Positive and Negative Domains: Standard representation

To solve the problem and determine for which values of $x$ the function $y = -x^2 + 4x - 3$ is greater than 0, we proceed with the following steps:

• Step 1: Identify the roots of the quadratic equation.
• Step 2: Analyze the sign of the function in the intervals determined by the roots.

Now, let us work through each step:

Step 1: Calculate the roots using the quadratic formula. The quadratic equation is $-x^2 + 4x - 3 = 0$. Using $a = -1$, $b = 4$, $c = -3$, we apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \times (-1) \times (-3)}}{2 \times (-1)}$

$x = \frac{-4 \pm \sqrt{16 - 12}}{-2} = \frac{-4 \pm \sqrt{4}}{-2}$

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

This gives roots: $x = 1$ and $x = 3$.

Step 2: With roots at $x = 1$ and $x = 3$, the real number line is divided into intervals: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

We test a point from each interval to determine the sign of the function:

• For $x \in (-\infty, 1)$, test $x = 0$:
  $y = -(0)^2 + 4(0) - 3 = -3$ (negative).
• For $x \in (1, 3)$, test $x = 2$:
  $y = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$ (positive).
• For $x \in (3, \infty)$, test $x = 4$:
  $y = -(4)^2 + 4(4) - 3 = -16 + 16 - 3 = -3$ (negative).

Therefore, the function is positive in the interval $(1, 3)$.

Thus, the solution is that the function $f(x) > 0$ for $1 < x < 3$.

Therefore, the correct choice is: $1 < x < 3$.

To solve the problem, we need to find where the function $f(x) = -x^2 - 6x - 8$ is negative. Let's proceed with a step-by-step solution:

• Step 1: Find the roots of the equation $-x^2 - 6x - 8 = 0$ using the quadratic formula.
• Step 2: Determine the intervals formed by these roots.
• Step 3: Test the intervals to see where $f(x) < 0$.

Step 1: The quadratic formula is given as follows:

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

In our equation, $a = -1$, $b = -6$, and $c = -8$.

Plugging these values into the formula:

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

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

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

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

This gives the roots:

• $x = \frac{6 + 2}{-2} = -4$
• $x = \frac{6 - 2}{-2} = -2$

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

Step 3: Test these intervals:

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

Thus, $f(x) < 0$ when $x < -4$ or $x > -2$.

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

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 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 solve for when the quadratic function $y = -x^2 + 2x + 35 > 0$, we must first find the roots of the function using the quadratic formula. The quadratic function is given as $y = -x^2 + 2x + 35$.

Step 1: Calculate the roots using the quadratic formula. For $ax^2 + bx + c = 0$, the formula is:

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

Here, $a = -1$, $b = 2$, and $c = 35$. Thus, we compute the discriminant:

$b^2 - 4ac = 2^2 - 4(-1)(35) = 4 + 140 = 144$

Since the discriminant is positive, there are two distinct real roots.

Step 2: Compute the roots using the quadratic formula:

$x = \frac{-2 \pm \sqrt{144}}{2(-1)} = \frac{-2 \pm 12}{-2}$

Calculating the two roots, we get:

$x_1 = \frac{-2 + 12}{-2} = \frac{10}{-2} = -5$ $x_2 = \frac{-2 - 12}{-2} = \frac{-14}{-2} = 7$

Step 3: Determine the intervals where $f(x) > 0$. The roots $x = -5$ and $x = 7$ partition the number line into intervals. A quadratic function with a negative leading coefficient $a$ opens downward, meaning it is positive between its roots:

The intervals are:

• $(-∞, -5)$
• $(-5, 7)$
• $(7, ∞)$

Test the interval between the roots: Choose a point, say $x = 0$, between $-5$ and $7$:

$f(0) = -(0)^2 + 2(0) + 35 = 35 > 0$

This confirms that the function is positive in the interval $(-5, 7)$.

Therefore, the solution to the inequality $f(x) > 0$ is $-5 < x < 7$.

The solution to the problem is $-5 < x < 7$.

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 solve the problem of identifying where the function $y = -x^2 + 10x - 16$ is greater than zero, follow these steps:

• Step 1: Calculate the roots of the quadratic equation. The roots are found using the quadratic formula:

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

Given: $a = -1$, $b = 10$, $c = -16$.

The discriminant is: $b^2 - 4ac = 10^2 - 4(-1)(-16) = 100 - 64 = 36$

Calculate the roots:

$x = \frac{-10 \pm \sqrt{36}}{2(-1)} = \frac{-10 \pm 6}{-2}$

Thus, the roots are:

• $x_1 = \frac{-10 + 6}{-2} = 2$
• $x_2 = \frac{-10 - 6}{-2} = 8$

Step 2: Determine where the function is positive. Since the parabola opens downward ($a = -1 < 0$), it is above the x-axis between the roots.

