Positive and Negative Domains: Standard representation

Let's analyze the function $y = 3x^2 - 6x + 4$ to determine when it is negative.

First, calculate the discriminant $\Delta$:

$\Delta = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$

A negative discriminant ($\Delta < 0$) indicates that there are no real roots, meaning the graph of the function does not intersect the x-axis. Since $a = 3 > 0$, the parabola opens upwards.

This means the vertex of the parabola represents the minimum point, and the entire graph is above the x-axis.

Consequently, the function $y = 3x^2 - 6x + 4$ does not attain any negative values for any real $x$.

The correct interpretation is that the function stays positive, confirming the conclusion:

The function has no negative values.

Let's analyze the function $y = -x^2 + 4x - 5$ and determine the interval where $y$ is negative.

1. **Find the roots using the quadratic formula**:
The function is given by $y = -x^2 + 4x - 5$. The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 4$, and $c = -5$. First, we calculate the discriminant:

$b^2 - 4ac = 4^2 - 4(-1)(-5) = 16 - 20 = -4$

Since the discriminant is negative, the quadratic equation has no real roots, implying that the parabola does not intersect the x-axis. The quadratic formula confirms there are no real solutions, confirming the function does not touch or cross the x-axis.

2. **Analyze the parabola's direction**:
Since $a = -1$, the parabola opens downwards. A downward-opening parabola with no real roots means it lies entirely below the x-axis. Hence, the function $y = -x^2 + 4x - 5$ is negative for all values of $x$.

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

The function given is $y = x^2 + 8x + 20$. This is a quadratic function where the coefficient of $x^2$ (which is $a = 1$) is positive, indicating the parabola opens upwards.

Let’s calculate the vertex to find the minimum value of $y$. The vertex of a parabola described by $y = ax^2 + bx + c$ is found at $x = -\frac{b}{2a}$.

Here, $a = 1$, $b = 8$. So the vertex is at:

$x = -\frac{8}{2 \times 1} = -4$

Substitute $x = -4$ into the function $y = x^2 + 8x + 20$ to calculate the minimum value of $y$.

$y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4$

The minimum value of the function is $y = 4$ at $x = -4$.

Given the opening direction of the parabola and the positive minimum value, the function $f(x) = x^2 + 8x + 20$ is always greater than 0.

Thus, the function is positive for all values of $x$.

The function $y = x^2 - 6x + 10$ represents a parabola opening upwards since its leading coefficient $a = 1$ is positive. Our task is to determine when the function is positive.

First, let's find the vertex of this parabola, which occurs at $x = -\frac{b}{2a}$.

Here, $b = -6$ and $a = 1$, so:

\begin{align*} x_{vertex} &= -\frac{-6}{2 \times 1} \\ &= \frac{6}{2} \\ &= 3. \end{align*}

Next, we evaluate the function at this vertex:

\begin{align*} f(3) &= 3^2 - 6 \cdot 3 + 10 \\ &= 9 - 18 + 10 \\ &= 1. \end{align*}

Since $f(3) = 1$, which is greater than zero, we observe that at the vertex the function is indeed positive.

Moreover, because the parabola opens upwards and the vertex value is positive, the entire parabola lies above the x-axis. Consequently, $f(x) > 0$ for all $x \in \mathbb{R}$.

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

The given function is $y = x^2 - 4x + 5$. To find where this function is positive, we'll first analyze the properties of this quadratic.

Let's start by completing the square. We have:

$y = x^2 - 4x + 5$

To complete the square, take the coefficient of $x$, which is $-4$, halve it to get $-2$, and then square it to get $4$. Add and subtract this inside the expression:

$y = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1$

Now, the expression is in vertex form $y = (x-2)^2 + 1$, which indicates a parabola with a vertex at $(2, 1)$ and opens upwards. The vertex is the minimum point of the function.

Since the minimum value of $y$ is 1 (when $x = 2$), and the parabola opens upwards, the function is positive for all real $x$, because $(x-2)^2 \geq 0$ for any real number $x$, making $(x-2)^2 + 1 > 0$.

Therefore, the answer is that the function is positive for all values of $x$.

In conclusion, the correct choice is:

The function is positive for all values of $x$.

