Problem 46 - Algebra 2

Question

Write an equation for a parabola with vertex at the origin and that passes through \((2,-8) .\)

Step-by-Step Solution

Answer

The equation is \( y = -2x^2 \).

Step 1: Understand the parabola's equation format

Parabolas can be expressed in the vertex form equation, which is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Since the vertex is at the origin \((0,0)\), the equation simplifies to \( y = ax^2 \).

Step 2: Substitute the point into the equation

We are given the point \((2, -8)\) that the parabola passes through. Substitute \(x = 2\) and \(y = -8\) into the equation \(y = ax^2\). This gives us \(-8 = a(2)^2\).

Step 3: Solve for 'a'

Simplify the equation \(-8 = 4a\) by dividing both sides by 4. This results in \(a = -2\).

Step 4: Write the final equation

Now that we've found \(a = -2\), substitute it back into the equation \(y = ax^2\) to get \(y = -2x^2\). This is the equation of the parabola.

Key Concepts

Vertex Form

The vertex form of a parabola is a handy way to express the equation of a parabola, especially when you know the vertex's coordinates. The vertex form is given by the equation \( y = a(x - h)^2 + k \). In this equation:

• \(a\) determines the "stretch" or "compression" of the parabola.
• \((h, k)\) represents the vertex, the highest or lowest point of the parabola, depending on its orientation.

This form is particularly useful because it easily tells you where the parabola turns and can help quickly identify its direction and shape. When the vertex is at the origin \((0,0)\), the equation simplifies to \( y = ax^2 \), making calculations simpler.

Parabola Passing Through a Point

A parabola can be identified not only by its vertex but also by knowing another point through which it passes. When given a point like \((2, -8)\), inserting it into the parabola’s equation helps determine specific parameters. For instance, if you plug \(x = 2\) and \(y = -8\) into the simplified vertex form \(y = ax^2\), you assist in solving for unknowns, such as \(a\). This point substitution is crucial in defining the uniqueness of the parabola's equation that fits additional given conditions effectively.

Solving for 'a' in a Parabola Equation

In a parabola's vertex form equation \(y = ax^2\), solving for \(a\) is essential to determine the equation's precise configuration. Here's how:

• Substitute the known values of \(x\) and \(y\) from the point that the parabola passes through.
• Once substituted, simplify the equation to solve for \(a\).

For example, if you have the point \((2, -8)\), substituting into \(y = ax^2\) gives you \(-8 = 4a\). By dividing both sides by 4, you find \(a = -2\). Solving for \(a\) allows us to understand how "open" or "close" the parabola is, depending on whether \(a\) is less than, greater than, or equal to 1.

Parabola with Vertex at the Origin

When a parabola has its vertex at the origin, the calculation of its equation undergoes simplification. A vertex at the origin means the vertex form \(y = a(x - h)^2 + k\) reduces to \(y = ax^2\) because both \(h\) and \(k\) become zero.This makes working with such parabolas straightforward since:

• The vertex or turning point is exactly at \((0,0)\).
• It’s simpler to solve for other parameters, like \(a\), using additional points like \((x, y)\).

For example, using a point like \((2, -8)\), it becomes very direct to calculate \(a\), knowing already that \(h = 0\) and \(k = 0\). This essential feature helps in quickly defining the parabola's orientation and opening direction.