• Check intervals: $(-\infty, 2)$, $(2, 8)$, and $(8, \infty)$.

Test a point in the interval $(2, 8)$, for example, $x = 5$:

$f(5) = -5^2 + 10 \times 5 - 16 = -25 + 50 - 16 = 9 > 0$

Thus, the function is positive for $2 < x < 8$.

Conclusion: The solution to $f(x) > 0$ is $2 < x < 8$.

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

• Step 1: Find the roots of the function using the quadratic formula.
• Step 2: Determine the intervals defined by the roots and test the sign of the function on these intervals.

First, let's calculate the discriminant $b^2 - 4ac$:

$b = 8$, $a = -2$, and $c = -6$.

$\text{Discriminant} = 8^2 - 4(-2)(-6) = 64 - 48 = 16$.

With a positive discriminant, the quadratic equation has two real roots. Apply the quadratic formula:

$x = \frac{-8 \pm \sqrt{16}}{2(-2)}$.

Calculate the roots:

$x_1 = \frac{-8 + 4}{-4} = 1$

$x_2 = \frac{-8 - 4}{-4} = 3$.

Now, divide the number line into intervals based on these roots: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

Test the sign of the function $f(x) = -2x^2 + 8x - 6$ in each interval:

• For $x < 1$, choose $x = 0$:

$f(0) = -2(0)^2 + 8(0) - 6 = -6$ (negative).

• For $1 < x < 3$, choose $x = 2$:

$f(2) = -2(2)^2 + 8(2) - 6 = 2$ (positive).

• For $x > 3$, choose $x = 4$:

$f(4) = -2(4)^2 + 8(4) - 6 = -6$ (negative).

Thus, $f(x) < 0$ for $x < 1$ or $x > 3$.

The solution is $x > 3$ or $x < 1$.

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

• Step 1: Determine the roots of the quadratic function.
• Step 2: Split the number line based on these roots.
• Step 3: Test intervals to check where the function is greater than zero.
• Step 4: Identify the correct interval corresponding to the solution.

Now, let's work through each step:

Step 1: Identify the roots of the quadratic equation $-2x^2 + 8x - 6 = 0$. Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -2$, $b = 8$, and $c = -6$.

Calculate the discriminant: $b^2 - 4ac = 8^2 - 4(-2)(-6) = 64 - 48 = 16$.

The roots are: $x = \frac{-8 \pm \sqrt{16}}{-4} = \frac{-8 \pm 4}{-4}$.

Thus, $x_1 = \frac{-8 + 4}{-4} = 1$ and $x_2 = \frac{-8 - 4}{-4} = 3$.

Step 2: The roots 1 and 3 split the number line into intervals: $(-\infty, 1)$, $(1, 3)$, $(3, \infty)$.

Step 3: Test a sample point from each interval:

• For $x = 0$ in $(-\infty, 1)$: $f(0) = -6$, which is not greater than 0.
• For $x = 2$ in $(1, 3)$: $f(2) = -2(2)^2 + 8(2) - 6 = 8 - 6 = 2$, which is greater than 0.
• For $x = 4$ in $(3, \infty)$: $f(4) = -14$, which is not greater than 0.

Step 4: We conclude that the function $-2x^2 + 8x - 6$ is greater than 0 only in the interval $(1, 3)$.

Therefore, the solution to the problem is $1 < x < 3$.

To solve the problem of finding where the function $y = -x^2 + 6x - 8$ is less than zero, we follow these steps:

• Step 1: Set the function equal to zero and use the quadratic formula to find the roots.
• Step 2: Analyze the sign of the function between and beyond the roots to identify the intervals where the function is negative.

Let's work through each step:
Step 1: The function $y = -x^2 + 6x - 8$ can be set to 0:
$-x^2 + 6x - 8 = 0$

Using the quadratic formula where $a = -1$, $b = 6$, and $c = -8$:
Discriminant $D = b^2 - 4ac = 6^2 - 4(-1)(-8) = 36 - 32 = 4$

The roots are:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{4}}{-2} = \frac{-6 \pm 2}{-2}$

The solutions are:
$x = \frac{-6 + 2}{-2} = 2$ and $x = \frac{-6 - 2}{-2} = 4$

Step 2: Determine where the function is negative. Since the parabola opens downwards, it will be negative outside of the roots.
Therefore, the function is negative for:
$x < 2$ and $x > 4$

Therefore, the solution to the problem is:

$x < 2$ or $x > 4$

To solve the problem, we need to determine the intervals where the quadratic function $y = -x^2 + 6x - 8$ is greater than zero.

