Positive and Negative Domains: Standard representation

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 solve the problem, we'll follow these steps:

• Step 1: Use the quadratic formula to find the roots of the equation $x^2 + 8x - 9 = 0$.
• Step 2: Analyze the intervals determined by these roots to find where the function $y = x^2 + 8x - 9$ is less than zero.
• Step 3: Identify the correct inequality and compare with the given multiple-choice options.

Now, let's work through each step:

Step 1: Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = 8$, and $c = -9$:
$x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-8 \pm \sqrt{64 + 36}}{2}$ $x = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}$ This results in the roots $x = 1$ and $x = -9$.

Step 2: Since the quadratic opens upwards (leading coefficient $a = 1$ is positive), the function will be less than zero between the roots. This gives us the interval:
$-9 < x < 1$

Step 3: Identifying the correct choice from the options, the solution is $(-9 < x < 1)$.

Therefore, the solution to the problem, where $y = x^2 + 8x - 9$ is less than zero, is $-9 < x < 1$.

To solve for the values of $x$ where $y = x^2 + 8x - 9$ is less than zero, we will follow these steps:

• Find the roots of the quadratic equation $x^2 + 8x - 9 = 0$ using the quadratic formula.
• Determine the intervals on the x-axis formed by these roots.
• Test values from each interval in the inequality to determine where the function is negative.

First, we calculate the roots of $x^2 + 8x - 9 = 0$ using the quadratic formula:

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

Here, $a = 1$, $b = 8$, $c = -9$.

The discriminant is calculated as:

$b^2 - 4ac = 8^2 - 4(1)(-9) = 64 + 36 = 100$

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

The roots are:

$x = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2}$

This gives us roots $x = 1$ and $x = -9$.

Now, we analyze the sign of $y = x^2 + 8x - 9$ around these root intervals:

• For $x < -9$, choose a number like $x = -10$.
• For $-9 < x < 1$, choose a number like $x = 0$.
• For $x > 1$, choose a number like $x = 2$.

Substituting these test points into the function:

• For $x = -10$: $y = (-10)^2 + 8(-10) - 9 = 100 - 80 - 9 = 11 > 0$
• For $x = 0$: $y = (0)^2 + 8(0) - 9 = -9 < 0$
• For $x = 2$: $y = (2)^2 + 8(2) - 9 = 4 + 16 - 9 = 11 > 0$

Therefore, the function $y = x^2 + 8x - 9$ is negative in the interval $-9 < x < 1$.

Considering the inequality $f(x) < 0$, we conclude:

The solution to the problem is $-9 < x < 1$, aligning with answer choice 4.

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 find the values of $x$ where the function $y = 2x^2 - 4x + 5$ is negative:

• Step 1: Identify the direction of the parabola:
  Since $a = 2$ is positive, the parabola opens upwards, indicating that any potential minimum will be at the vertex.

• Step 2: Find the vertex:
  The vertex is at $x = -\frac{b}{2a} = -\frac{-4}{2 \times 2} = 1$.
  Substituting $x = 1$ back into the function gives: $f(1) = 2(1)^2 - 4(1) + 5 = 3$.

• Step 3: Determine if the function can be negative:
  Since the vertex provides the minimum value of the parabola and this value is positive $f(1) = 3 > 0$, the function does not have any $x$ values for which $f(x) < 0$.

Thus, the correct answer is that the function has no negative values.

To solve this problem, we'll check where the quadratic function $y = -2x^2 + 8x - 10$ is greater than zero.

The steps to solve are as follows:

• Step 1: Identify the direction of opening and calculate the vertex.
  For the quadratic function $y = ax^2 + bx + c$ with $a = -2$, $b = 8$, $c = -10$, since $a = -2 < 0$, the parabola opens downwards.
• Step 2: Find the vertex, which is the peak of the parabola.
  The x-coordinate of the vertex $x_v$ is given by the formula $x_v = -\frac{b}{2a} = -\frac{8}{2 \times (-2)} = 2$.
  Substitute $x = 2$ into the function to find the y-coordinate: $y = -2(2)^2 + 8(2) - 10 = -8 + 16 - 10 = -2$.
  The vertex is at $(2, -2)$. This indicates that the maximum value of the function is $-2$, which is less than zero.
• Step 3: Analyze the function's values.
  Since the maximum point $(2, -2)$ is less than zero and the parabola opens downwards, the function values are always less than or equal to the vertex's y-value, which is $-2$.
• Step 4: Conclusion.
  It is clear that the function is never positive at any point in its domain for real values of $x$.

