Positive and Negative Domains: With fractions

Let's begin by finding the roots of the quadratic function $y = -x^2 + \frac{3}{4}x - 2$. This will help us determine the intervals where the function is positive or negative.

The coefficients of the quadratic are $a = -1$, $b = \frac{3}{4}$, and $c = -2$. Applying the quadratic formula:

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

Substitute in the values:

$x = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4(-1)(-2)}}{2(-1)}$ $x = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - 8}}{-2}$

Calculate inside the square root:

$\frac{9}{16} - 8 = \frac{9}{16} - \frac{128}{16} = \frac{-119}{16}$

The discriminant $b^2 - 4ac$ is negative ($\frac{-119}{16}$), indicating there are no real roots. Therefore, the quadratic function doesn't intersect the x-axis.

Since the parabola is downward (the coefficient of $x^2$ is negative), it is negative for all $x \in \mathbb{R}$.

We conclude that:

• The function $y$ does not have a positive domain for any real $x$.
• The function $y$ is negative for all $x$.

Therefore, the positive and negative domains, based on choices given, are:

$x > 0$: none

$x < 0$: all $x$

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

• Step 1: Convert mixed number $22\frac{2}{9}$ to an improper fraction.
• Step 2: Use the quadratic formula to find the roots of the function.
• Step 3: Determine the intervals where $y > 0$ and $y < 0$.

Now, let's work through each step:

Step 1: Convert $22\frac{2}{9}$ to an improper fraction:

$22\frac{2}{9}$ is $\frac{200}{9} = \frac{200}{9}$.

Step 2: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.

Here, $a = -2$, $b = 0$, and $c = \frac{200}{9}$.

Calculate the discriminant:

$b^2 - 4ac = 0 - 4(-2)\left(\frac{200}{9}\right) = \frac{1600}{9}$.

The roots become:

$x = \frac{0 \pm \sqrt{\frac{1600}{9}}}{-4} = \frac{\pm \frac{40}{3}}{-4} = \frac{\pm 10}{3}$.

Thus, the roots are $x_1 = -\frac{10}{3}$ and $x_2 = \frac{10}{3}$.

Step 3: Examine where the quadratic is positive:

- The function is shaped as a downward-opening parabola. Its positive zone (above x-axis) will be between the roots.

- So, the positive domain is: $-\frac{10}{3} < x < \frac{10}{3}$.

- Outside these roots, the function is negative:

- The negative domain corresponds to intervals $x < -\frac{10}{3}$ or $x > \frac{10}{3}$.

Therefore, the solution to the problem is:

$x > 3\frac{1}{3}$ or $x < 0 : x < -3\frac{1}{3}$

and

$x > 0 : -3\frac{1}{3} < x < 3\frac{1}{3}$.

To solve the given problem, we will perform the following steps:

• Calculate the roots of the quadratic equation $y = -2x^2 + 3x + 2$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Given $a = -2$, $b = 3$, and $c = 2$, use the formula:

$x = \frac{-3 \pm \sqrt{3^2 - 4(-2)(2)}}{2(-2)} = \frac{-3 \pm \sqrt{9 + 16}}{-4} = \frac{-3 \pm \sqrt{25}}{-4}$

• The roots are $x = \frac{-3 + 5}{-4} = -\frac{1}{2}$ and $x = \frac{-3 - 5}{-4} = 2$.
• The roots $-\frac{1}{2}$ and $2$ divide the x-axis into three intervals: $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 2)$, and $(2, \infty)$.
• Test a point from each interval in the function to determine the sign of the function in those intervals:

Choose $x = -1$ for interval $(- \infty, -\frac{1}{2})$:
Substitute into the function: $y = -2(-1)^2 + 3(-1) + 2 = -2 - 3 + 2 = -3$ (negative).

Choose $x = 0$ for interval $(- \frac{1}{2}, 2)$:
Substitute into the function: $y = -2(0)^2 + 3(0) + 2 = 2$ (positive).

