Positive and Negative Domains: Representation of missing c value

The given quadratic function is $y = -3x^2 + 12x$. We are tasked with finding for which values of $x$ this function is negative, i.e., $y < 0$.

First, identify the roots of the quadratic by solving the equation:

$-3x^2 + 12x = 0$

Factor out common terms:

$x(-3x + 12) = 0$

This gives us two solutions or critical points:

$x = 0 \quad \text{and} \quad -3x + 12 = 0$

Solve for $x$ in the second equation:

$-3x + 12 = 0 \Rightarrow -3x = -12 \Rightarrow x = \frac{-12}{-3} \Rightarrow x = 4$

The roots of the quadratic are $x = 0$ and $x = 4$. These roots divide the real number line into three intervals:

• $x < 0$
• $0 < x < 4$
• $x > 4$

To find where the function is negative, evaluate the sign of $y$ in these intervals:

• For $x < 0$: Choose $x = -1$. Then, $-3(-1)^2 + 12(-1) = -3 - 12 = -15$, which is negative.
• For $0 < x < 4$: Choose $x = 2$. Then, $-3(2)^2 + 12(2) = -12 + 24 = 12$, which is positive.
• For $x > 4$: Choose $x = 5$. Then, $-3(5)^2 + 12(5) = -75 + 60 = -15$, which is negative.

The function $y = -3x^2 + 12x$ is negative for $x < 0$ and $x > 4$.

Therefore, the values of $x$ that satisfy $f(x) < 0$ are:

$x < 0$ and $x > 4$.

To determine for which values of $x$ the quadratic function is positive, follow these steps:

• Step 1: Set the quadratic equation to zero: $-x^2 - 5x = 0$.
• Step 2: Factor the equation: $-x(x + 5) = 0$.
• Step 3: Solve for the roots: $x = 0$ and $x = -5$.
• Step 4: Determine the intervals defined by the roots: $(-\infty, -5)$, $(-5, 0)$, and $(0, \infty)$.
• Step 5: Test each interval:
• For interval $(-\infty, -5)$, choose $x = -6$: $-(-6)^2 - 5(-6) = -36 + 30 = -6$ (negative).
• For interval $(-5, 0)$, choose $x = -1$: $-(-1)^2 - 5(-1) = -1 + 5 = 4$ (positive).
• For interval $(0, \infty)$, choose $x = 1$: $-(1)^2 - 5(1) = -1 - 5 = -6$ (negative).
• Step 6: The function is positive in the interval $(-5, 0)$.

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

To determine the values of $x$ for which the function $y = -4x^2 + 24x$ is greater than zero, we will proceed as follows:

Step 1: Find the roots of the quadratic equation.

We start by solving $-4x^2 + 24x = 0$ to find the critical points. This can be factored as:

$x(-4x + 24) = 0$

This equation gives us two roots:

1. $x = 0$
2. $-4x + 24 = 0 \Rightarrow x = 6$

Step 2: Determine intervals for positivity.

The roots divide the number line into three intervals: $(-\infty, 0)$, $(0, 6)$, and $(6, \infty)$.

Since the parabola opens downwards (as indicated by the negative leading coefficient), the function will be positive between the roots:

$0 < x < 6$

Conclusion: To ensure the function $y = -4x^2 + 24x$ is greater than zero, the value of $x$ must be between 0 and 6.

Therefore, the solution to the problem is $0 < x < 6$.

To determine when the function $f(x) = x^2 + 4x$ is positive, we start by analyzing the quadratic expression. The expression can be factored as:

$x^2 + 4x = x(x + 4)$

To find when this is greater than zero, identify the roots of the equation $x(x + 4) = 0$. Solving this, we find the roots to be:

• $x = 0$
• $x = -4$

These roots split the real number line into three intervals, which we must analyze to determine where the function is positive:

• Interval 1: $x < -4$
• Interval 2: $-4 < x < 0$
• Interval 3: $x > 0$

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

• For $x < -4$: Choose $x = -5$. Then $f(-5) = (-5)((-5) + 4) = 5$, which is positive.
• For $-4 < x < 0$: Choose $x = -2$. Then $f(-2) = (-2)((-2) + 4) = -4$, which is negative.
• For $x > 0$: Choose $x = 1$. Then $f(1) = (1)((1) + 4) = 5$, which is positive.