Therefore, the solution to the problem is The function has no positive domain.

To solve for where $y = -2x^2 - 8x - 10 > 0$, we must analyze the quadratic equation.

First, identify the coefficients: $a = -2$, $b = -8$, and $c = -10$.
The parabola opens downwards since $a < 0$.

Calculate the discriminant $\Delta = b^2 - 4ac = (-8)^2 - 4(-2)(-10) = 64 - 80 = -16$.
Since the discriminant is negative, there are no real roots.

As a result, the quadratic does not intersect the x-axis, meaning it has no intervals where it is positive.
Because the parabola opens downward and lies entirely below the x-axis, the function $y = -2x^2 - 8x - 10$ has no positive domain.

Thus, the function $f(x) = -2x^2 - 8x - 10$ is never greater than zero.

Therefore, the solution to the problem is The function has no positive domain.

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

• Step 1: Analyze the quadratic function's graph from given coefficients.
• Step 2: Calculate discriminant to find root characteristics.
• Step 3: Discuss the implications on positivity or negativity of $f(x)$.

Now, let's work through each step:

Step 1: The quadratic function $y = -2x^2 - 8x - 10$ has a leading coefficient $a = -2$, which is negative. This indicates that the parabola opens downwards, potentially sitting below the $x$-axis.

Step 2: Calculate the discriminant $\Delta = b^2 - 4ac$.

Here, $b = -8$, $a = -2$, and $c = -10$. Plug these into the formula:

$\Delta = (-8)^2 - 4(-2)(-10) = 64 - 80 = -16$

Since the discriminant $\Delta$ is negative, there are no real roots. The parabola does not intersect the $x$-axis.

Step 3: Discuss implications.

Because the parabola opens downward and has no real roots, it lies entirely below the $x$-axis. This means for all real values of $x$, $f(x) < 0$.

Therefore, the function $y = -2x^2 - 8x - 10$ is negative for all $x$.

The function is negative for all $x$.

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, we'll follow these steps:

• Step 1: Solve the quadratic equation $3x^2 - 6x + 4 = 0$ using the quadratic formula.
• Step 2: Determine the intervals based on the roots and analyze the sign of the quadratic in each interval.
• Step 3: Use the results to conclude for which values of $x$ the quadratic is positive.

Step 1: The quadratic formula yields the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 3$, $b = -6$, and $c = 4$.

Calculating the discriminant:

$b^2 - 4ac = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$.

Since the discriminant is negative ($-12$), the quadratic equation has no real roots. This means the parabola does not intersect the x-axis, and the entire graph is above or below it.

Step 2: Since $a = 3$ (positive), the parabola opens upwards. A quadratic function with no real roots and a positive leading coefficient will be entirely above the x-axis, indicating it is always greater than zero.

Step 3: Since it is positive across all $x$-values, the solution is:

The function is positive for all $x$.

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.

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

• Step 1: Identify the given quadratic function and its form.
• Step 2: Determine the nature of the parabola based on the leading coefficient.
• Step 3: Calculate the discriminant to find possible roots.
• Step 4: Solve for the roots using the quadratic formula if applicable.
• Step 5: Analyze the intervals determined by the roots to assess where the function is positive.
• Step 6: Conclude whether the function is positive over any interval based on the parabola's direction.

Now, let's work through each step:

Step 1: The given function is $y = -3x^2 + 6x - 9$, which is a quadratic in the standard form $ax^2 + bx + c$, where $a = -3$, $b = 6$, and $c = -9$.

Step 2: The coefficient of $x^2$ is negative ($a = -3$), indicating the parabola opens downward.

Step 3: The discriminant $\Delta$ for the quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. Calculating this:

$\Delta = (6)^2 - 4(-3)(-9) = 36 - 108 = -72$.

Step 4: A negative discriminant ($\Delta < 0$) shows that the quadratic equation $-3x^2 + 6x - 9 = 0$ has no real roots. This means the parabola does not intersect the x-axis.

Step 5: Knowing the downward opening parabola and lack of real roots, the parabola lies entirely below the x-axis, and it never becomes positive anywhere.

Step 6: Since the function is always non-positive, we conclude that the function has no positive domain.

Therefore, the solution to the problem is The function has no positive domain.

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 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 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 $x < 2$.

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 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 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:

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

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

Step 2: Compute the roots using the quadratic formula:

Calculating the two roots, we get:

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 $7$:

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

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