Choose $x = 3$ for interval $(2, \infty)$:
Substitute into the function: $y = -2(3)^2 + 3(3) + 2 = -18 + 9 + 2 = -7$ (negative).

Therefore, the positive domain where the function is positive is $-\frac{1}{2} < x < 2$, and the negative domains are $x < -\frac{1}{2}$ or $x > 2$.

The solution to the problem is:

$x > 0 : -\frac{1}{2} < x < 2$

$x > 2$ or $x < 0 : x < -\frac{1}{2}$

To solve this problem, we'll find the intervals where the given quadratic function $y = -2x^2 + 7x - 3$ is greater than zero (positive) and less than zero (negative).

Step 1: Find the roots of the quadratic function.

The general form of the quadratic equation is $ax^2 + bx + c = 0$. Here, $a = -2$, $b = 7$, and $c = -3$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we calculate the roots.

First, calculate the discriminant:

$b^2 - 4ac = 7^2 - 4 \times (-2) \times (-3) = 49 - 24 = 25$.

Thus, the roots are:

$x = \frac{-7 \pm \sqrt{25}}{2 \times (-2)} = \frac{-7 \pm 5}{-4}$.

Calculating for the two roots:

• $x_1 = \frac{-7 + 5}{-4} = \frac{-2}{-4} = \frac{1}{2}$
• $x_2 = \frac{-7 - 5}{-4} = \frac{-12}{-4} = 3$

The roots are $x = \frac{1}{2}$ and $x = 3$.

Step 2: Determine the sign of the function in each interval.

The function is defined as:

$(-\infty, \frac{1}{2}), \left(\frac{1}{2}, 3\right), (3, \infty)$.

Test each interval to determine where the function is positive or negative:

• For $x < \frac{1}{2}$, choose $x = 0$:
  $y = -2(0)^2 + 7(0) - 3 = -3$ (negative)
• For $\frac{1}{2} < x < 3$, choose $x = 1$:
  $y = -2(1)^2 + 7(1) - 3 = 2$ (positive)
• For $x > 3$, choose $x = 4$:
  $y = -2(4)^2 + 7(4) - 3 = -3$ (negative)

Conclusion: The positive domain is $\frac{1}{2} < x < 3$, and the negative domain is $x < \frac{1}{2}$ or $x > 3$.

Therefore, the correct option is:

$x > 0 :\frac{1}{2} < x < 3$

$x > 3$ or $x < 0 : x <\frac{1}{2}$

To solve this problem, we need to find when the quadratic function changes from positive to negative and vice versa.

• Step 1: Find the roots using the quadratic formula. For $y = 2x^2 + 7x - 9$, identify $a = 2$, $b = 7$, and $c = -9$.
• Step 2: Apply the quadratic formula:
  $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Step 3: Calculate:
  $x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-9)}}{2 \cdot 2}$
  $x = \frac{-7 \pm \sqrt{49 + 72}}{4}$
  $x = \frac{-7 \pm \sqrt{121}}{4}$
  $x = \frac{-7 \pm 11}{4}$
• Step 4: Solve for the roots:
  $x = \frac{-7 + 11}{4} = 1$
  $x = \frac{-7 - 11}{4} = -\frac{18}{4} = -4.5$
• Step 5: The roots are $x = 1$ and $x = -4.5$. These roots divide the real number line into intervals: $(-\infty, -4.5)$, $(-4.5, 1)$, and $(1, \infty)$.
• Step 6: Test each interval:
• Interval $(-\infty, -4.5)$: Choose a test point like $x = -5$.
  Calculate $y$ using $x = -5$:
  $y = 2(-5)^2 + 7(-5) - 9 = 50 - 35 - 9 = 6 \, (> 0)$. The function is positive in this interval.
• Interval $(-4.5, 1)$: Choose a test point like $x = 0$.
  Calculate $y$ using $x = 0$:
  $y = 2(0)^2 + 7(0) - 9 = -9 \, (< 0)$. The function is negative in this interval.
• Interval $(1, \infty)$: Choose a test point like $x = 2$.
  Calculate $y$ using $x = 2$:
  $y = 2(2)^2 + 7(2) - 9 = 8 + 14 - 9 = 13 \, (> 0)$. The function is positive in this interval.

