Problem 81 - Algebra 2

Question

Each point lies on a parabola with vertex \((0,2) .\) Write the equation of the parabola. $$ (2,-2) $$

Step-by-Step Solution

Answer

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

Step 1: Identify and substitute vertex into the parabola equation

Observe that we have a vertex at \((h,k)=(0,2)\). So, we substitute these into the equation of a parabola, and it simplifies to \(y=a(x-0)^2 +2\), or, more simply, \(y=ax^2 +2\).

Step 2: Substitute the given point and solve for a

We know that the given point \((2,-2)\) lies on the parabola, so we substitute this point into our simplified equation to solve for \(a\). We get: \(-2=a(2^2) +2\). Simplify this to \(-2=4a +2\). After that, solve for \(a\): \(-2-2=4a\), or \(-4=4a\). Thus, \(a=-1\).

Step 3: Substitute a back into the equation

Now that we know \(a=-1\), we plug it back into our simplified parabolic equation to get the final result which is \(y=-x^2 +2\).

Key Concepts

Vertex Form

The vertex form of a parabola is a special expression that helps us understand and graph the parabola more easily. It is written as \( y = a(x-h)^2 + k \). Here, \((h, k)\) represents the vertex of the parabola, which is its highest or lowest point, depending on the orientation.

This form is very convenient because it directly tells us where the vertex is located. In our exercise, the vertex is \((0, 2)\). By substituting \(h = 0\) and \(k = 2\) into the formula, we simplify the equation to \( y = a(x-0)^2 + 2 \), or \( y = ax^2 + 2 \).

Understanding the vertex form is crucial because it offers a quick insight into graph characteristics, without needing additional calculation. It helps in easily determining the direction in which the parabola opens and its width based on the value of \(a\).

Substitution

Substitution is a powerful algebraic tool used to solve equations by replacing variables with given values. It simplifies solving processes by letting us input known information into an equation.

In the context of our parabola exercise, we have a simplified equation from the vertex form: \( y = ax^2 + 2 \). The given point \((2, -2)\) lies on the parabola. This means the coordinates satisfy the parabola equation.

We substitute \(x = 2\) and \(y = -2\) into the equation:

• \(-2 = a(2)^2 + 2 \)

This substitution allows us to focus on solving for the unknown \(a\), which defines the particular shape and direction of the parabola.

Solving for a

Once we have substituted the known values into our parabola equation, the next step is solving for the unknown parameter \(a\). This parameter determines how "wide" or "narrow" the parabola is, and its direction.
In our exercise, substituting \((2, -2)\) into \( y = ax^2 + 2 \) gives

• \(-2 = a(2)^2 + 2 \)
• Simplifying gives \(-2 = 4a + 2\)
• Rearrange to \(-2 - 2 = 4a\), which simplifies to \(-4 = 4a \)
• To find \(a\), divide each side by 4: \( a = -1 \).

Thus, finding \(a\) is critical as it directly impacts how the parabola is shaped. In this case, since \(a = -1\), the parabola opens downwards and is inverted.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves describing geometric figures using a coordinate system. This field of mathematics merges algebra and geometry, useful for solving spatial problems.

For a parabola, coordinate geometry helps in understanding the placement and features of the curve on the coordinate plane.
In our function \(y = -x^2 + 2\), this indicates:

• The vertex at \((0, 2)\), showing where the curve reaches its maximum point.
• The parabola opens downward, as \(a = -1\), indicating a flip from the standard upward parabola.
• The point \((2, -2)\) lies on the curve, representing a real solution in the coordinate plane.

By applying coordinate geometry, we can visualize how the parabola interacts with the grid of the plane, making it easier to understand and predict its behavior across varied intervals.