Positive and Negative Domains: Representation of value a only

The function $y = -\frac{3}{5}x^2$ is quadratic with a negative leading coefficient, meaning the parabola opens downward. This implies that the function cannot be greater than zero for any real value of $x$ because it reaches its maximum (zero) at $x = 0$ and decreases as $|x|$ increases.

Therefore, there are no $x$ values for which $f(x) > 0$.

The correct choice reflecting this conclusion is: No x.

The given function is $y = 5x^2$. This is a quadratic function where the coefficient of $x^2$ is positive, making the parabolic shape open upwards.

The expression $5x^2$ is always non-negative. For this function to be greater than zero, it should not be equal to zero. We find the expression equals zero when $x = 0$.

When $x \neq 0$, $5x^2$ is positive. Therefore, $y = 5x^2 > 0$ whenever $x \neq 0$. This applies to both positive and negative $x$ values, except for zero.

Thus, the correct answer is: $x \neq 0$.

To determine for which values of $x$ the function $y = x^2$ satisfies $f(x) < 0$, we need to analyze the nature of the quadratic function.

Step 1: Recognize the function $y = x^2$ is parabolic and opens upwards. For any real number $x$, $x^2$ is always non-negative, i.e., $x^2 \geq 0$.

Step 2: Since squaring any real number results in a value greater than or equal to zero, it is not possible for $x^2$ to be less than zero.

Conclusion: Therefore, there are no real values of $x$ for which $x^2 < 0$. The correct conclusion is that no $x$ satisfies $x^2 < 0$.

To determine where the function $y = \frac{3}{4}x^2$ is positive, observe:

• **Step 1:** Identify the nature of the expression $x^2$.
  • The expression $x^2$ is always non-negative for any real number $x$, meaning $x^2 \geq 0$.
• **Step 2:** Consider when $x^2$ equals zero.
  • We see that $x^2 = 0$ only when $x = 0$, as any other value will yield a positive $x^2$.
• **Step 3:** Analyze the entire function.
  • The term $\frac{3}{4}$ is positive. Hence, $\frac{3}{4}x^2$ becomes positive whenever $x^2 > 0$, which implies $x \ne 0$.
  • Therefore, for all non-zero $x$, $y = \frac{3}{4}x^2 > 0$.

Thus, the function $y = \frac{3}{4}x^2$ is positive for all $x \ne 0$.

Therefore, the correct answer is $x \ne 0$.

To solve the problem of finding for which values of $x$ the function $y = -7x^2$ is negative, we perform the following analysis:

Step 1: We are given a quadratic function $y = -7x^2$. The function is quadratic because it is of the form $ax^2 + bx + c$ where $a = -7$, $b = 0$, and $c = 0$.

Step 2: Observe the form, $y = -7x^2$, which implies that $y$ depends solely on $x^2$. Since $x^2 \geq 0$ for all real numbers $x$, multiplying by -7 (a negative constant) ensures that $y$ will always be less than or equal to zero.

Step 3: Specifically, $f(x) = -7x^2$ is negative ($f(x) < 0$) wherever $x^2 > 0$. The only time $x^2 = 0$ occurs is when $x = 0$. Therefore, $y = 0$ when $x = 0$.

Step 4: We conclude that the only condition under which $f(x) \geq 0$ is precisely when $x = 0$, meaning for all other real numbers $x \neq 0$, $f(x) < 0$.

Thus, the function $y = -7x^2$ is negative for all real numbers except for when $x = 0$.

Therefore, the solution is $x \ne 0$.

To solve the problem of finding when $f(x) = -\frac{3}{5}x^2 < 0$, we utilize properties of quadratic functions:

• The given function is $y = -\frac{3}{5}x^2$, where the coefficient of $x^2$ is $-\frac{3}{5}$.
• The negative coefficient indicates the parabola opens downwards.
• For any quadratic function of the form $y = ax^2$ where $a < 0$, the function is negative for all $x$ except at $x = 0$, where it equals zero.

Therefore, the function $f(x) = -\frac{3}{5}x^2$ is negative for all $x$ except $x = 0$.

Thus, the set of $x$ satisfying the condition $f(x) < 0$ is $x \neq 0$.