The positive domain of $y$ is $(-\infty, -4.5) \cup (1, \infty)$.
The negative domain of $y$ is $(-4.5, 1)$.

Thus, the domains are as follows:
$x < 0 : -4\frac{1}{2} < x < 1$
$x > 1$ or $x > 0 : x < -4\frac{1}{2}$

To solve this problem, we examine the function $y = -\frac{10}{3}x^2 - 0.4$.

Step 1: Identify $a = -\frac{10}{3}$, which is negative. This means the parabola opens downwards.

Step 2: Find the zeros by setting $y = 0$:

$-\frac{10}{3}x^2 - 0.4 = 0$. Solving gives $-\frac{10}{3}x^2 = 0.4$, yielding

$x^2 = -\frac{0.4 \times 3}{10}$, which is negative. Therefore, no real solutions exist for zero crossing.

Step 3: Recognize since it doesn't cross the x-axis, the entire parabola is below the x-axis.

Conclusion: For $x < 0$, $y$ is negative for all $x$; for $x > 0$, $y$ remains negative.

Therefore, the solution is:
$x < 0 :$ all $x$
$x > 0 :$ none

To find the positive and negative domains of the function $y = -3x^2 + 4\frac{11}{16}$, we need to first find the roots of the equation.

Step 1: Set the function equal to zero to find the roots:

$-3x^2 + 4\frac{11}{16} = 0$

Simplify the equation:

First, express $4\frac{11}{16}$ as a fraction:

$4\frac{11}{16} = \frac{75}{16}$

Substitute into the equation:

$-3x^2 + \frac{75}{16} = 0$

Multiply through by 16 to clear the fraction:

$-48x^2 + 75 = 0$ $48x^2 = 75$

Divide both sides by 48:

$x^2 = \frac{75}{48}$

Simplify the fraction:

$x^2 = \frac{25}{16}$

Take the square root of both sides:

$x = \pm \frac{5}{4}$

We have two roots: $x = \frac{5}{4}$ and $x = -\frac{5}{4}$.

Step 2: Identify the intervals to test for positivity and negativity.

The critical points create intervals: $(-\infty, -\frac{5}{4})$, $(-\frac{5}{4}, \frac{5}{4})$, and $(\frac{5}{4}, \infty)$.

Step 3: Determine the sign of $y$ in each interval by testing sample points:

• For $x \in (-\infty, -\frac{5}{4})$ (e.g., $x = -2$): $y = -3(-2)^2 + \frac{75}{16}$ yields a negative value.
• For $x \in (-\frac{5}{4}, \frac{5}{4})$ (e.g., $x = 0$): $y = -3(0)^2 + \frac{75}{16} = \frac{75}{16}$, which is positive.
• For $x \in (\frac{5}{4}, \infty)$ (e.g., $x = 2$): $y = -3(2)^2 + \frac{75}{16}$ yields a negative value.

Thus, the function is positive for $x \in \left(-\frac{5}{4}, \frac{5}{4}\right)$ and negative for $x < -\frac{5}{4}$ and $x > \frac{5}{4}$.

Therefore, the solution to the problem is:

$x > 1\frac{1}{4}$ or $x < 0 : x < -1\frac{1}{4}$

$x > 0 : -1\frac{1}{4} < x < 1\frac{1}{4}$

This matches choice 4.

