sides results in:

$-8 < x < -4$.

Thus, the interval where the function is below the x-axis is $-8 < x < -4$.

This corresponds to answer choice 3 from the given options.

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

To find the negative area for the function $y = -(x+4)^2$, consider when the function is negative or equals zero:

• The parabola $y = -(x+4)^2$ opens downward with the vertex at $x = -4$, when $x = -4$, $y = 0$.
• For $x \neq -4$, the expression $-(x+4)^2$ is negative because $(x + 4)^2$ is always positive or zero, and the negative sign flips it to non-positive, i.e., negative unless zero.
• Hence, everywhere except at the vertex $x = -4$, the function is negative.

The function $y = -(x+4)^2$ is negative for every point except where $x = -4$.

Therefore, the correct answer from the choice given is: For each $x \neq -4$.