Increasing and Decreasing Domain of a Parabola: True / false

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.

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 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 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