To find the positive and negative domains of the function $y = -3x^2 + 5x - 2$, we follow these steps:

• Step 1: Find the roots of the equation by solving $-3x^2 + 5x - 2 = 0$.
• Step 2: The quadratic equation $ax^2 + bx + c = 0$ has roots $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant, $b^2 - 4ac = 5^2 - 4(-3)(-2) = 25 - 24 = 1$.
• Step 4: Find the roots: $x = \frac{-5 \pm \sqrt{1}}{2(-3)} = \frac{-5 \pm 1}{-6}$.
• Step 5: This gives roots $x = \frac{-4}{-6} = \frac{2}{3}$ and $x = \frac{-6}{-6} = 1$.
• Step 6: Determine intervals: $(-\infty, \frac{2}{3})$, $(\frac{2}{3}, 1)$, $(1, \infty)$.
• Step 7: Test a point from each interval in the original equation to determine sign:
• For $x \in (-\infty, \frac{2}{3})$, choose $x = 0$: $y = -3(0)^2 + 5(0) - 2 = -2$ (negative).
• For $x \in (\frac{2}{3}, 1)$, choose $x = 0.8$: $y = -3(0.8)^2 + 5(0.8) - 2 = -1.28 + 4 - 2 = 0.72$ (positive).
• For $x \in (1, \infty)$, choose $x = 2$: $y = -3(2)^2 + 5(2) - 2 = -12 + 10 - 2 = -4$ (negative).

Therefore, the positive domains of the function are when $\frac{2}{3} < x < 1$, and the negative domains are when $x < \frac{2}{3}$ or $x > 1$.

Thus, the solution to the problem is:

$x > 1$ or $x < 0 : x < \frac{2}{3}$

$x > 0 : \frac{2}{3} < x < 1$

To determine the positive and negative domains of the quadratic function $y = 3x^2 + 7x + 2$, we will first find its roots using the quadratic formula.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

For this function, $a = 3$, $b = 7$, and $c = 2$.

First, calculate the discriminant $\Delta$:

$\Delta = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25$.

Since the discriminant is positive, the function has two distinct real roots.

Now, calculate the roots:

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$.

$x = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}$.

This results in:

• Root 1: $x = \frac{-7 + 5}{6} = \frac{-2}{6} = -\frac{1}{3}$
• Root 2: $x = \frac{-7 - 5}{6} = \frac{-12}{6} = -2$

The roots are $x = -2$ and $x = -\frac{1}{3}$, dividing the x-axis into three intervals: $x < -2$, $-2 < x < -\frac{1}{3}$, and $x > -\frac{1}{3}$.

For $x \to -\infty$, the quadratic $y = 3x^2 + 7x + 2$ is positive, as the leading coefficient $a = 3$ is positive, indicating the parabola opens upwards.

Evaluate the sign of $y$ within the intervals:

• For $x < -2$, pick $x = -3$: $y = 3(-3)^2 + 7(-3) + 2 = 27 - 21 + 2 = 8$. Hence, $y > 0$.
• For $-2 < x < -\frac{1}{3}$, pick $x = -1$: $y = 3(-1)^2 + 7(-1) + 2 = 3 - 7 + 2 = -2$. Hence, $y < 0$.
• For $x > -\frac{1}{3}$, pick $x = 0$: $y = 3(0)^2 + 7(0) + 2 = 2$. Hence, $y > 0$.

Therefore, the positive domain of the function is $x < -2$ and $x > -\frac{1}{3}$, and the negative domain is $-2 < x < -\frac{1}{3}$.

Thus, the solution matches the given correct answer:

$x < 0 : -2 < x < -\frac{1}{3}$

$x > -\frac{1}{3}$ or $x > 0 : x < -2$

To solve this problem, let's start by finding the roots of the quadratic equation:

The given function is $y = -4x^2 + x + 3$. We set it to zero to find the roots:

$-4x^2 + x + 3 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -4$, $b = 1$, and $c = 3$:

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

$x = \frac{-1 \pm \sqrt{1 + 48}}{-8}$

$x = \frac{-1 \pm \sqrt{49}}{-8}$

$x = \frac{-1 \pm 7}{-8}$

The roots are:

• $x_1 = \frac{-1 + 7}{-8} = \frac{6}{-8} = -\frac{3}{4}$
• $x_2 = \frac{-1 - 7}{-8} = \frac{-8}{-8} = 1$

These roots divide the number line into intervals: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

We test each interval to determine where the function is positive or negative:

Interval $(-\infty, -\frac{3}{4})$: Choose $x = -1$.

