Problem 79 - Algebra 2

Question

Each point lies on a parabola with vertex \((0,2) .\) Write the equation of the parabola. $$ (-1,5) $$

Step-by-Step Solution

Answer

The equation of the parabola is \(y=3x^2 +2\).

Step 1: Insert The Vertex Into The Equation

Replace (h,k) in the equation \(y=a(x-h)^2 +k\) with the vertex coordinates \((0,2)\). So, the equation becomes \(y=a*x^2 +2\).

Step 2: Find The Value Of 'a'

Substitute the values from the provided point (-1,5) into the equation from step 1. That means replace x with -1 and y with 5. So, we have \(5=a*(-1)^2 +2 \). Solving for a, we get \(a=5-2=3\).

Step 3: Write The Equation Of the Parabola

Replace 'a' in the equation from Step 1 with 3, to get the equation of the parabola as \(y = 3x^2 + 2\).

Key Concepts

Vertex Form

The vertex form of a parabola is a special way of writing its equation that easily shows you the location of the vertex. The basic structure of the vertex form is:\[ y = a(x-h)^2 + k \]In this formula:

• \(a\) determines the width and direction of the parabola.
• \((h, k)\) is the vertex of the parabola.

Understanding vertex form is crucial because it simplifies finding the vertex directly from the equation. You can easily see both the horizontal and vertical shifts from the standard position where the vertex would be at the origin (0,0).
For our exercise with the vertex at (0, 2), substituting this into the equation, we get:\[ y = a(x - 0)^2 + 2 = ax^2 + 2 \]This new form directly tells us the position of the vertex, enhancing graphing and interpretation of the parabola.

Quadratic Function

A quadratic function represents a second-degree polynomial. Its general equation is:\[ y = ax^2 + bx + c \]Quadratic functions are represented graphically as parabolas. Depending on the value of \(a\):

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

The quadratic function can also be expressed in vertex form, as discussed earlier, which is particularly useful in finding the axis of symmetry and vertex.
These functions are essential in various fields like physics, engineering, and economics because their graphs show maximum or minimum points, which can be crucial in optimization problems.

Finding Coefficients

To find the coefficients, especially \(a\) in the vertex form, you subsitute known points on the parabola into the equation. Given our problem, we already have the equation in vertex form:\[ y = ax^2 + 2 \]To find \(a\), use the point (-1, 5) on the parabola. Plugging in these values:
\[ 5 = a(-1)^2 + 2 \]Solving for \(a\):

• Calculate \((-1)^2 = 1\)
• Set the equation as \(5 = a \times 1 + 2\)
• Subtract 2 from both sides: \(3 = a\)

Finally, substitute this value into the equation to get the complete quadratic equation:\[ y = 3x^2 + 2 \]This process efficiently allows us to determine the stretch or compression and orientation of the parabola based on its points on the graph.