Problem 69 - A Graphical Approach to Precalculus with Limits

Question

Find the equation of the parabola (with vertical axis that passes through the data points shown or specified. Check your answer. $$(-1,4),(1,2),(3,8)$$

Step-by-Step Solution

Answer

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

Step 1: General Equation of a Parabola

The general form of a parabola with a vertical axis is given by the equation \(y = ax^2 + bx + c\). Our goal is to find the coefficients \(a\), \(b\), and \(c\) such that the parabola passes through the points \((-1,4)\), \((1,2)\), and \((3,8)\).

Step 2: Substitute Point (-1,4)

Substitute \(x = -1\) and \(y = 4\) into the general equation: \(4 = a(-1)^2 + b(-1) + c\). Simplifying gives the equation \(4 = a - b + c\).

Step 3: Substitute Point (1,2)

Substitute \(x = 1\) and \(y = 2\) into the general equation: \(2 = a(1)^2 + b(1) + c\). This simplifies to \(2 = a + b + c\).

Step 4: Substitute Point (3,8)

Substitute \(x = 3\) and \(y = 8\) into the general equation: \(8 = a(3)^2 + b(3) + c\). This simplifies to \(8 = 9a + 3b + c\).

Step 5: Solve the System of Equations

We now have a system of equations:1. \(4 = a - b + c\)2. \(2 = a + b + c\)3. \(8 = 9a + 3b + c\)Subtract equation (1) from equation (2), we get: \(2 - 4 = (a + b + c) - (a - b + c) \Rightarrow -2 = 2b \Rightarrow b = -1\).Subtract equation (2) from equation (3), we get:\(8 - 2 = (9a + 3b + c) - (a + b + c) \Rightarrow 6 = 8a + 2b \Rightarrow 6 = 8a - 2\), solve for \(a\):\(8a = 8 \Rightarrow a = 1\).Using \(a = 1\) and \(b = -1\) in equation (2), \(2 = 1 - 1 + c \Rightarrow 2 = c\).

Step 6: Equation of the Parabola

Substitute the values of \(a = 1\), \(b = -1\), and \(c = 2\) into the general form of the parabolic equation: \(y = ax^2 + bx + c\). Thus, the parabola's equation is \(y = x^2 - x + 2\).

Step 7: Verify the Solution

Substitute each given point into the parabolic equation \(y = x^2 - x + 2\) to verify:1. For \((-1, 4)\): \(y = (-1)^2 - (-1) + 2 = 1 + 1 + 2 = 4\).2. For \((1, 2)\): \(y = (1)^2 - 1 + 2 = 1 - 1 + 2 = 2\).3. For \((3, 8)\): \(y = (3)^2 - 3 + 2 = 9 - 3 + 2 = 8\).All points satisfy the equation, confirming the solution is correct.

Key Concepts

Solving Systems of Equations

To find the equation of a parabola that goes through specific points, solving a system of equations is crucial. This involves deriving equations from each point and solving them simultaneously. The goal is to find the values of unknown variables, which in the context of a parabola, are the coefficients. Each chosen point on the parabola, such as

• (-1, 4)
• (1, 2)
• (3, 8),

provides an equation when substituted into the general form of a parabola, \(y = ax^2 + bx + c\). Solving these equations together, often by subtracting them from each other, allows us to isolate each variable to find its value. Remember, solving systems of equations can sometimes feel like solving a puzzle. Every step taken brings you closer to the complete picture of what the parabola's equation is.

Substitution Method

The substitution method is an important technique in solving systems of equations. It's straightforward and useful, especially when you can quickly isolate and substitute variables. Here's how it works in essence:- Start with a set equation derived from a point.- Solve for one of the variables, usually the simplest to isolate.- Substitute this expression in place of the variable in another equation.- Solve the new equation, which will now have fewer variables.In our exercise, applying the substitution method made solving for \(b\) easier. Once \(b\) was determined, it was substituted back to find \(a\), and so on. This step-by-step substitution ensures each variable is neatly solved, allowing us to gradually unveil the full equation.

Quadratic Functions

Quadratic functions describe parabolas and take the form \(y = ax^2 + bx + c\). These curves open upwards or downwards, depending on the value of \(a\). Understanding the different roles of the coefficients:- \(a\) decides the direction and width of the parabola. A positive \(a\) means it opens upwards.- \(b\) influences the axis of symmetry and the vertex's horizontal position.- \(c\) sets the parabola's position along the y-axis (it's the y-intercept).These equations are not only about the examinations. They appear in real life: physics, calculating profits, or even plotting trajectories. Mastering quadratic functions enlightens the path to understanding the dynamic world of algebra. The fascinating part of quadratics is how they visually represent the interplay of constants and variables in a curve, which is often mesmerizing to students of mathematics.