• $y = -4(-1)^2 + (-1) + 3 = -4 - 1 + 3 = -2$ (Negative)

Interval $(-\frac{3}{4}, 1)$: Choose $x = 0$.

• $y = -4(0)^2 + 0 + 3 = 3$ (Positive)

Interval $(1, \infty)$: Choose $x = 2$.

• $y = -4(2)^2 + 2 + 3 = -16 + 2 + 3 = -11$ (Negative)

Therefore, the function is positive in the interval $(-\frac{3}{4}, 1)$ and negative in the intervals $(-\infty, -\frac{3}{4})$ and $(1, \infty)$.

Thus, the positive and negative domains of the function are:

$x > 0 : -\frac{3}{4} < x < 1$

$x > 1$ or $x < 0 : x < -\frac{3}{4}$

The correct answer choice corresponds to:

$x > 1$ or $x<0:x<-\frac{3}{4}$

$x > 0 : -\frac{3}{4} < x <1$

To solve for the positive and negative domains of the function $y = 4x^2 - x - 3$, follow these steps:

• Step 1: Identify and apply the quadratic formula to find the roots.
• Step 2: Calculate the discriminant to ensure real roots.
• Step 3: Analyze the sign changes in the resulting intervals.

Step 1: The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 4$, $b = -1$, and $c = -3$.

Step 2: Calculate the discriminant: $b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-3) = 1 + 48 = 49$. Since the discriminant is positive, two distinct real roots exist.

Step 3: Calculate the roots using the formula:

$x = \frac{-(-1) \pm \sqrt{49}}{2 \times 4} = \frac{1 \pm 7}{8}$

Thus, the roots $x_1 = \frac{1 + 7}{8} = 1$ and $x_2 = \frac{1 - 7}{8} = -\frac{3}{4}$.

Now we examine the sign of $y = 4x^2 - x - 3$ across the intervals determined by these roots: $(-\infty, -\frac{3}{4})$, $(-\frac{3}{4}, 1)$, and $(1, \infty)$.

• For $x < -\frac{3}{4}$: Choose $x = -1$. Substitute into the function: $y = 4(-1)^2 - (-1) - 3 = 4 + 1 - 3 = 2$. Positive.
• For $-\frac{3}{4} < x < 1$: Choose $x = 0$. Substitute: $y = 4(0)^2 - (0) - 3 = -3$. Negative.
• For $x > 1$: Choose $x = 2$. Substitute: $y = 4(2)^2 - (2) - 3 = 16 - 2 - 3 = 11$. Positive.

Therefore, the positive domains are $x < -\frac{3}{4}$ and $x > 1$, and the negative domain is $-\frac{3}{4} < x < 1$.

The positive and negative domains are: $x < 0 : -\frac{3}{4} < x < 1$ and $x > 1$ or $x > 0 : x < -\frac{3}{4}$.

The given function is $y = -\frac{1}{2}x^2 + 2\frac{25}{72}$. We need to find when this function is equal to zero to determine the positive and negative domains.

First, identify the coefficients from the function:

• $a = -\frac{1}{2}$, $b = 0$, $c = 2\frac{25}{72}$ which we convert to improper fraction: $\frac{169}{72}$.

We then use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the roots.

Substituting in our values:

$x = \frac{-0 \pm \sqrt{0 - 4\left(-\frac{1}{2}\right)\left(\frac{169}{72}\right)}}{-1}$.