Hence, the solution is $x \ne 0$.

To solve the problem, we need to understand when the function $y = -x^2$ is greater than 0.

1. **Analysis of the Function**: The given function is $y = -x^2$. Here, $y$ represents the output of the quadratic expression, where the coefficient of $x^2$ is negative ($-1$). This means that for every input $x$, the output is the negative of $x^2$.

2. **Properties of the Square**: The expression $x^2$ is always non-negative, i.e., $x^2 \geq 0$ for all real numbers $x$. This implies:

• For $x = 0$, $x^2 = 0$.
• For any non-zero $x$, $x^2 > 0$.

3. **Impact of the Negative Sign**: Since $y = -x^2$:

• If $x^2 = 0$ (which happens when $x = 0$), then $y = 0$.
• If $x^2 > 0$ (which happens when $x \ne 0$), then $y < 0$.

4. **Conclusion on Positivity**: There are no values of $x$ for which $y = -x^2$ is greater than 0, as $y \leq 0$ for all real $x$. Thus, the function is never positive.

Therefore, the solution to this problem is No $x$.

To solve this problem, consider the nature of the quadratic function $y = 5x^2$. The function has a leading coefficient of 5, which is positive, indicating that the parabola opens upwards.

A parabola opening upwards, such as this one, has its minimum value at the vertex. For the function $y = 5x^2$, the minimum value occurs at $x = 0$, where $y = 5 \cdot 0^2 = 0$. Since $y = 5x^2$ is a non-negative quadratic for all real $x$, the function $f(x) = 5x^2 \geq 0$ for all $x$.

This means that there are no values of $x$ for which $f(x) < 0$ holds. The function is only zero when $x = 0$ and positive otherwise for any non-zero $x$.

Conclusively, there are no values of $x$ where $f(x) < 0$. Therefore, the solution is that no $x$ satisfies $f(x) < 0$.

Hence, the answer is that there are

No x.

To solve this problem, let's apply the analysis and reasoning as follows:

Step 1: Analyze the function $y = -0.9x^2$. This is a quadratic function of the form $ax^2$ where $a = -0.9$. Since $a < 0$, the parabola opens downwards.

Step 2: Consider the values of $y$. For a parabola opening downwards, the peak (vertex) is at its maximum, and from this point, the parabola decreases, stretching indefinitely in the negative direction of $y$.

Step 3: Determine the maximum value. In the quadratic function $y = -0.9x^2$, the vertex at $x = 0$ gives the maximum value of $y$, which is $y = 0$ since $y = -0.9 \times 0^2 = 0$.

Step 4: Examine the entire function's range. Since beyond the vertex $y = 0$, the values of $y$ are strictly negative, there are no values of $x$ for which $f(x) > 0$.

Conclusion: Because the function $y = -0.9x^2$ has its only non-negative point at $x = 0$ (where it equals zero) and decreases for all other values of $x$, there are no $x$-values that make the function positive (i.e., $f(x) > 0$ is never true). Therefore, no $x$ satisfies the condition $f(x) > 0$.

The correct choice is 3: No x.

To solve this problem, let's break down the given quadratic function:

• Step 1: The function is $y = 0.4x^2$. This is a basic quadratic function where $a = 0.4$, $b = 0$, and $c = 0$.
• Step 2: Since $a = 0.4 > 0$, the parabola opens upwards. The vertex of this parabola is at the point $(0, 0)$.
• Step 3: Address the condition $f(x) > 0$. The function value $y$ is zero exactly at the vertex, $x = 0$. For any other real number value of $x$, the term $x^2$ is positive, and therefore $0.4x^2$ is also positive.

Since $y = 0.4x^2$ will always be greater than zero for every $x \neq 0$, the correct set of values for $x$ where $f(x) > 0$ is all $x$ except $x = 0$. Thus, the solution is expressed as:

$x \ne 0$

To solve this problem, let's follow our planned approach:

• Consider the function $y = x^2$.
• The quadratic function $y = x^2$ is non-negative for all real numbers.
• Specifically, $y = x^2 = 0$ only at $x = 0$. Therefore, $y > 0$ when $x \neq 0$.