From this analysis, the function $f(x)$ is positive in the intervals:

$x < -4$ and $x > 0$.

Therefore, the correct choice is:

x < -4 or x > 0

To determine where the function $f(x) = -3x^2 + 12x$ is greater than zero, we need to find the roots of the equation.

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

$-3x^2 + 12x = 0$

Step 2: Factor the equation:

$-3x(x - 4) = 0$

Setting each factor equal to zero gives us the roots:

• $-3x = 0 \rightarrow x = 0$
• $x - 4 = 0 \rightarrow x = 4$

Step 3: Since the parabola opens downwards (as the coefficient of $x^2$ is negative), $f(x) > 0$ for $x$ values between the roots. Thus, the function is positive between $x = 0$ and $x = 4$.

Therefore, the solution is the interval $0 < x < 4$.

In conclusion, the values of $x$ for which the function is greater than zero are $0 < x < 4$.

To determine where the function $y = -x^2 - 5x$ is negative, follow these steps:

• Step 1: Find the roots of the equation $-x^2 - 5x = 0$.
• Step 2: Factor the equation: $-x(x + 5) = 0$.
• Step 3: Solve for $x$:
  • $x = 0$
  • $x = -5$
• Step 4: Analyze the sign of the quadratic function in the intervals determined by the roots:
  • Interval 1: $x < -5$
  • Interval 2: $-5 < x < 0$
  • Interval 3: $x > 0$

Consider an example point in each interval to determine if $y$ is negative:

• For $x < -5$, choose $x = -6$:
  $y = -(-6)^2 - 5(-6) = -36 + 30 = -6$, which is negative.
• For $-5 < x < 0$, choose $x = -1$:
  $y = -(-1)^2 - 5(-1) = -1 + 5 = 4$, which is positive.
• For $x > 0$, choose $x = 1$:
  $y = -(1)^2 - 5(1) = -1 - 5 = -6$, which is negative.

Therefore, the intervals where $f(x) < 0$ are $x < -5$ and $x > 0$.

The correct answer is $x > 0$ or $x < -5$.

To solve for the values of $x$ where $y = 2x^2 - 24x > 0$, we begin with the quadratic equation:

$y = 2x^2 - 24x$

First, factor the quadratic expression:

$y = 2x(x - 12)$

To find where this expression is greater than zero, first determine the zeros of the function by setting the equation to zero:

$2x(x - 12) = 0$

Solving for $x$, we find:

• $x = 0$
• $x = 12$

These zeros divide the number line into three intervals to test: $x < 0$, $0 < x < 12$, and $x > 12$.

Choose test points from each interval, such as $x = -1$, $x = 1$, and $x = 13$, to evaluate the sign of the expression $2x(x - 12)$:

• For $x = -1$: $2(-1)((-1) - 12) = 26$, thus positive.
• For $x = 1$: $2(1)((1) - 12) = -22$, thus negative.
• For $x = 13$: $2(13)((13) - 12) = 26$, thus positive.

From the above test results, $y = 2x(x - 12) > 0$ when $x < 0$ or $x > 12$.

Thus, the values of $x$ that satisfy $f(x) > 0$ are:

$x > 12$ or $x < 0$

To solve for which values of $x$ the function $y = 4x^2 + 8x$ is negative:

• Step 1: Start by factoring the quadratic expression:
  $y = 4x^2 + 8x = 4x(x + 2)$.
• Step 2: Set each factor to zero to find the roots:
  $4x = 0$ or $x + 2 = 0$, giving roots $x = 0$ and $x = -2$.
• Step 3: Analyze the intervals determined by these roots:
  - For $x < -2$, pick a test point like $x = -3$; plug into $4x(x + 2)$:
  $4(-3)((-3) + 2) = 4(-3)(-1) = 12 > 0$
  - For $-2 < x < 0$, pick a test point like $x = -1$; plug into $4x(x + 2)$:
  $4(-1)((-1) + 2) = 4(-1)(1) = -4 < 0$
  - For $x > 0$, pick a test point like $x = 1$; plug into $4x(x + 2)$:
  $4(1)(1 + 2) = 4(1)(3) = 12 > 0$.

Hence, the quadratic is negative in the interval $-2 < x < 0$.

The correct answer is therefore $-2 < x < 0$.

