Problem 41 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 41-42: Find an equation for the parabola that has a horizontal axis and passes through the given points. $$ P(-1,1), \quad Q(11,-2), \quad R(5,-1) $$

Step-by-Step Solution

Answer

The equation of the parabola is \( y = \frac{1}{72}x^2 - \frac{11}{12}x + \frac{1031}{864} \).

Step 1: Understand the Parabola's Orientation

Since the parabola has a horizontal axis, its equation is of the form \( y = ax^2 + bx + c \). The vertex will not be a solution since we are provided points and no vertex information.

Step 2: Set Up the System of Equations

Use each point \( P(-1, 1) \), \( Q(11, -2) \), and \( R(5, -1) \) to form equations: \((1)\, 1 = a(-1)^2 + b(-1) + c\), \((2)\, -2 = a(11)^2 + b(11) + c\), \((3)\, -1 = a(5)^2 + b(5) + c\). This translates to:1. \( a - b + c = 1 \)2. \( 121a + 11b + c = -2 \)3. \( 25a + 5b + c = -1 \)

Step 3: Solve the System of Equations - Eliminate c

Subtract equation (1) from equation (2) to eliminate \(c\):\(120a + 12b = -3 \) (Equation 4)Subtract equation (3) from equation (2) to eliminate \(c\):\(96a + 6b = -1 \) (Equation 5)

Step 4: Solve for a and b

Multiply Equation 5 by 2:\(192a + 12b = -2 \). Subtract Equation 4 from this result:\(72a = 1 \), thus \(a = \frac{1}{72}\). Substitute \(a\) back into Equation 4:\(120(\frac{1}{72}) + 12b = -3 \).Simplify to find \(b\):\(\frac{10}{6} + 12b = -3 \)\(12b = -3 - \frac{10}{6} \)\(b = -\frac{11}{12} \).

Step 5: Solve for c

Substitute \(a\) and \(b\) back into equation (1) to solve for \(c\):\(\frac{1}{72} + \frac{11}{12} + c = 1 \).Calculate:\(c = 1 - \frac{1}{72} - \frac{11}{12} \)\(c = \frac{1031}{864} \) after simplification.

Step 6: Write the Equation of the Parabola

Now that we have \(a = \frac{1}{72}\), \(b = -\frac{11}{12}\), and \(c = \frac{1031}{864}\), the equation of the parabola is:\[ y = \frac{1}{72}x^2 - \frac{11}{12}x + \frac{1031}{864} \]

Key Concepts

Horizontal Axis

A parabola can open in two different orientations: vertically (up or down) or horizontally (left or right). When a parabola has a horizontal axis of symmetry, it means it opens to the left or right. The general form of a parabola with a horizontal axis is expressed as:
\( y = ax^2 + bx + c \).
This equation differs from the vertical orientation, \( x = ay^2 + by + c \), which opens up or down. Since our task involves a horizontal parabola, we'll use the first form.

In the equation \( y = ax^2 + bx + c \), the coefficients \( a \), \( b \), and \( c \) determine the parabola's shape and position:

• The coefficient \( a \) determines the direction and the width of the parabola's opening.
• The coefficient \( b \) affects the tilt of the parabola.
• The coefficient \( c \) represents the y-intercept.

Understanding the role of each coefficient helps us correctly graph the parabola based on its equation.

System of Equations

To find the specific equation of a parabola that passes through given points, we set up a system of equations. Each point on a parabola satisfies its equation \( y = ax^2 + bx + c \). Given points \( P(-1, 1) \), \( Q(11, -2) \), and \( R(5, -1) \), we substitute these coordinates into the equation to form three different equations:

• For \( P(-1, 1) \): \( a(-1)^2 + b(-1) + c = 1 \)
• For \( Q(11, -2) \): \( a(11)^2 + b(11) + c = -2 \)
• For \( R(5, -1) \): \( a(5)^2 + b(5) + c = -1 \)

This results in a system of linear equations with three unknowns: \( a \), \( b \), and \( c \). Solving this system allows us to determine the specific values for these coefficients, which define the unique parabola passing through the given points.

Points

Points are specific locations on the plane represented by coordinates, like \( (x, y) \). In this exercise, the points \( P(-1, 1) \), \( Q(11, -2) \), and \( R(5, -1) \) lie on the parabola. This means each point satisfies the parabola's equation.

Each set of coordinates tells us an exact spot on the graph where the parabola intersects. By substituting these coordinates into the parabola's equation, we form a system of equations that expresses the specific relationship of the parabola with its path through these points.

Using points as a way to solve for the parabola's equation highlights the utility of coordinates in algebra. It connects abstract mathematical equations to concrete geometric representations in graphing.

Solving Equations

Solving equations is the process of finding the unknown variables in our system. For a parabola, solving involves determining the coefficients \( a \), \( b \), and \( c \).

From our system of equations previously derived:

• Equation 1: \( a - b + c = 1 \)
• Equation 2: \( 121a + 11b + c = -2 \)
• Equation 3: \( 25a + 5b + c = -1 \)

The goal is to simplify and reduce these equations to solve for \( a \), \( b \), and \( c \). Begin by using elimination or substitution processes to reduce the number of variables.

In this case, eliminate \( c \) by subtracting pairs of equations, leaving us with a simpler two-variable problem:

• Subtraction of Equation 1 from Equation 2 and Equation 3 helps simplify calculations.
• Further simplification using elimination reveals \( a \).
• Once \( a \) is known, substitute back to find \( b \) and then \( c \).

This method systematically breaks down the problem, providing a solution for each coefficient, and finally leading to the parabolic equation.