Question

Describe the equations of the quadratic functions that go through the points (-1,3) and \((2,6) .\) Give 2 examples.

Short answer

Quadratic functions: \( f(x) = x^2 - x + 3 \) and \( f(x) = 2x^2 + 1 \).

Step-by-step solution

Understand the General Form of a Quadratic Function

A quadratic function can generally be expressed in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The goal is to find these constants such that the quadratic function passes through the given points \((-1, 3)\) and \((2, 6)\).

Set Up System of Equations

Using the general quadratic formula, substitute each point into the equation. For point (-1, 3):\[ a(-1)^2 + b(-1) + c = 3 \] simplifies to \[ a - b + c = 3 \].For point (2, 6):\[ a(2)^2 + b(2) + c = 6 \] simplifies to \[ 4a + 2b + c = 6 \].

Use Different Quadratic Functions

To find two different quadratic equations, let's assume one common coefficient. Example 1: Assume \( a = 1 \). Solve the system:1. \( 1 - b + c = 3 \)2. \( 4 + 2b + c = 6 \)Find \( b = -1 \) and \( c = 3 \), giving equation \( f(x) = x^2 - x + 3 \).Example 2: Assume \( a = 2 \). Solve system:1. \( 2 - b + c = 3 \)2. \( 8 + 2b + c = 6 \)Find \( b = 0 \) and \( c = 1 \), giving equation \( f(x) = 2x^2 + 1 \).

Verification

Check that both quadratic functions pass through the points. For \( f(x) = x^2 - x + 3 \):\((x, y) = (-1, 3)\) and \((2, 6)\) check out because substituting gives the correct \( y \)-values.For \( f(x) = 2x^2 + 1 \), \((-1, 3)\) and \((2, 6)\) also give correct \( y \)-values.

Key concepts

System of Equations

A system of equations is a set of two or more equations that have common variables. When solving a system, you are looking for values of the variables that satisfy all equations simultaneously. In the case of quadratic functions, we often set up systems of equations to determine specific coefficients in the polynomial. For instance, if we have a quadratic function expressed in the general form \( f(x) = ax^2 + bx + c \), and we know it goes through certain points, we can substitute these points into the function to create a system of equations. For example, if a quadratic goes through points \((-1, 3)\) and \((2, 6)\), we substitute these into the equation to get two separate equations:

• \(a(-1)^2 + b(-1) + c = 3 \), which simplifies to \( a - b + c = 3 \).
• \(a(2)^2 + b(2) + c = 6 \), which simplifies to \( 4a + 2b + c = 6 \).

By solving this system, we find specific values for \( a \), \( b \), and \( c \) that allow the quadratic to pass through the given points.To find distinct quadratic functions, we can manipulate one of the coefficients to demonstrate how different forms of the same equation can satisfy the given conditions.

General Form of a Quadratic Function

The general form of a quadratic function is a standard way to express quadratic equations, given by \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). This general form shows that:

• The term \( ax^2 \) represents the quadratic term, which affects the curvature or the "openness" of the parabola.
• The term \( bx \) is the linear term, which affects the tilt of the parabola.
• The constant \( c \) shifts the parabola up or down along the \( y \)-axis.

Understanding the roles of these coefficients is essential when you need to derive the specific form of the quadratic function that fits certain criteria, such as passing through given points. It's the manipulation of these constants that helps create various quadratic functions, showing their flexibility and diversity. You can adjust these coefficients to sketch different shapes and locations of parabolas on a graph.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \( x \) that satisfy the equation, commonly \( ax^2 + bx + c = 0 \). Various methods can be used, such as factoring, using the quadratic formula, or completing the square.For specific problems, such as ensuring a quadratic function passes through two points, solving involves algebraically determining the coefficients \( a \), \( b \), and \( c \).One of the most flexible methods to solve a quadratic equation is the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]This formula provides the roots of any quadratic equation, demonstrating the connection between algebraic manipulation and geometric representation on the graph.When constructing a quadratic function that passes through specific points, solving also involves verifying that the values obtained satisfy the original conditions; that is, substituting back into the equation to ensure it matches the known \( y \)-values at those \( x \)-positions. This confirms that the derived equation is correct and reliable.