First, let's find the roots of the equation by setting $y = 0$:
$-x^2 + 6x - 8 = 0$.

We apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,
where $a = -1$, $b = 6$, and $c = -8$.

Calculating the discriminant:
$b^2 - 4ac = 6^2 - 4(-1)(-8) = 36 - 32 = 4$.

Finding the roots:
$x = \frac{-6 \pm \sqrt{4}}{2(-1)}$,
$x = \frac{-6 \pm 2}{-2}$.

This gives us two roots:
$x_1 = \frac{-6 + 2}{-2} = 2$,
$x_2 = \frac{-6 - 2}{-2} = 4$.

Now, examine the sign of the function in the intervals determined by these roots: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$. We plug test points from each interval into the original function to determine where it is positive.

• For $x \in (-\infty, 2)$, choose $x = 0$: $f(0) = -0^2 + 6 \times 0 - 8 = -8$ (negative)
• For $x \in (2, 4)$, choose $x = 3$: $f(3) = -(3)^2 + 6 \times 3 - 8 = -9 + 18 - 8 = 1$ (positive)
• For $x \in (4, \infty)$, choose $x = 5$: $f(5) = -(5)^2 + 6 \times 5 - 8 = -25 + 30 - 8 = -3$ (negative)

The interval where $f(x) > 0$ is $(2, 4)$.

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

To solve the problem, we need to determine the conditions under which the quadratic function $y = x^2 + 4x + 5$ satisfies $y < 0$.

We start by analyzing the discriminant of the quadratic equation. The standard form of a quadratic equation is:

$ax^2 + bx + c$, where here $a = 1$, $b = 4$, and $c = 5$.

The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$.

Calculating the discriminant, we have:

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

Since the discriminant is negative ($\Delta < 0$), the quadratic function has no real roots. This means the parabola does not intersect the x-axis and opens upwards (since $a = 1 > 0$).

Next, we find the vertex of the parabola to determine its minimum point. The vertex $(h, k)$ of a parabola given by $ax^2 + bx + c$ is:

$h = -\frac{b}{2a} = -\frac{4}{2 \times 1} = -2$.

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

Thus, the vertex is at $(-2, 1)$, and since the vertex is the minimum point of the upward-opening parabola and its value ($k$) is positive, the parabola $y = x^2 + 4x + 5$ is always above the x-axis.

Therefore, the function is never negative, and the solution is:

The function has no negative values.

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 $x$.

To solve this problem, we need to rewrite the quadratic function $y = x^2 + 2x + 2$ into its vertex form. This process allows us to find the vertex and understand the behavior of the function.

First, we complete the square for the quadratic expression. Starting with:
$y = x^2 + 2x + 2$

We take the $x$-terms $x^2 + 2x$ and complete the square as follows:

• Take half of the coefficient of $x$, which is 2, resulting in 1.
• Square this value to get 1.
• Add and subtract this square inside the equation to maintain equality.

Therefore, $x^2 + 2x + 1 - 1 + 2$ can be rewritten as:
$y = (x+1)^2 + 1$

Now, the function is in the form $y = (x+1)^2 + 1$, which shows the vertex at $(-1, 1)$. The vertex is the minimum point because the parabola opens upwards (as the coefficient of $x^2$ is positive).

This vertex indicates that the minimum value of $y$ is 1, which means the function never reaches below zero. As a result, the function never assumes negative values.

Based on this analysis, we conclude that the function has no negative values.

The correct answer is therefore: The function has no negative values.

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 solve this problem, we first consider the function $y = x^2 - 4x + 5$. This is a quadratic function, and we can analyze the parabola it represents.

The standard form of a quadratic function is $y = ax^2 + bx + c$, where in our case $a = 1$, $b = -4$, and $c = 5$.

To determine if the function has any negative values, we first find the vertex of the parabola. The vertex form of a quadratic function is determined by:

• The $x$-coordinate of the vertex is given by $x = -\frac{b}{2a}$.

Plugging in our values:

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

The $y$-coordinate of the vertex can be found by substituting $x = 2$ back into the equation:

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

The vertex of the parabola is $(2, 1)$, which means the minimum point of the parabola is above the x-axis at $y = 1$.

Next, we assess whether there are any real roots by finding the discriminant $\Delta$:

$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4$.

Since the discriminant is negative ($\Delta < 0$), this indicates the parabola does not intersect the x-axis at any real point. Therefore, it never dips below the x-axis.

Given that the vertex is above the x-axis and the discriminant is negative, the quadratic function $y = x^2 - 4x + 5$ is never negative for any real $x$.

The function has no negative values.

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 $x$.