Increasing and Decreasing Domain of a Parabola: Worded problems

Given any quadratic function given by $f(x) = ax^2 + bx + c$, it describes a parabola. Whether it opens upwards or downwards depends on the coefficient $a$.

A parabola opens upwards when $a > 0$, where it reaches a minimum at the vertex. It opens downwards when $a < 0$, where the vertex is a maximum point. Since the problem specifies a "decreasing parabola," we can infer that $a < 0$, indicating an opening downwards.

Importantly, if a parabola does not touch the $x$-axis, its discriminant $b^2 - 4ac$ is less than zero, indicating no real roots, and the parabola never crosses or touches the x-axis.

An opening downwards parabola has segments that are increasing on the left of the vertex and decreasing on the right. Therefore, it is incorrect to claim a downwards-opening parabola is always increasing, even if it doesn't touch the $x$-axis.

Therefore, the statement, "the parabola is always increasing," in this context is False.

Thus, the correct answer is:

False

Let's solve this problem step-by-step:

The problem addresses a quadratic function or "parabola" with the equation typically given as $f(x) = ax^2 + bx + c$, where $a, b,$ and $c$ are real numbers. For a parabola that opens upwards, the coefficient $a$ must be positive ($a > 0$).

Given that the parabola does not intersect or touch the x-axis, this means there are no real roots. This condition is satisfied when the discriminant $\Delta = b^2 - 4ac$ is less than zero ($\Delta < 0$).

For a parabola described by $ax^2 + bx + c$:

• The vertex forms the minimum point since $a > 0$. The vertex $(h, k)$ is calculated using $h = -\frac{b}{2a}$.
• The function is decreasing on the interval $(-\infty, h)$ and increasing on the interval $(h, \infty)$.

For it to be always decreasing would mean that $(-\infty, \infty)$, but we identified that it decreases up to $h$ and then increases beyond $h$. Thus, it cannot be always decreasing.

Conclusively, the statement that a positive parabola that does not intersect or touch the x-axis is always decreasing is False.

The problem provides insufficient information to determine the interval of increase as it lacks information regarding the parabola's orientation.

Therefore, the solution to the problem is: Cannot be determined.

To determine if a parabola is always decreasing when it does not intersect or touch the x-axis, we analyze the properties of quadratic functions.

• Step 1: For a parabola to not intersect the x-axis, its discriminant must be negative: $b^2 - 4ac < 0$. This condition ensures the quadratic equation has no real roots.
• Step 2: Depending on $a$, the parabola either opens upward ($a > 0$) or downward ($a < 0$).
• Step 3: A parabola is "always decreasing" only if it opens downward and does not have a minimum turning point in its domain, which is impossible for a standard quadratic function.
• Step 4: If $a > 0$, the parabola opens upwards, and cannot always decrease as it eventually increases after the vertex.
• Step 5: If $a < 0$, the parabola opens downwards; hence, it decreases after the vertex.

Therefore, regardless of whether the parabola opens upward or downward, it cannot "always be decreasing" because it either increases or decreases after the vertex. Thus, the statement is incorrect.

The correct answer is Incorrect.

To determine the intervals of increase and decrease for a parabola, we primarily rely on the coefficient of $x^2$ in the quadratic function, noted as $a$ in either the standard form $ax^2 + bx + c$ or vertex form $a(x-h)^2 + k$. The vertex of the parabola, given by $(h, k)$, plays a crucial role as the turning point.

Steps to find intervals of increase and decrease:

• Step 1: The coefficient $a$ tells us whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$). A parabola opening upwards decreases to the vertex and increases thereafter, while a parabola opening downwards increases to the vertex and decreases thereafter.
• Step 2: The vertex's $x$-coordinate, $h$, provides the dividing line between these intervals. The parabola is increasing on one side and decreasing on the other.

The intervals of increase and decrease depend on both $a$ and $h$ - not $k$ alone. Therefore, knowing just the $y$-coordinate of the vertex ($k$) is insufficient to determine these intervals, as it does not influence the $x$-intercepts or the opening direction.

Conclusively, knowledge of only the coefficient $a$ and the $y$-coordinate of the vertex is insufficient to fully determine the intervals of increase and decrease of a parabola. The intervals are primarily determined by the sign of $a$ and the vertex’s $x$-coordinate.

Therefore, the correct choice is: Incorrect.

To determine the intervals of increase for the given "smiling" parabola with a vertex at $x = 4$, follow these steps:

• Since the parabola is described as "smiling," it opens upwards, which means it has its minimum at the vertex.
• The function decreases in the interval to the left of the vertex and increases to the right of it.
• Thus, the parabola is increasing for $x > 4$.

Therefore, the correct interval of increase for the function is $x > 4$.