The problem asks us to determine when the quadratic function $f(x) = x^2 + 4x + 5$ is greater than zero. Here's how we solve it:

Step 1: Analyze the Vertex
The quadratic function $f(x) = x^2 + 4x + 5$ is in the standard form $y = ax^2 + bx + c$, where $a = 1$, $b = 4$, and $c = 5$. Since $a > 0$, the parabola opens upwards, and thus the vertex represents its minimum point.
To find the x-coordinate of the vertex, use the formula $x = -\frac{b}{2a}$:

$x = -\frac{4}{2 \times 1} = -2$

Substitute $x = -2$ back into the function to find the y-coordinate:

$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$

The vertex of the parabola is $(-2, 1)$, which implies the minimum value of the function is 1.

Step 2: Analyze the Discriminant
The discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots:

$\Delta = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$

Since $\Delta < 0$, the quadratic equation has no real roots, meaning it doesn't intersect the x-axis. Therefore, $f(x) > 0$ for all $x$.

Conclusion
Because the vertex is the minimum point and the function does not intersect the x-axis, the function $f(x) = x^2 + 4x + 5$ is positive for all values of $x$.

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

The problem asks us to determine where the function $y = x^2 + 10x + 16$ is less than zero.

• Step 1: Find the roots of the equation $x^2 + 10x + 16 = 0$ using the quadratic formula.
• Step 2: Calculate the discriminant $\Delta = b^2 - 4ac = 10^2 - 4 \times 1 \times 16 = 100 - 64 = 36$.
• Step 3: Find the roots using $x = \frac{{-b \pm \sqrt{\Delta}}}{2a} = \frac{{-10 \pm \sqrt{36}}}{2}$.
• Step 4: Calculate the roots: $x = \frac{{-10 + 6}}{2} = -2$ $x = \frac{{-10 - 6}}{2} = -8$

The roots are $x = -2$ and $x = -8$. These roots will divide the number line into intervals.

• Step 5: Analyze the sign of the quadratic on the intervals $-\infty, -8$, $-8, -2$, and $-2, \infty$.
• Step 6: Choose test points: - For $x < -8$, choose $x = -9$, - For $-8 < x < -2$, choose $x = -5$, - For $x > -2$, choose $x = 0$.

Test each interval:

• For $x = -9$, $y = (-9)^2 + 10 \times (-9) + 16 = 81 - 90 + 16 = 7$ (positive).
• For $x = -5$, $y = (-5)^2 + 10 \times (-5) + 16 = 25 - 50 + 16 = -9$ (negative).
• For $x = 0$, $y = (0)^2 + 10 \times 0 + 16 = 16$ (positive).

The function is negative between $x = -8$ and $x = -2$. Therefore, the solution to $f(x) < 0$ is $-8 < x < -2$.

Therefore, the correct answer is $-8 < x < -2$.

To determine for which values of $x$ the function $f(x) = -3x^2 + 6x - 9 < 0$, follow these steps:

• Step 1: Identify the coefficient information. The function is $-3x^2 + 6x - 9$ with $a = -3$, $b = 6$, $c = -9$.
• Step 2: Determine the orientation of the parabola. The leading coefficient $a = -3$ is negative, so the parabola opens downward.
• Step 3: Calculate the vertex using the formula $x = -\frac{b}{2a} = -\frac{6}{2 \times (-3)} = 1$. The vertex is at $x = 1$.
• Step 4: Find the function's value at the vertex: $f(1) = -3(1)^2 + 6(1) - 9 = -3 + 6 - 9 = -6$. The function is negative at the vertex.
• Step 5: Calculate the discriminant $\Delta = b^2 - 4ac = 6^2 - 4(-3)(-9) = 36 - 108 = -72$. The discriminant is negative, indicating no real roots.

Since the quadratic opens downward and does not cross or touch the x-axis, it remains entirely below the x-axis for all values of $x$. Therefore, the function is negative for all $x$.

Thus, the solution is: The function is negative for all $x$.

To determine for which values of $x$ the function $y = 2x^2 - 4x + 5$ is positive, we will analyze its characteristics.

Step 1: Determine the direction of the parabola.
The given quadratic function $y = 2x^2 - 4x + 5$ has a leading coefficient $a = 2$, which is positive. Therefore, the parabola opens upwards.

Step 2: Check for real roots.
To identify where the function might be zero, calculate the discriminant $\Delta = b^2 - 4ac$.
Here, $a = 2$, $b = -4$, $c = 5$.
The discriminant $\Delta = (-4)^2 - 4 \cdot 2 \cdot 5 = 16 - 40 = -24$.
Since the discriminant is negative, the quadratic has no real roots, meaning it doesn't intersect the x-axis.

Step 3: Analyze positivity over the entire domain.
Since the parabola opens upwards and has no real roots, the function does not touch or cross the x-axis. Therefore, $y = 2x^2 - 4x + 5$ is always positive.

