Positive and Negative Domains: Standard representation

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 the problem of determining for which values of $x$ the quadratic function $y = -x^2 + 4x - 3$ is less than zero, we should first find the roots of the equation $-x^2 + 4x - 3 = 0$.

Using the quadratic formula, where $a = -1$, $b = 4$, and $c = -3$, we have:

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

Calculating the discriminant:

$b^2 - 4ac = 4^2 - 4(-1)(-3) = 16 - 12 = 4$

Since the discriminant is positive, we will have two distinct real roots:

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

This gives us:

$x = \frac{-6}{-2} = 3$ and $x = \frac{-2}{-2} = 1$

This tells us the quadratic function crosses the x-axis at $x = 1$ and $x = 3$.

To determine the sign of the function, consider test values in the intervals determined by the roots, which are: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

• For the interval $(-\infty, 1)$, test a point such as $x = 0$: $f(0) = -0^2 + 4(0) - 3 = -3$, which is less than 0.
• For the interval $(1, 3)$, test a point such as $x = 2$: $f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1$, which is greater than 0.
• For the interval $(3, \infty)$, test a point such as $x = 4$: $f(4) = -(4)^2 + 4(4) - 3 = -16 + 16 - 3 = -3$, which is less than 0.

Therefore, the solution where $f(x) < 0$ is when the variable $x$ satisfies $x < 1$ or $x > 3$.

Hence, the values of $x$ for which $f(x) < 0$ are $x > 3$ or $x < 1$.

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 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 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 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 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.

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 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 $-5 < x < 7$.