Problem 92 - A Graphical Approach to College Algebra

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

Equation: \( y = \frac{1}{9}(x + 1)^2 + 2 \)

Step 1: Understand the Conditions

The parabola has a horizontal axis, which means it can be expressed in the form \( y = a(x-h)^2 + k \). The vertex is \((-1, 2)\), so \(h = -1\) and \(k = 2\). The parabola passes through the point \((2, 3)\).

Step 2: Set up the Parabola Equation

Substitute the vertex into the equation of a parabola to get: \( y = a(x + 1)^2 + 2 \).

Step 3: Use the Point the Parabola Passes Through

Substitute the point \((2, 3)\) into the equation to find \(a\):\[ 3 = a(2 + 1)^2 + 2 \].

Step 4: Solve for \(a\)

Solve the equation from Step 3: \[ 3 = a(3)^2 + 2 \]Subtract 2 from both sides: \[ 1 = 9a \]Divide both sides by 9: \[ a = \frac{1}{9} \].

Step 5: Write the Final Equation

Having found \(a\), substitute back into the parabola equation:\[ y = \frac{1}{9}(x + 1)^2 + 2 \].

Key Concepts

Understanding the Horizontal Axis

The horizontal axis in a parabola refers to the direction in which the parabola opens. While many parabolas open vertically (upward or downward), some open horizontally (to the left or to the right). In this problem, the parabola has a horizontal axis, implying it opens either to the left or the right.

• This requires the use of a horizontal parabola equation, which is typically written as \( y = a(x-h)^2 + k \).
• The condition of the horizontal axis affects the equation form, reflecting that the graph is not centered on the vertical, but rather shifted sideways.

This nuance is crucial when decoding the behavior of quadratic functions and determining their graphical representation.

Role of the Vertex

The vertex of a parabola is its peak point, which is either the maximum or minimum point depending on the opening direction. For a parabola with a horizontal axis, the vertex serves as the leftmost or rightmost point.

• In this exercise, the vertex is given by the coordinates \((-1, 2)\).
• This vertex \((h, k)\) is a direct input into the standard form of a horizontal parabola, influencing its placement on the coordinate plane.

The location of the vertex provides a starting point for sketching or analyzing the parabola's shape, helping to ensure it meets given conditions in graphing or geometric problem solving.

Coordinate Geometry in Parabolas

Coordinate geometry provides a system for representing geometric figures like parabolas using algebraic equations. It enables us to use points and lines to describe curves and shapes.

• For parabolas, coordinate geometry helps define the precise location of key features such as vertices and axes.
• The exercise exemplifies using points such as the vertex \((-1,2)\) and a passing-through point \((2,3)\) to find the correct parabola equation.

This method bridges algebra and geometry, allowing for a clear method of plotting or manipulating parabola graphs based on given criteria.

Exploring the Quadratic Function

Quadratic functions form the backbone of parabolas. They typically appear in the standard form of \( y = ax^2 + bx + c \) for vertical parabolas, or in this exercise, \( y = a(x-h)^2 + k \) for horizontal ones.

• These functions are characterized by the squared expression, resulting in the characteristic 'U' or 'n' shape of parabolas.
• Identifying \( a \), \( h \), and \( k \) from vertex coordinates and an additional point allows us to concretely form the parabola's equation.

Understanding how to adjust these parameters makes it possible to manipulate and predict the parabola's behavior under various conditions.