Conclusion.
The function $y = 2x^2 - 4x + 5$ is positive for all values of $x$.

Therefore, the solution to the problem is The function is positive for all values of $x$.

To determine the values of $x$ for which the quadratic function $y = -x^2 - 6x - 8$ is greater than 0, we will first find the roots of the quadratic equation where it equals zero.

We apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = -1$, $b = -6$, and $c = -8$ into the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-8)}}{2(-1)}$

Simplifying inside the square root and the rest of the expression:

$x = \frac{6 \pm \sqrt{36 - 32}}{-2}$ $x = \frac{6 \pm \sqrt{4}}{-2}$

Since $\sqrt{4} = 2$, the equation becomes:

$x = \frac{6 \pm 2}{-2}$

This gives us two potential solutions:

- $x = \frac{8}{-2} = -4$

- $x = \frac{4}{-2} = -2$

The roots divide the x-axis into three intervals: $x < -4$, $-4 < x < -2$, and $x > -2$.

To find where the function is positive, choose test points from these intervals:

• For $x < -4$ (e.g., $x = -5$): $f(-5) = -(-5)^2 - 6(-5) - 8 = -25 + 30 - 8 = -3$
• For $-4 < x < -2$ (e.g., $x = -3$): $f(-3) = -(-3)^2 - 6(-3) - 8 = -9 + 18 - 8 = 1$
• For $x > -2$ (e.g., $x = 0$): $f(0) = -(0)^2 - 6(0) - 8 = -8$

From this, the function is positive on the interval $-4 < x < -2$.

Therefore, the solution to the problem is $-4 < x < -2$.

To determine the values of $x$ where the function $y = x^2 + x - 20$ is greater than zero, we follow these steps:

• First, let us find the roots of the equation $x^2 + x - 20 = 0$.
• We attempt to factor the quadratic. We need two numbers that multiply to $-20$ and add up to $1$ (the coefficient of $x$).
• The numbers $5$ and $-4$ work, since $5 \times (-4) = -20$ and $5 + (-4) = 1$.
• Thus, we can factor the quadratic expression as $(x - 4)(x + 5) = 0$.
• Setting each factor equal to zero gives us the roots $x = 4$ and $x = -5$.
• The function changes its sign at these roots.
• To determine the sign in each interval, choose test points in the intervals defined by these roots: $(-\infty, -5)$, $(-5, 4)$, and $(4, \infty)$.
• For $x < -5$, say $x = -6$: $(-6 - 4)(-6 + 5) = (-10)(-1) > 0$.
• For $-5 < x < 4$, say $x = 0$: $(0 - 4)(0 + 5) = (-4)(5) < 0$.
• For $x > 4$, say $x = 5$: $(5 - 4)(5 + 5) = (1)(10) > 0$.
• The function is positive for $x < -5$ and $x > 4$, corresponding to the intervals where $(x - 4)(x + 5) > 0$.

Therefore, the solution to the problem, for which values of $x$ make $f(x) > 0$, is $x < -5$ or $x > 4$.

To determine when the function $f(x) = -x^2 - 2x - 3$ is negative, we begin by stating that it is a quadratic function in the form $ax^2 + bx + c$ where $a = -1$, $b = -2$, and $c = -3$. Since $a < 0$, the parabola opens downwards, indicating that it is concave down. This means that the function will be negative at all points unless it touches or crosses the x-axis.

First, we need to determine the vertex of the quadratic to ascertain where the maximum occurs. For any quadratic function in the form $ax^2 + bx + c$, the x-coordinate of the vertex is given by the formula:

$x = -\frac{b}{2a}$

Substitute $b = -2$ and $a = -1$ into the formula:

$x = -\frac{-2}{2(-1)} = -\frac{2}{-2} = 1$

The x-coordinate of the vertex is $x = 1$. The vertex lies at $(1, f(1))$.

Substitute $x = 1$ into the equation to find $f(1)$:

$f(1) = -(1)^2 - 2(1) - 3 = -1 - 2 - 3 = -6$

The vertex of the parabola is at $(1, -6)$, showing that the maximum point is negative.

Since the parabola opens downwards, all other values are below this vertex, hence **the parabola never crosses or touches the $x$-axis** implying the function is always below the $x$-axis, confirming that the function is negative for all values of $x$.

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

To determine where $f(x) < 0$ for the given quadratic function $y = -x^2 + 2x + 35$, we'll perform the following steps:

• Step 1: Identify the roots of the function using the quadratic formula.
• Step 2: Analyze the intervals around the roots to establish where the function is negative.

