Parabola of the Form y=x²+c: Identify positive and negative domains on a graph

To solve this problem, follow these steps:

• Step 1: Understand that the function $f(x) = -x^2$ describes a downward-facing parabola with its vertex at the origin (0,0).
• Step 2: Analyze where the function has negative values. For $f(x) = -x^2$, any value of $x \neq 0$ will yield negative output since squaring any nonzero value is positive and multiplying by $-1$ gives a negative result.
• Step 3: Recognize that the point $x = 0$ results in $f(0) = 0$, meaning exactly at the vertex there is no negative value.

From these observations, you can conclude that $f(x)$ is negative for all $x \neq 0$.

Therefore, the solution to the problem is $x \neq 0$ .

To determine where the function $f(x) = x^2$ is positive, we consider the nature of this parabolic graph, which opens upwards.

Step 1: Recognize that the function $f(x) = x^2$ outputs non-negative values for any real number $x$. The graph of this function is a U-shaped parabola.

Step 2: Analyze the values of the function:
- For $x = 0$, $f(0) = 0^2 = 0$.
- For $x \neq 0$, $f(x) = x^2 > 0$, because squaring any non-zero real number results in a positive value.

Therefore, the function is positive for all $x$ except at $x = 0$, where it is zero.

Step 3: Based on the comparison given in the choices, and our calculation, the area of interest is positive for $x \neq 0$.

Thus, the solution to the problem is that the positive area occurs for $x \neq 0$.

To solve find the positive area of the function $f(x) = x^2 - 4$, we proceed as follows:

• Step 1: Find the x-intercepts by setting $f(x) = 0$:

$x^2 - 4 = 0$

$x^2 = 4$

Take the square root of both sides:

$x = \pm 2$

• Step 2: Identify the intervals formed by the intercepts:

The intercepts $x = -2$ and $x = 2$ divide the x-axis into three intervals: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.

• Step 3: Test each interval to determine where $f(x) > 0$:

- For $x \in (-\infty, -2)$, pick $x = -3$:
$f(-3) = (-3)^2 - 4 = 9 - 4 = 5 > 0$, so the function is positive.

- For $x \in (-2, 2)$, pick $x = 0$:
$f(0) = (0)^2 - 4 = -4 \leq 0$, so the function is not positive.

- For $x \in (2, \infty)$, pick $x = 3$:
$f(3) = (3)^2 - 4 = 9 - 4 = 5 > 0$, so the function is positive.

Conclusively, the function $f(x)$ is positive in the intervals $x < -2$ and $x > 2$.

The correct answer is: $x < -2$ or $2 < x$.

To determine the positive area of the function $f(x) = -x^2 + 9$, we follow these steps:

• Step 1: Find the roots of the equation $-x^2 + 9 = 0$.
• Step 2: Solve the equation for the roots:

$-x^2 + 9 = 0$

$x^2 = 9$

$x = \pm 3$

• Step 3: Identify the intervals defined by these roots: $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$.
• Step 4: Test the sign of $f(x)$ in each interval:

For interval $(-3, 3)$: Choose $x = 0$, $f(0) = 9$ (positive)

For interval $(-\infty, -3)$: Choose $x = -4$, $f(-4) = -7$ (negative)

For interval $(3, \infty)$: Choose $x = 4$, $f(4) = -7$ (negative)

• Step 5: Conclude the positive area is for $-3 < x < 3$.

The positive area of the function is described by $-3 < x < 3$, which corresponds to the second choice.

Therefore, the solution to the problem is $-3 < x < 3$.

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

• Step 1: Solve the equation $2x^2 - 50 = 0$ to find the roots.
• Step 2: Analyze intervals on the number line based on the roots to find where $f(x) > 0$.

Now, let's work through each step:
Step 1: Set the function to zero: $2x^2 - 50 = 0$. Solving gives:

Step 2: The roots divide the number line into intervals: $x < -5$, $-5 < x < 5$, and $x > 5$. We test these intervals in $f(x) = 2x^2 - 50$.

• For $x < -5$, pick $x = -6$: $f(-6) = 2(-6)^2 - 50 = 72 - 50 = 22 > 0$ This interval is positive.
• For $-5 < x < 5$, pick $x = 0$: $f(0) = 2(0)^2 - 50 = -50 < 0$ This interval is negative.
• For $x > 5$, pick $x = 6$: $f(6) = 2(6)^2 - 50 = 72 - 50 = 22 > 0$ This interval is positive.

Therefore, the function is positive for $x < -5$ and $x > 5$.

Hence, the solution to the problem is $x < -5$ or $5 < x$.

To determine the interval where the function $f(x) = x^2 - 4$ is negative, follow these steps:

• Step 1: Identify the roots of the function by solving $x^2 - 4 = 0$.
  $x^2 = 4$ yields $x = 2$ and $x = -2$.
• Step 2: Consider the intervals created by these roots: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.
• Step 3: Determine the sign of $f(x)$ in each interval by selecting a test point from each:
• For $x = 0$ in $(-2, 2)$:
  $f(0) = 0^2 - 4 = -4$ (Negative)
• For $x = -3$ in $(-\infty, -2)$:
  $f(-3) = (-3)^2 - 4 = 9 - 4 = 5$ (Positive)
• For $x = 3$ in $(2, \infty)$:
  $f(3) = 3^2 - 4 = 9 - 4 = 5$ (Positive)
• Step 4: The function $f(x)$ is negative only in the interval $(-2, 2)$.

Consequently, the interval where the function has negative values is $-2 < x < 2$, which aligns with choice 2 in the provided options.

To solve this problem, we find where the function $f(x) = x^2 - 16$ is negative.

• Step 1: Set $f(x) = 0$ to find where the function has zero values, solving $x^2 - 16 = 0$.
• Step 2: Factor the equation as $(x - 4)(x + 4) = 0$.
• Step 3: Find the roots: $x = 4$ and $x = -4$.
• Step 4: Test intervals: $x < -4$, $-4 < x < 4$, and $x > 4$. Inside the interval $-4 < x < 4$, $x^2 - 16$ is negative because the square of any number less than 4 but greater than -4 will be less than 16.

Therefore, the function $f(x)$ is negative on the interval $-4 < x < 4$.

To solve this problem, we'll follow the steps outlined in our analysis.

• Step 1: Analyze the function's form $f(x) = x^2 + 16$. Here, $a=1, b=0,$ and $c=16$.

• Step 2: Find the vertex to see if the function ever takes negative values. The vertex $x$ is calculated by $x = -\frac{b}{2a} = 0$.

• Step 3: Evaluate $f(x)$ at this vertex: $f(0) = 0^2 + 16 = 16$.

• Step 4: Determine when f(x) < 0 . Since $f(x) = x^2 + 16 \geq 16$ for all real numbers $x$, the function is always positive.

• Step 5: Compare the finding against multiple-choice options. The choice indicating that $f(x)$ is always positive is the correct one: "Always positive".

The conclusion, therefore, is as follows: the function $f(x) = x^2 + 16$ is always positive, and there is no negative area under the graph relative to the x-axis.