To solve the inequality $3x^2 + 6x > 0$, we first factor the quadratic equation.

• Step 1: Factor the quadratic expression:
  $3x^2 + 6x = 3x(x + 2)$.
• Step 2: Find the roots by setting the factored expression equal to zero:
  - $3x(x + 2) = 0$ gives $x = 0$ and $x = -2$ as roots.
• Step 3: Determine the intervals dictated by the roots on a number line:
  - The critical points divide the number line into three intervals: $(-\infty, -2)$, $(-2, 0)$, and $(0, \infty)$.
• Step 4: Analyze the sign of $3x(x + 2)$ in each interval:
• Interval $(-\infty, -2)$: Pick a test point like $x = -3$.
  $3(-3)((-3) + 2) = 3(-3)(-1) = 9 > 0$. Hence, $f(x) > 0$.
• Interval $(-2, 0)$: Pick a test point like $x = -1$.
  $3(-1)((-1) + 2) = 3(-1)(1) = -3 < 0$. Hence, $f(x) < 0$.
• Interval $(0, \infty)$: Pick a test point like $x = 1$.
  $3(1)((1) + 2) = 3(1)(3) = 9 > 0$. Hence, $f(x) > 0$.

Therefore, the solution to the inequality $3x^2 + 6x > 0$ is:
$x > 0$ or $x < -2$.

The correct choice from the given options is .

Thus, when $x > 0$ or $x < -2$, the function $f(x) = 3x^2 + 6x$ is greater than zero.

To solve the inequality $x^2 + 4x < 0$, we first need to determine the roots of the quadratic equation $x^2 + 4x = 0$.

Step 1: Find roots of the equation:

Factor the quadratic expression: $x(x + 4) = 0$.

Setting each factor to zero gives us the roots:

• $x = 0$
• $x + 4 = 0 \Rightarrow x = -4$

Step 2: Analyze intervals between the roots:

The roots divide the real number line into intervals: $(-\infty, -4)$, $(-4, 0)$, and $(0, \infty)$.

Step 3: Test the sign of $f(x) = x^2 + 4x$ in each interval:

• For $x \in (-\infty, -4)$, pick $x = -5$: $f(-5) = (-5)^2 + 4(-5) = 25 - 20 = 5$ (positive).
• For $x \in (-4, 0)$, pick $x = -2$: $f(-2) = (-2)^2 + 4(-2) = 4 - 8 = -4$ (negative).
• For $x \in (0, \infty)$, pick $x = 1$: $f(1) = 1^2 + 4(1) = 1 + 4 = 5$ (positive).

Step 4: Conclusion:

The function $f(x) = x^2 + 4x$ is negative in the interval $(-4, 0)$, specifically $-4 < x < 0$.

Therefore, the values of $x$ for which $f(x) < 0$ are in the interval $-4 < x < 0$.

To solve the problem, follow these steps:

• Step 1: Solve the quadratic equation $2x^2 - 24x = 0$.
• Step 2: Factor the equation as $2x(x - 12) = 0$.
• Step 3: Determine the roots, which are $x = 0$ and $x = 12$.
• Step 4: Analyze the intervals determined by these roots: $x < 0$, $0 < x < 12$, and $x > 12$.
• Step 5: Test each interval to see where $f(x) < 0$.

Now let's work through each step:

Step 1: Solve the equation $2x(x - 12) = 0$. This gives us roots at $x = 0$ and $x = 12$.

Step 2: The quadratic can be negative between the roots, so we consider the interval $0 < x < 12$.

Step 3: Test the sign of $2x(x - 12)$ in each interval:

• Interval $(-\infty, 0)$: Choose a test point $x = -1$. The expression is $2(-1)((-1) - 12) = 2(-1)(-13) = 26$, which is positive.
• Interval $(0, 12)$: Choose a test point $x = 1$. The expression is $2(1)(1 - 12) = 2(1)(-11) = -22$, which is negative.
• Interval $(12, \infty)$: Choose a test point $x = 13$. The expression is $2(13)(13 - 12) = 2(13)(1) = 26$, which is positive.

Thus, the quadratic is negative in the interval $0 < x < 12$.

Therefore, the solution to the problem is $0 < x < 12$.

To solve this problem, we first need to find the roots of the quadratic function $y = 4x^2 + 8x$.

