Problem 81 - A Graphical Approach to Precalculus with Limits

Question

Write an equation for each parabola with vertex at the origin. Through \((\sqrt{10},-5) ;\) opening downward

Step-by-Step Solution

Answer

The equation of the parabola is \( y = -\frac{1}{2}x^2 \).

Step 1: Understand the general form of the equation

Since the vertex of the parabola is at the origin (0, 0), we know the general equation for the parabola is in the form \[ y = ax^2 \] since it is a vertical parabola. When it opens downward, the coefficient \( a \) must be negative.

Step 2: Substitute the point into the equation

We know that the point \((\sqrt{10}, -5)\) lies on the parabola. Substitute \( x = \sqrt{10} \) and \( y = -5 \) into the equation \( y = ax^2 \): \[ -5 = a(\sqrt{10})^2 \] or \[ -5 = a \cdot 10 \].

Step 3: Solve for the coefficient \( a \)

Solve the equation \(-5 = 10a\) for \( a \): \[ a = \frac{-5}{10} = -\frac{1}{2} \].

Step 4: Write the equation of the parabola

Now that we have the value of \( a \), substitute it back into the general form of the equation: \[ y = -\frac{1}{2}x^2 \]. This is the equation of the parabola that passes through the given point and opens downward.

Key Concepts

Vertex Form of a Parabola

The vertex form of a parabola is an important concept as it provides a neat way to identify key features of a parabola quickly. It is expressed as: \[ y = a(x-h)^2 + k \] Here,

• \(a\) indicates the direction and "width" of the parabola
• \((h, k)\) are the coordinates of the vertex.

However, when the vertex is at the origin, this equation simplifies significantly because both \(h\) and \(k\) are zero. Thus, the equation becomes: \[ y = ax^2 \]This simplification makes calculations easier since we don't have additional variables to consider. Identifying the effect of \(a\) can further help determine if the parabola opens upwards or downwards and whether it is "stretched" or "compressed" compared to the standard parabola \(y = x^2\). The vertex form also helps when transitioning from one parabola form to another.

Vertex at Origin

A parabola with its vertex at the origin, point \((0, 0)\), has special properties that simplify its equation. In this case, the vertex form becomes: \[ y = ax^2 \]The origin is a significant point in mathematics because it serves as a natural reference point with coordinates

• \((h, k) = (0, 0)\)

This makes the equation of the parabola much tidier, allowing us to focus mainly on the value of \(a\) to determine the parabola's shape and direction. For practical applications, recognizing when a parabola has this form can aid in graphing it by connecting the vertex and symmetry to other crucial points, streamlining the process of visualization and analysis. The parabola's axis of symmetry will naturally align with the y-axis, leading to easier calculations and a clearer understanding of the graph's behavior, especially when dealing with other graph transformations.

Downward Opening Parabola

When a parabola opens downward, the term that describes its property in the equation is the coefficient \(a\). For it to be a downward opening parabola, \(a\) must be negative. The general equation in this case, when based at the origin, appears as: \[ y = ax^2 \]However, \(a < 0\). This negative value indicates that as \(x\) moves away from 0 on either side, \(y\) decreases. As a result, the parabola "opens" downward forming a "U" shape upside down. Understanding the effect of this negative \(a\) is crucial because it directly affects the vertex's nature as the maximum point of the parabola instead of a minimum, the latter of which would occur if \(a\) were positive. In solving our particular problem, understanding that the parabola opens downward helped us determine that \(a = -\frac{1}{2}\), and hence, the equation\( y = -\frac{1}{2}x^2 \) was confirmed. This property of downward opening also influences problems related to maximum heights, trajectories, and optimization, commonly examined in physics and calculus.