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

The function given is $y = -(x+3)^2 - 4$. This function represents a parabola opening downwards, with the vertex at $(-3, -4)$. Because $y = -(x+3)^2 - 4$ is a downward-opening parabola with a vertex below the x-axis at -4, it means every point on the parabola has a y-value less than zero. Thus, for all values of $x$, the function is negative.

The entire graph of this quadratic function lies below the x-axis; therefore, any form of "area" discussed here would necessarily be negative since the function never crosses into positive y-values (it is not asking for the integration under the curve relative to x-axis as such).

Consequently, the situation describes that the parabola completely lies under the x-axis across its domain: for all real $x$.

Therefore, the solution to the problem is: For all $x$, the area is negative.

To solve this problem, we'll consider the steps as follows:

• Step 1: Recognize the given function as a standard parabola centered at $(h, k) = (1, -4)$ that opens upwards.
• Step 2: The formula $y = (x-1)^2 - 4$ represents the parabola with vertex at $(1, -4)$.
• Step 3: Determine roots by solving the equation $(x-1)^2 - 4 = 0$.
• Step 4: Rearrange to $(x-1)^2 = 4$ leading to $x - 1 = \pm 2$.
• Step 5: Solve these to find roots: $x = 3$ and $x = -1$.
• Step 6: Calculate regions where the function is negative by testing intervals between the roots.

Now, we apply these steps:

Step 1: The function, expressed in its vertex form, indicates a parabola opening upwards (since the coefficient of $(x-1)^2$ is positive).

Step 2: The equation for determining where the parabola touches the x-axis is $(x−1)^2−4=0$. Solving this, we rearrange it to $(x-1)^2 = 4$.

Step 3: Solving for $(x-1)^2 = 4$, gives $x - 1 = 2$ or $x - 1 = -2$, leading to solutions $x = 3$ and $x = -1$.

Step 4: The parabola is negative between these roots. So, the inequality $(x-1)^2 < 4$ holds true for $-1 < x < 3$.

Therefore, the negative area of the function lies in the interval $-1 < x < 3$.

Conclusively, the negative domain of the function is $-1 < x < 3$.

To solve this problem, we will analyze the given quadratic function:

The function is $y = (x-4)^2 + 1$.

Step 1: Identify the vertex of the parabola.
The function $y = (x-4)^2 + 1$ is in the form $y = (x-h)^2 + k$, where $h = 4$ and $k = 1$. This gives the vertex at the point $(4, 1)$.

Step 2: Analyze the shape and direction of the parabola.
This parabola opens upwards because the coefficient of $(x-4)^2$ is positive. Hence, the vertex is the minimum point of the parabola.

Step 3: Determine when the function is positive.
Since the minimum value of the function at the vertex is $1$, and all quadratic functions in the form of $(x-h)^2 + k$ are non-negative, this means the function $y = (x-4)^2 + 1$ is always positive for any real number $x$.

Step 4: Comparison with the choices:
From the explanation, the function is always positive for all $x$. Thus, the correct choice is: For all X.

The positive area of the function covers all real numbers $x$.

Therefore, the function is positive for all $x$.

To find the positive area of the function $y = (x+5)^2-4$, follow these steps:

• Step 1: Identify when the function is greater than zero.
  We need $(x+5)^2 - 4 > 0$.
• Step 2: Find the roots of $(x+5)^2 - 4 = 0$.
  Solving, we set: $(x+5)^2 = 4$.
• Step 3: Solve for $(x+5)^2 = 4$.
  Take the square root on both sides: $x + 5 = \pm 2$.
  This gives: $x + 5 = 2$ or $x + 5 = -2$.
  Thus, $x = -3$ or $x = -7$.
• Step 4: Identify intervals.
  We need to look at intervals determined by the roots: $(-\infty, -7)$, $(-7, -3)$, and $(-3, \infty)$.
• Step 5: Determine where the function is positive by testing each interval:
  - For $x < -7$, choose $x = -8$, then $((-8)+5)^2 - 4 = 9 - 4 = 5$ (positive).
  - For $-7 < x < -3$, choose $x = -5$, then $((-5)+5)^2 - 4 = 0 - 4 = -4$ (negative).
  - For $x > -3$, choose $x = 0$, then $(5)^2 - 4 = 25 - 4 = 21$ (positive).