Let's factor the quadratic equation:

$y = 4x^2 + 8x = 4x(x + 2)$

The roots of this equation are found by setting each factor to zero:

$4x = 0$ gives $x = 0$

$x + 2 = 0$ gives $x = -2$

Thus, the roots are $x = 0$ and $x = -2$. These roots divide the number line into three intervals: $(-\infty, -2)$, $(-2, 0)$, and $(0, \infty)$.

Next, we test a value from each interval to determine where the function is positive:

• For the interval $(-\infty, -2)$, test $x = -3$:
  $y = 4(-3)^2 + 8(-3) = 36 - 24 = 12$ (positive)
• For the interval $(-2, 0)$, test $x = -1$:
  $y = 4(-1)^2 + 8(-1) = 4 - 8 = -4$ (negative)
• For the interval $(0, \infty)$, test $x = 1$:
  $y = 4(1)^2 + 8(1) = 4 + 8 = 12$ (positive)

Based on these tests, $y = 4x^2 + 8x$ is positive in the intervals $x < -2$ and $x > 0$.

Therefore, the solution to the problem is $x > 0$ or $x < -2$.

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

• Step 1: Find the roots by setting the function equal to zero: $-4x^2 + 24x = 0$.
• Step 2: Factor the equation. We get $-4x(x - 6) = 0$.
• Step 3: Solve for $x$ to find the roots: $x = 0$ and $x = 6$.
• Step 4: Analyze intervals defined by these roots: these intervals are $(-\infty, 0)$, $(0, 6)$, and $(6, \infty)$.
• Step 5: Test the sign of $y = -4x^2 + 24x$ in each interval:
• In $(-\infty, 0)$, choose $x = -1$: $-4(-1)^2 + 24(-1) = -4 - 24 = -28$, so $y < 0$.
• In $(0, 6)$, choose $x = 3$: $-4(3)^2 + 24(3) = -36 + 72 = 36$, so $y > 0$.
• In $(6, \infty)$, choose $x = 7$: $-4(7)^2 + 24(7) = -196 + 168 = -28$, so $y < 0$.
• Step 6: Compile the solution set based on where $y < 0$: the intervals are $(-\infty, 0)$ and $(6, \infty)$.

Therefore, the solution is that $f(x) < 0$ for $x > 6$ or $x < 0$.

To solve this problem, we'll perform the following steps:

• Step 1: Determine the roots of the equation $-2x^2 - 16x = 0$ using the quadratic formula.
• Step 2: Identify the intervals determined by these roots.
• Step 3: Test a value from each interval to determine where the quadratic function is positive.

Step 1: Finding the roots of the quadratic equation.

The quadratic equation is $-2x^2 - 16x = 0$. We can simplify this by factoring:

Factor out the common term: $-2x(x + 8) = 0$.

Setting each factor to zero, we find the roots:

• $-2x = 0 \implies x = 0$
• $x + 8 = 0 \implies x = -8$

Step 2: Use these roots to determine intervals on the number line: $(-\infty, -8)$, $(-8, 0)$, and $(0, \infty)$.

Step 3: Test each interval to see where the function is positive:

• Choose $x = -9$ in the interval $(-\infty, -8)$:
  $y = -2(-9)^2 -16(-9) = -162 + 144 = -18$. Negative, so the function is not positive here.
• Choose $x = -4$ in the interval $(-8, 0)$:
  $y = -2(-4)^2 -16(-4) = -32 + 64 = 32$. Positive, so the function is positive here.
• Choose $x = 1$ in the interval $(0, \infty)$:
  $y = -2(1)^2 -16(1) = -2 - 16 = -18$. Negative, so the function is not positive here.

Thus, the function $y = -2x^2 - 16x$ is positive for $x$ in the interval $(-8, 0)$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-8 < x < 0$.

To solve this problem, we need to determine when the quadratic function $y = -2x^2 - 16x$ is less than zero. Let us follow these steps:

First, identify the roots of the quadratic equation by setting $y = 0$:

$-2x^2 - 16x = 0$

Factor out the common factor:

$-2x(x + 8) = 0$

The solutions to this equation give the x-values where the function equals zero (its roots):

So, $x = 0$ or $x = -8$

Now, analyze the intervals determined by these roots:

• Interval 1: $x < -8$
• Interval 2: $-8 < x < 0$
• Interval 3: $x > 0$

The quadratic $y = -2x^2 - 16x$ is a downward-opening parabola. We know it is zero at the roots.