The discriminant calculation is as follows:

$4 \cdot \frac{1}{2} \cdot \frac{169}{72} = \frac{169}{36}$.

So our roots are:

$x = \pm \sqrt{\frac{169}{36}} = \pm \frac{13}{6} = \pm 2\frac{1}{6}$.

The roots are $x = 2\frac{1}{6}$ and $x = -2\frac{1}{6}$. These divide the x-axis into intervals to be tested.

- When $x < -2\frac{1}{6}$, test point $x = -3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x < -2\frac{1}{6}$ gives negative values.

- When $-2\frac{1}{6} < x < 2\frac{1}{6}$, test $x = 0$:
$y = -\frac{1}{2}(0)^2 + \frac{169}{72} > 0$: Positive.
Hence, $-2\frac{1}{6} < x < 2\frac{1}{6}$ gives positive values.

- When $x > 2\frac{1}{6}$, test point $x = 3$:
$y = -\frac{1}{2}(3)^2 + \frac{169}{72} < 0$: Negative.
Hence, $x > 2\frac{1}{6}$ gives negative values.

Therefore, the positive domain is $x > 0 : -2\frac{1}{6} < x < 2\frac{1}{6}$ and the negative domain is $x < 0 : x < -2\frac{1}{6}$ or $x > 2\frac{1}{6}$.

In comparing to the provided choices, the correct choice is Choice 2: $x > 2\frac{1}{6}$ or $x < 0 : x < -2\frac{1}{6}$

$x > 0 : - 2\frac{1}{6} < x < 2\frac{1}{6}$

To find the positive and negative domains of the function $y = -\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4}$, we must determine where the function is above or below the x-axis.

Step 1: Find the roots of the quadratic equation. This requires solving:

$-\frac{1}{2}x^2 + \frac{1}{3}x - \frac{1}{4} = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = -\frac{1}{2}$, $b = \frac{1}{3}$, and $c = -\frac{1}{4}$, we calculate:

• Discriminant: $b^2 - 4ac = \left(\frac{1}{3}\right)^2 - 4\left(-\frac{1}{2}\right)\left(-\frac{1}{4}\right) = \frac{1}{9} - \frac{1}{2} = \frac{1}{9} - \frac{4}{8} = \frac{1}{9} - \frac{9}{18} = \frac{1 - 9}{18} = -\frac{8}{18} = -\frac{4}{9}$

The discriminant is negative, indicating no real roots.

Step 2: Analyze the parabola's orientation. Because the leading term is negative, the parabola opens downwards. With no x-intercepts, this implies the entire graph is below the x-axis.

Therefore, the function is negative for all x-values. In the context of positive and negative domains:

$x > 0 :$ none, as the function doesn't cross the x-axis in positive domain.

$x < 0 :$ all $x$, as the function is always negative.

To determine the positive and negative domains of the quadratic function $y = \frac{1}{2}x^2 + \frac{3}{4}x + \frac{5}{6}$, we start by considering the possibility of real roots using the discriminant.

The discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac = \left(\frac{3}{4}\right)^2 - 4\left(\frac{1}{2}\right)\left(\frac{5}{6}\right)$

Calculating gives:

$\Delta = \frac{9}{16} - \frac{20}{12} = \frac{9}{16} - \frac{5}{3} = \frac{9}{16} - \frac{80}{48}$

Convert $\frac{9}{16}$ to a common denominator:

$\Delta = \frac{27}{48} - \frac{80}{48} = -\frac{53}{48}$

The discriminant $\Delta$ is negative, indicating that this quadratic equation has no real roots.

Since the coefficient $a = \frac{1}{2}$ is positive and there are no real roots, the parabola opens upwards and never crosses the x-axis.

This means that the function is always positive for all $x$.

Thus, the positive domain is all $x$, and there is no negative domain.

Therefore, the correct choice is:

$x > 0 :$ for all $x$

$x < 0 :$ none