Therefore, the positive area of the function is for $x < -7$ and $-3 < x$.

We begin by analyzing the given function $y = (x-6)^2 + 4$. This is a parabola in vertex form, with the vertex at the point $(6, 4)$. The expression $(x-6)^2$ represents the square of $(x-6)$, and as a square, it can only be zero or positive, thus $(x-6)^2 \geq 0$.

Adding 4, the smallest value that $y$ can take is therefore $4$, which occurs when $(x-6)^2 = 0$. That means at the vertex, the function achieves its minimum value of $y = 4$, which is clearly positive.

This implies the function $y = (x-6)^2 + 4$ is never negative for any real value of $x$. Hence, the area above the x-axis is positive or non-negative for all values of $x$, which means no restrictions on the domain are necessary based on positivity.

Thus, the solution concludes that the function remains non-negative for all $x$, satisfying the condition of positive area.

Therefore, the correct choice is For all X.

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

• Step 1: Set up the inequality $-(x-4)^2 + 9 < 0$.
• Step 2: Simplify to get $(x-4)^2 > 9$.
• Step 3: Solve $(x-4)^2 > 9$ into two linear inequalities.
• Step 4: Identify the solution intervals.

Let's work through these steps:

Step 1: Set up the inequality:
We have $-(x-4)^2 + 9 < 0$.

Step 2: Simplify the inequality:
This becomes $(x-4)^2 > 9$.

Step 3: Solve the quadratic inequality:
The inequality $(x-4)^2 > 9$ can be split into two cases:
$x-4 > 3$ or $x-4 < -3$.

Simplifying these gives:
$x > 7$ or $x < 1$.

Step 4: Identify the solution intervals:
Thus, the intervals where the function is negative are $x < 1$ or $x > 7$.

Therefore, the correct solution is the interval $x < 1 , 7 < x$.

This matches choice 3.

To solve this problem, follow these steps:

• Step 1: Express the function in a more workable form.
• Step 2: Solve for the points where the function equals zero.
• Step 3: Determine where the function is negative within the identified intervals.

Let's begin:

Step 1: The given function is $y = (x+3)^2 - 36$. This can be rewritten as a difference of squares:

$y = \left((x+3) - 6\right)\left((x+3) + 6\right) = (x-3)(x+9)$.

Step 2: Set the function equal to zero to find the roots:

$(x-3)(x+9) = 0$.

The roots are $x = 3$ and $x = -9$.

Step 3: Test the sign of the quadratic in each interval determined by the roots:

- For $x < -9$, choose $x = -10$: $(x-3)(x+9)$ becomes negative, because $(-)(-)$ = positive.

- For $-9 < x < 3$, choose $x = 0$: $(x-3)(x+9)$ becomes negative because $(-)(+)$ = negative.

- For $x > 3$, choose $x = 10$: $(x-3)(x+9)$ becomes positive because $(+)(+)$ = positive.

Therefore, the function is negative within the interval $-9 < x < 3$.

Therefore, the correct answer is $-9 < x < 3$.

To solve this problem, follow these steps:

• Step 1: Find the x-intercepts by solving the equation $-(x+3)^2 + 25 = 0$.
• Step 2: Determine when the expression $-(x+3)^2 + 25$ is positive.
• Step 3: Verify the correct answer choice based on the solution.

Now let's work through each step:

Step 1: Solve $-(x+3)^2 + 25 = 0$.

Rewriting, we have:

$-(x+3)^2 = -25$

$(x+3)^2 = 25$

This gives $x+3 = \pm 5$.

Solving these, we find $x = 2$ and $x = -8$.

Step 2: Determine when $-(x+3)^2 + 25 > 0$.

We know that the parabola is downward opening, so it will be positive between the roots found, $-8$ and $2$.

Thus, the positive interval is $-8 < x < 2$.

Step 3: Verify the correct answer choice.

The correct answer, matching the solution above, is choice 1: $-8 < x < 2$.

Therefore, the positive area of the function occurs in the interval $-8 < x < 2$.

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: $3 < x < 5$.