Let's analyze the sign of $y$ in each interval:

• In Interval 1 ($x < -8$): Choose a test point $x = -10$
  $y = -2(-10)^2 - 16(-10) = -200 + 160 = -40$ (negative).
• In Interval 2 ($-8 < x < 0$): Choose a test point $x = -4$
  $y = -2(-4)^2 - 16(-4) = -32 + 64 = 32$ (positive).
• In Interval 3 ($x > 0$): Choose a test point $x = 1$
  $y = -2(1)^2 - 16(1) = -2 - 16 = -18$ (negative).

Hence, the function is negative in Intervals 1 and 3: where $x < -8$ or $x > 0$.

Therefore, the solution to the problem is $x < -8$ or $x > 0$.

Referring to the multiple-choice options, the correct answer is: Option 2.

Thus, the solution to the problem is $x > 0$ or $x < -8$.

We will work through the problem as follows:

• Step 1: Start by factoring the quadratic equation $y = 3x^2 + 6x$.
• Step 2: Solve for $x$ where $y = 0$.
• Step 3: Analyze the intervals between the roots to find where $y < 0$.

Let's proceed to the solution:
Step 1: Factor the quadratic expression:

The equation $y = 3x^2 + 6x$ can be factored by taking out the common factor:

$y = 3x(x + 2)$.

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

$3x(x + 2) = 0$ gives the roots $x = 0$ and $x = -2$.

Step 3: Analyze the intervals determined by these roots:

• Interval $(-\infty, -2)$: Choose a test point like $x = -3$. Substituting into $3x(x + 2)$ gives a positive value.
• Interval $(-2, 0)$: Choose a test point like $x = -1$. Substituting into $3x(x + 2)$ gives a negative value.
• Interval $(0, \infty)$: Choose a test point like $x = 1$. Substituting into $3x(x + 2)$ gives a positive value.

Therefore, the function $y = 3x^2 + 6x$ is less than zero on the interval $-2 < x < 0$.

Therefore, the solution to the problem is $-2 < x < 0$.

First, we need to find the roots of the quadratic equation:

The quadratic is given by:
$y = -\frac{1}{7}x^2 - \frac{17}{7}x$

Setting $y = 0$ to find the $x$-intercepts (roots):
$-\frac{1}{7}x^2 - \frac{17}{7}x = 0$

Factor out the common factor, $-\frac{1}{7}x$:
$-\frac{1}{7}x(x + 17) = 0$

This gives the roots:
$x = 0$ and $x + 17 = 0 \rightarrow x = -17$

These roots divide the number line into intervals. We need to determine where $y > 0$. Because the coefficient of $x^2$ is negative, the parabola opens downward. The function will be positive between the roots.

Thus, we test the interval:
$(-17, 0)$

Since the parabola opens downward, the function $y > 0$ is true in the interval $-17 < x < 0$.

Therefore, the solution to the problem is $-17 < x < 0$, which corresponds to choice 3.

Let's solve the problem step by step:

The function given is $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. The first step is to convert the mixed number into an improper fraction:

$1\frac{1}{3} = \frac{4}{3}$, so the function becomes:

$y = \frac{1}{5}x^2 + \frac{4}{3}x$.

This can be written in standard form $ax^2 + bx + c = 0$ where $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$ (not visible, but necessary for proper representation).

Next, we use the quadratic formula to find the roots:

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

Plug in the values:

$x = \frac{-\frac{4}{3} \pm \sqrt{\left(\frac{4}{3}\right)^2 - 4 \times \frac{1}{5} \times 0}}{2 \times \frac{1}{5}}$.

Simplify where possible:

$x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{\frac{2}{5}}$, leading to:

$x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}}$.

This gives roots:

$x = 0$ and $x = -6\frac{2}{3}$.

The parabola opens upward (since $a = \frac{1}{5} > 0$). The quadratic function is positive between the roots and for values outside them:

This results in two intervals where $f(x) > 0$:

• $x < -6\frac{2}{3}$
• $x > 0$

Thus, the solution to the problem is:

$x > 0$ or $x < -6\frac{2}{3}$.