Therefore, for $f(x) = x^2$ to be greater than zero, the condition is that $x \neq 0$.

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

To solve this problem, let's work through the following steps:

Step 1: Analyze the inequality $-7x^2 > 0$.
Since $x^2 \geq 0$ for all real $x$ and $x^2 = 0$ when $x = 0$, we know $x^2 > 0$ when $x \neq 0$. However, the expression $-7x^2$ is always less than or equal to zero because multiplying a non-negative $x^2$ by $-7$ gives a non-positive result.
Therefore, $-7x^2 > 0$ cannot be true for any real $x$.

Step 2: Conclude based on this analysis.
The only scenario where $-7x^2$ could have been greater than zero is if we had a positive term offsetting it, which is not the case.
Hence, there are no values of $x$ for which $f(x) > 0$.

The correct answer based on the provided choices is No x.

To solve this problem, we'll analyze the quadratic function $y = -x^2$.

• Step 1: Determine when $y = -x^2$ is less than zero.
  The function value is zero at $x = 0$ and less than zero for any other $x$. Since $-x^2$ will be negative for all non-zero $x$, this means $f(x) < 0$ for all $x \ne 0$.
• Step 2: Interpretation of results.
  The graph of $y = -x^2$ opens downward and only touches the x-axis at $x = 0$. This means, except at the point where $x = 0$, the function outputs negative values.
• Step 3: Comparison with given choices.
  We want $f(x) < 0$. Since this holds for all $x \ne 0$, we will select the answer $x \ne 0$.

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

To solve this problem, we'll consider the given quadratic function $y = \frac{3}{4}x^2$ and analyze when it could be negative.

Let's break this down:

• The expression $x^2$ represents the square of $x$, which is always non-negative for any real number $x$.
• Since the coefficient of $x^2$, which is $\frac{3}{4}$, is positive, multiplying $x^2$ by this constant results in a non-negative value.

Combining these observations:

• Since $x^2 \geq 0$ for all $x$, it follows that $\frac{3}{4}x^2 \geq 0$ for all $x$.
• Thus, the function $y = \frac{3}{4}x^2$ will always be non-negative, meaning it can never be less than zero.

Therefore, there are no values of $x$ for which $f(x) < 0$.

Based on this analysis, the correct answer is that there are No $x$ for which the expression is negative.

To solve this problem, we need to determine when the quadratic function $y = -0.9x^2$ is less than zero. This requires analyzing the entire set of x-values.

• Given that the function $y = -0.9x^2$ is a parabola opening downward because $a = -0.9 < 0$, the function will take negative values for all $x$ except when $x^2 = 0$.
• The vertex of this parabola is at the origin, $x = 0$, since terms for linear ($bx$) and constant ($c$) are zero in this function.
• At $x = 0$, $y = -0.9(0)^2 = 0$, meaning at $x = 0$, the function is exactly zero.
• For any $x \neq 0$, the term $x^2 > 0$, hence $-0.9x^2 < 0$, indicating the function is negative.

Therefore, the function $y = -0.9x^2$ is less than zero for all values except at $x = 0$.

Consequently, the solution to the problem is that the function is negative for all $x \ne 0$.

This corresponds to choice: $x\ne0$.

To solve this problem, we're given a quadratic function $y = 0.4x^2$. We need to determine for which values of $x$, $f(x) < 0$. Let's start by examining the function.

The given function is quadratic, and it takes the form $y = ax^2$, where $a = 0.4$. It's important to note that the coefficient $a$ is positive.

For quadratic functions in the form $ax^2$, where $a > 0$, the graph of the function is a parabola that opens upwards. This means that the values of the quadratic function are non-negative for all real $x$. Specifically, the value $f(x)$ is always greater than or equal to zero, reaching zero exactly when $x = 0$.

For the inequality $f(x) < 0$ to hold, $y$ would have to be negative. Since the parabola opens upwards and the vertex (the minimum point) is at $y = 0$, there are no real $x$ that satisfy $f(x) < 0$.

Therefore, for the given function $y = 0.4x^2$, there are no values of $x$ for which $f(x) < 0$.

Thus, based on the analysis, the correct choice is that there are no values of $x$ for which $f(x) < 0$ holds, which is No x.