Step 1: Find the roots using the quadratic formula:

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our function $y = -x^2 + 2x + 35$, we have $a = -1$, $b = 2$, and $c = 35$. Substituting into the formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(35)}}{2(-1)}$

$x = \frac{-2 \pm \sqrt{4 + 140}}{-2}$

$x = \frac{-2 \pm \sqrt{144}}{-2}$

$x = \frac{-2 \pm 12}{-2}$

This gives two roots:

- $x_1 = \frac{-2 + 12}{-2} = -5$ - $x_2 = \frac{-2 - 12}{-2} = -7$

Step 2: Analyze the intervals created by the roots:

The roots divide the number line into the intervals $x < -7$, $-7 < x < -5$, and $x > -5$.

Since the parabola $y = -x^2 + 2x + 35$ opens downwards, it will be less than 0 outside the region between the roots. Therefore, the intervals where $y < 0$ are:

• $x > -7$
• $x < -5$

Therefore, the correct answer is:

$x > -7$ or $x < -5$

To determine where the function $f(x) = -x^2 + 10x - 16$ is less than zero, we should first find the roots by solving $f(x) = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = 10$, and $c = -16$, we can find the roots:

• Calculate the discriminant: $b^2 - 4ac = 10^2 - 4(-1)(-16) = 100 - 64 = 36$.
• Find the roots: $x = \frac{-10 \pm \sqrt{36}}{2(-1)} = \frac{-10 \pm 6}{-2}$.
• The roots are $x = \frac{-10 + 6}{-2} = 2$ and $x = \frac{-10 - 6}{-2} = 8$.

These roots divide the number line into intervals: $x < 2$, $2 < x < 8$, and $x > 8$.

To determine where $f(x) < 0$, test a point in each interval:

• For $x < 2$, choose $x = 0$. $f(0) = -0^2 + 10(0) - 16 = -16$ (which is less than zero).
• For $2 < x < 8$, choose $x = 5$. $f(5) = -(5)^2 + 10(5) - 16 = -25 + 50 - 16 = 9$ (which is greater than zero).
• For $x > 8$, choose $x = 10$. $f(10) = -(10)^2 + 10(10) - 16 = -100 + 100 - 16 = -16$ (which is less than zero).

Therefore, the function $f(x)$ is negative for $x < 2$ and $x > 8$.

Thus, the values of $x$ for which $f(x) < 0$ are $x < 2$ or $x > 8$.

The correct choice corresponding to this solution is: $x > 8$ or $x < 2$.

To determine where the function $f(x) = x^2 + 5x + 4$ is less than zero, we will first factor the quadratic expression.

Step 1: Factor the quadratic function.

• The quadratic $x^2 + 5x + 4$ can be factored into $(x + 4)(x + 1)$.

Step 2: Find the roots of the quadratic equation.

• Set each factor equal to zero: $x + 4 = 0$ and $x + 1 = 0$.
• The solutions are $x = -4$ and $x = -1$.

Step 3: Determine the sign of the quadratic in the intervals defined by these roots.

• Consider the intervals $(-\infty, -4)$, $(-4, -1)$, and $(-1, \infty)$.
• Pick test points from each interval: for $(-\infty, -4)$, choose $x = -5$; for $(-4, -1)$, choose $x = -2$; for $(-1, \infty)$, choose $x = 0$.
• Evaluate the sign of $(x+4)(x+1)$ at each test point:
• At $x = -5$, $(x+4)(x+1) = (-1)(-4) = 4$ (positive).
• At $x = -2$, $(x+4)(x+1) = (2)(-1) = -2$ (negative).
• At $x = 0$, $(x+4)(x+1) = (4)(1) = 4$ (positive).
• Therefore, the function is negative in the interval $(-4, -1)$.

Consequently, the solution is $-4 < x < -1$.

The correct choice from the options given is choice 4.

To determine where the function $y = x^2 + 5x + 4$ is greater than zero, we first find its roots by setting $x^2 + 5x + 4 = 0$.

Step 1: Factor the quadratic equation.

The expression $x^2 + 5x + 4$ can be factored as $(x + 1)(x + 4) = 0$.

Step 2: Solve for the roots.

Setting each factor to zero gives the roots as follows:
$x + 1 = 0 \Rightarrow x = -1$
$x + 4 = 0 \Rightarrow x = -4$

Step 3: Determine the sign of the quadratic on the intervals defined by the roots.