To determine whether a parabola that does not intersect or touch the x-axis is always increasing, we need to analyze its general behavior:

• A parabola described by the quadratic function $y = ax^2 + bx + c$ will open upwards if $a > 0$ and downwards if $a < 0$.
• The point $(h, k)$, derived from the parabola's vertex form $y = a(x-h)^2 + k$, defines its vertex. The vertex is the point of symmetry in a parabola.
• The condition that it does not touch or intersect the x-axis implies its vertex is either completely above or below the x-axis.
• If the parabola opens upwards ($a > 0$), there are sections where the graph is both increasing and decreasing, divided by the vertex, hence it cannot be always increasing.
• Similarly, if the parabola opens downwards ($a < 0$), it is both increasing and decreasing around the vertex, and thus it cannot be always increasing.

In both scenarios, the understanding that a parabola does not always increase stems from the symmetry of its form about its vertex.

Therefore, the claim that the parabola is always increasing is incorrect.

To find where the parabola is decreasing, remember that the parabola is symmetric around the vertical line passing through its vertex. The vertex given is at $x = 4$, meaning it is the point at which the direction changes from decreasing to increasing if the parabola opens upwards, which we'll assume unless told otherwise.

For a standard upward-opening parabola, to the left of the vertex (i.e., $x < 4$), the parabola is decreasing. This is because as $x$ approaches 4 from the left, the function's value increases until it reaches the vertex.

Therefore, the interval where the function is decreasing is:

$x < 4$

To solve this problem and find the intervals of increase for the function, let's follow these steps:

• Step 1: Identify the nature of the parabola.
• Step 2: Determine the intervals where the function is increasing.
• Step 3: Provide a clear explanation of the result.

Now, let's dive into each step:
Step 1: We know the quadratic function opens downwards because the coefficient of $x^2$ is $-2$, which is negative. This implies that the vertex is at a maximum point, and the parabola decreases on either side of the vertex.

Step 2: The vertex given is $(4, 5)$. For parabolas that open downwards, the function is increasing to the left of the vertex. Therefore, the function is increasing for $x < 4$.

Step 3: Therefore, the interval of increase for the parabola is $x < 4$.

In conclusion, the interval where the function is increasing is $x < 4$.

To solve this problem, let's begin by analyzing the given pieces of information:

• The coefficient of $x^2$: For a parabola in the form $f(x) = ax^2 + bx + c$, this coefficient $a$ determines whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$).
• The $x$-coordinate of the vertex: The vertex $x$-coordinate of the parabola is given by $x = -\frac{b}{2a}$. This point is crucial because it marks the transition point where the parabola shifts from increasing to decreasing or vice versa.

Based on the direction of the parabola determined by $a$ and the $x$-coordinate of the vertex, we can conclude:

• If $a > 0$ (parabola opens upwards), the function decreases on the interval $(-\infty, -\frac{b}{2a})$ and increases on the interval $(- \frac{b}{2a}, \infty)$.
• If $a < 0$ (parabola opens downwards), the function increases on the interval $(-\infty, -\frac{b}{2a})$ and decreases on the interval $(- \frac{b}{2a}, \infty)$.

Therefore, knowing both the coefficient of $x^2$ and the $x$-coordinate of the vertex allows us to determine the intervals of increase and decrease of the parabola.

Correct

To solve this problem, we'll analyze the behavior of a parabola based on its vertex:

• Step 1: Identify the vertex. We are given that the vertex is at $x = -6$.
• Step 2: Understand that a standard upward-opening parabola decreases to the left of its vertex. Thus, it decreases on the interval $(-\infty, -6)$.
• Step 3: Since we need the interval where it is decreasing: Given the parabola decreases on $(-\infty, -6)$, the function is increasing on $x > -6$.

Therefore, for a standard parabola, the function is decreasing on the interval $x < -6$ which means:

Therefore, the solution to the problem is $x > -6$.

To solve this problem, we'll consider the following:

• Understand that the vertex form of a parabola, located at point $x = 4$, gives critical information about its symmetry line and vertex location.
• The parabola can either open upwards or downwards, determined by the sign of $a$ in the quadratic equation $y = ax^2 + bx + c$.
• If $a > 0$, the parabola opens upwards, meaning it decreases for $x < 4$ and increases for $x > 4$.
• If $a < 0$, the parabola opens downwards, meaning it increases for $x < 4$ and decreases for $x > 4$.
• Given only the vertex $x = 4$, we lack the information about the parabola's opening without the value of $a$.

Therefore, it cannot be determined whether the function is decreasing on any specific interval without knowing the sign of $a$.

The correct choice is Cannot be determined.

We know that the equation for the parabola is given in vertex form $f(x) = -2(x-4)^2 + 5$. This form reveals the position of the vertex $(4, 5)$ and the opening direction of the parabola, which opens downward due to the negative coefficient of $-2$.

For a parabola, the intervals of increase and decrease are determined by its symmetry around the vertex. In this case, because the parabola opens downward, it will be increasing on the interval $(-\infty, 4)$ and decreasing on the interval $(4, \infty)$.

The vertex at $x = 4$ is the point where the rate of change (slope) shifts. Thus, to find where the function is decreasing, we look to the right side of the vertex.

Therefore, the function is decreasing for $x > 4$.

In conclusion, the interval where the function is decreasing is $x > 4$.