Parabola Intersection: Must-Be-True Statements

Let's dive into the fascinating world of parabolas and explore what happens when they intersect! We're given two parabolas, and understanding their properties can help us deduce some crucial information about their relationship. Our main focus will be on figuring out which statement must be true when these two parabolas intersect at exactly two points. This isn't just about solving for intersection points; it's about understanding the fundamental geometry and algebraic representation of parabolas. We'll be dissecting the implications of their vertex positions and opening directions. So, grab your thinking caps, and let's embark on this mathematical journey together. We'll start by analyzing the given information about the vertices and then move on to exploring the possible scenarios that lead to their intersection. Remember, when we talk about parabolas, we're dealing with curves that have a very specific shape, defined by a quadratic equation. The vertex is a key point, representing the maximum or minimum point of the parabola. The direction it opens – upward or downward – is determined by the sign of the coefficient of the squared term in its equation. Understanding these basics is fundamental to solving this problem. We'll also consider how the distance between the vertices and their relative positions can influence whether they intersect and how many intersection points they have. This problem challenges us to think beyond just calculating values and to grasp the conceptual underpinnings of quadratic functions.

Understanding Parabola Basics: Vertex and Opening Direction

The vertex of a parabola is a pivotal point. It's the point where the parabola changes direction. For a parabola that opens upward, the vertex is the minimum point, while for a parabola opening downward, it's the maximum point. The coordinates of the vertex are often denoted as $(h, k)$. In our problem, the first parabola has its vertex at $(0, 4)$, and we are told it opens downward. This means its highest point is at $(0, 4)$, and it extends infinitely downwards from there. The standard form of a parabola opening downward with vertex $(h, k)$ is $y = a(x-h)^2 + k$, where $a$ is a negative number. For our first parabola, with vertex $(0, 4)$ and opening downward, the equation would be of the form $y = ax^2 + 4$, where $a < 0$. The second parabola has its vertex at $(0, -4)$. We are not initially told which way it opens. A parabola opening upward has a vertex that is its minimum point, and its equation is of the form $y = a(x-h)^2 + k$, where $a$ is a positive number. If the second parabola opens downward, its vertex $(0, -4)$ would be its maximum point, and its equation would be $y = a(x-0)^2 - 4$, with $a < 0$. If it opens upward, its vertex $(0, -4)$ would be its minimum point, and its equation would be $y = a(x-0)^2 - 4$, with $a > 0$. The fact that the two parabolas intersect at two points is a critical piece of information. This tells us there's a specific relationship between their shapes and positions that allows for these intersections. Let's consider the implications of the vertex positions. The first parabola is centered vertically around $y=4$ and opens down. The second parabola is centered vertically around $y=-4$. Since the first parabola opens down from $y=4$, its y-values will be $4$ or less. Since the second parabola's vertex is at $y=-4$, if it opens upward, its y-values will be $-4$ or greater. If it opens downward, its y-values will be $-4$ or less. The relative vertical positioning of the vertices and the direction of opening are key to determining if and how they intersect.

Analyzing Intersection Scenarios

Now, let's analyze the condition that the two parabolas intersect at exactly two points. We know the first parabola has its vertex at $(0, 4)$ and opens downward. This means its equation is of the form $y = ax^2 + 4$, where $a < 0$. The second parabola has its vertex at $(0, -4)$. Let's consider the two possibilities for the second parabola's opening direction. Scenario 1: The second parabola opens downward. If the second parabola also opens downward, its equation is of the form $y = cx^2 - 4$, where $c < 0$. For the parabolas to intersect at two points, the equations must have two common solutions for $x$ when we set them equal to each other: $ax^2 + 4 = cx^2 - 4$. Rearranging this, we get $(a-c)x^2 = -8$. For there to be two distinct real solutions for $x$ (which means two intersection points), $(a-c)$ must be negative, and $-8/(a-c)$ must be positive. Since $a$ and $c$ are both negative, $a-c$ could be positive or negative. For example, if $a = -1$ and $c = -2$, then $a-c = -1 - (-2) = 1 > 0$. In this case, $x^2 = -8/1 = -8$, which has no real solutions. If $a = -2$ and $c = -1$, then $a-c = -2 - (-1) = -1 < 0$. In this case, $x^2 = -8/(-1) = 8$, which gives two real solutions for $x$: $x = \pm\sqrt{8}$. So, it is possible for two downward-opening parabolas with these vertices to intersect at two points, but it depends on the specific values of $a$ and $c$. Scenario 2: The second parabola opens upward. If the second parabola opens upward, its equation is of the form $y = cx^2 - 4$, where $c > 0$. Setting the equations equal to find intersections: $ax^2 + 4 = cx^2 - 4$. Rearranging gives $(a-c)x^2 = -8$. Since $a < 0$ (first parabola opens down) and $c > 0$ (second parabola opens up), the term $(a-c)$ will always be negative. Specifically, if $a = -1$ and $c = 1$, then $a-c = -1 - 1 = -2$. Then, $(-2)x^2 = -8$, which means $x^2 = 4$. This gives two real solutions: $x = \pm 2$. This scenario guarantees two intersection points because one parabola is