Problem 43 - Precalculus with Calculus Previews

Question

Find an equation of parabola that satisfies the given conditions. Vertex \((0,0),\) through \((-2,8),\) axis along the \(y\) -axis

Step-by-Step Solution

Answer

The equation of the parabola is \(y = 2x^2\).

Step 1: Understand the Vertex Form of a Parabola

For a parabola with vertex at \((h, k)\), the general vertex form of the equation is \(y = a(x - h)^2 + k\). In this problem, the vertex is \((0,0)\), so the equation simplifies to \(y = ax^2\). We need to determine the value of \(a\).

Step 2: Substitute the Given Point into the Equation

Since the parabola passes through the point \((-2,8)\), we can substitute \(x = -2\) and \(y = 8\) into the equation \(y = ax^2\) to find the value of \(a\). This gives us: \(8 = a(-2)^2\).

Step 3: Solve for 'a'

In the equation \(8 = a(-2)^2\), simplify the expression. The equation becomes \(8 = a(4)\), which is \(8 = 4a\). Solve for \(a\) by dividing both sides by 4, yielding \(a = 2\).

Step 4: Write the Final Equation of the Parabola

Substitute \(a = 2\) back into the equation \(y = ax^2\). Therefore, the equation of the parabola is \(y = 2x^2\).

Key Concepts

Vertex Form

The vertex form is a special way of writing the equation of a parabola. It focuses on the vertex, which is a crucial part of the parabola, as it represents the highest or lowest point depending on the orientation of the parabola. The general equation is formulated as:

• \( y = a(x - h)^2 + k \)

Here:

• \( a \) adjusts the "stretch" or "compression" and the direction (upward or downward) of the parabola.
• \( (h, k) \) is the vertex of the parabola.

This form is especially useful when you know the vertex of the parabola and need to construct its equation. In the original problem, the vertex is at \( (0,0) \), making the vertex-oriented equation very simple. Our equation becomes \( y = ax^2 \), focusing our task on finding \( a \) using another point on the curve.

Axis of Symmetry

The axis of symmetry is an imaginary line that neatly divides the parabola into two identical halves. For any parabola, this line runs vertically through the vertex.
Given that the vertex is \( (0,0) \) in our problem and the parabola’s axis is along the \( y \)-axis, the axis of symmetry is simply the line \( x = 0 \). This means for any point \((x, y)\) on the parabola, \(( -x, y)\) will also be a point on the parabola, exploiting the symmetry effectively.
This axis of symmetry helps in plotting points around it and examining how the parabola opens.
Understanding this concept improves our grasp of parabola behavior and facilitates work with different forms and equations.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves using algebraic equations to represent and study geometric figures. For parabolas, and indeed all conic sections, coordinate geometry serves as a powerful tool.
In this scenario, it allows us to take points and express the parabola's behavior using the vertex form equation. We saw how a point through which the parabola passes, like \((-2,8)\), helps find the important \( a \) value in our equation.

• By substituting the point \((-2,8)\) into the simplified vertex form \(y = ax^2\), we derived \(a = 2\).
• This points to the necessity of coordinate geometry for relating geometric locations directly to algebraic expressions.

The coordinate geometry framework provides an organized way to navigate this transition from geometric figures to algebraic representation, crucial for effectively solving many geometrical problems.