• Interval 1: $x < -4$. Pick $x = -5$, then $(x + 1)(x + 4) = (-5 + 1)(-5 + 4) = (-4)(-1) = 4 > 0.$
• Interval 2: $-4 < x < -1$. Pick $x = -3$, then $(x + 1)(x + 4) = (-3 + 1)(-3 + 4) = (-2)(1) = -2 < 0.$
• Interval 3: $x > -1$. Pick $x = 0$, then $(x + 1)(x + 4) = (0 + 1)(0 + 4) = 1 \times 4 = 4 > 0.$

Conclusion: The function $y = x^2 + 5x + 4$ is positive when $x < -4$ or $x > -1$.

Thus, the solution is $x > -1$ or $x < -4$.

To find for which values of $x$ the function $f(x) = x^2 + 9x + 18$ is less than 0, we first find the roots of the quadratic equation:

Step 1: Calculate the discriminant $b^2 - 4ac$ from the quadratic formula:
For $f(x) = x^2 + 9x + 18$, we have $a = 1$, $b = 9$, and $c = 18$.
The discriminant is $9^2 - 4 \times 1 \times 18 = 81 - 72 = 9$.

Step 2: Find the roots using the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9}}{2 \times 1} = \frac{-9 \pm 3}{2}$

Thus, the roots are:
$x_1 = \frac{-9 + 3}{2} = -3$
$x_2 = \frac{-9 - 3}{2} = -6$

Step 3: Analyze the sign of $f(x)$ around these roots:

The parabola opens upwards (since the coefficient of $x^2$ is positive), so it will be below the x-axis between the roots. This means $f(x) < 0$ for $-6 < x < -3$.

Therefore, the solution is:

$-6 < x < -3$.

To find for which values of $x$ the function $y = x^2 + 2x + 2$ is positive, we'll begin by analyzing the quadratic expression.

First, let's complete the square for the quadratic function:

$y = x^2 + 2x + 2$

To complete the square, take the coefficient of the $x$ term (which is 2), divide it by 2 to get 1, and then square it to obtain 1. Add and subtract this value inside the expression:

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

This simplifies to:

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

Now, this function $y = (x+1)^2 + 1$ shows a parabola in the vertex form $(x-h)^2 + k$, where the vertex is at $(-1, 1)$ and opens upwards, as the coefficient of the squared term is positive.

This indicates that the minimum point, $\text{(vertex)}$, is at $y = 1$, which is above the x-axis. As such, the function $y = (x+1)^2 + 1$ will be positive for all $x$ since the entire curve lies above the x-axis and the value of $y$ is always greater than zero.

Therefore, the function $y = x^2 + 2x + 2$ is positive for all real values of $x$.

Hence, the solution is that the function is positive for all values of $x$.

To find the values of $x$ for which the function $y = x^2 + x - 20$ is less than zero, we proceed as follows:

Step 1: Identify the roots of the quadratic equation.

• The quadratic function is $y = x^2 + x - 20$.
• Set the quadratic equation equal to zero: $x^2 + x - 20 = 0$.
• Use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -20$.
• The discriminant is $b^2 - 4ac = 1^2 - 4 \times 1 \times (-20) = 1 + 80 = 81$.
• Since the discriminant is positive, the roots are real and distinct: $x = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$.
• The roots are $x = \frac{-1 + 9}{2} = 4$ and $x = \frac{-1 - 9}{2} = -5$.

Step 2: Determine intervals based on the roots.

• The roots split the real number line into intervals: $x < -5$, $-5 < x < 4$, and $x > 4$.
• For each interval, test if $f(x) < 0$.
• Choose a test point in each interval: $x = -6$, $x = 0$, and $x = 5$.
• Calculate $f(x)$ at each point:
• For $x = -6$, $f(-6) = (-6)^2 + (-6) - 20 = 36 - 6 - 20 = 10$ (not less than 0).
• For $x = 0$, $f(0) = 0^2 + 0 - 20 = -20$ (less than 0).
• For $x = 5$, $f(5) = 5^2 + 5 - 20 = 25 + 5 - 20 = 10$ (not less than 0).

Step 3: Conclusion

From these tests, $f(x) < 0$ on the interval $-5 < x < 4$, corresponding to choices where the quadratic lies below the x-axis between its roots.

Based on the function's nature, it changes sign between and outside its roots, indicating the function is negative in intervals $x < -5 \text{ or } x > 4$.

Thus, the solution is $x > 4$ or $x < -5$, corresponding to the correct answer choice.