Problem 92 - A Graphical Approach to Precalculus with Limits

Question

Find an equation of a parabola that satisfies the given conditions. Horizontal axis; vertex \((-1,2) ;\) passing through \((2,3)\)

Step-by-Step Solution

Answer

The equation of the parabola is \((y-2)^2 = \frac{1}{3}(x+1)\)."

Step 1: Understand the Parabola's Properties

The problem states that the parabola has a horizontal axis. This means the parabola opens either to the left or right. The general form of such a parabola is \( (y-k)^2 = 4p(x-h) \), where \((h, k)\) is the vertex.

Step 2: Apply Known Vertex to the Equation

Given the vertex \((-1, 2)\), substitute \(h = -1\) and \(k = 2\) into the equation: \[ (y-2)^2 = 4p(x+1) \]

Step 3: Use the Point on the Parabola

Substitute the point \((2,3)\) into the equation to find \(p\). This gives:\[ (3-2)^2 = 4p(2+1) \]\[ 1 = 12p \]

Step 4: Solve for \(p\)

Solve \(1 = 12p\) to get \(p = \frac{1}{12}\).

Step 5: Write the Final Equation

Substitute \(p = \frac{1}{12}\) back into the equation \((y-2)^2 = 4p(x+1)\) to get:\[ (y-2)^2 = \frac{1}{3}(x+1) \]

Key Concepts

Vertex form

The vertex form of a parabola helps us understand its structure by focusing on its vertex, a crucial point that represents the peak or the lowest point on the graph. For parabolas with a horizontal axis, this form is given by \( (y-k)^2 = 4p(x-h) \), where

• \((h, k)\) represents the vertex of the parabola.
• \(p\) is a parameter that affects the width of the parabola, influencing how 'open' or 'closed' it appears.

In this exercise, the vertex is \((-1, 2)\), which means that it is the point from where the parabola turns or changes direction. This form allows us to easily identify the basic shape and orientation of the parabola by simply knowing its vertex.

Horizontal axis

A parabola with a horizontal axis opens either to the left or right rather than up or down, as seen with a vertical axis. In our given problem, we recognize a horizontal axis by the equation's structure:\( (y-k)^2 = 4p(x-h) \). When a parabola has a horizontal axis:

• Its equation involves a squared term on the \(y\) variable as opposed to the \(x\) variable.
• Depending on the sign of \(p\), the parabola will open to the right if \(p > 0\), or to the left if \(p < 0\).

This characteristic is crucial as it dictates the orientation and shape of the parabola. Here, since we find \(p = \frac{1}{12}\), this tells us the parabola opens to the right.

Equation of a parabola

The equation of a parabola in a specific form can tell us a lot about its characteristics. In this case, we start with the vertex form equation:\( (y-2)^2 = 4p(x+1) \).From the exercise, after identifying the vertex and substituting other parameters, we arrive at\[ (y-2)^2 = \frac{1}{3}(x+1) \].

• \((y-2)^2\) indicates that the vertex’s \(y\)-coordinate is 2.
• \(\frac{1}{3}(x+1)\) adjusts based on \(x\), with \(+1\) shifting the vertex to \(x = -1\).

This equation format reveals that for every increase or shift along the \(x\)-axis, changes on the \(y\)-axis are parabolically distributed, hence describing how the parabola bends and curves.

Graphing conic sections

Graphing conic sections, like parabolas, involves plotting their equation on a coordinate plane to visualize their shape. The conic sections are defined by quadratic equations which include parabolas, ellipses, circles, and hyperbolas. For parabolas, graphing becomes intuitive by using points like the vertex and another point on the curve:

• The vertex gives a starting point and indicates the direction the parabola will open.
• A known additional point, like the \((2, 3)\) given, helps plot the parabola accurately as it establishes the distance and relation between points.

This exercise shows a horizontal parabola that opens to the right as determined by our calculations. Understanding how these points and equations translate into a graph forms the basis for accurately drawing and interpreting the geometric representation of quadratic equations.