Find the quadratic function y = ax2+bx+c whose graph passes through the given points. (−1, 6), (1, 4), (2, 9)

Ch. 5 - Systems of Equations and Inequalities

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Blitzer 8th Edition

Ch. 5 - Systems of Equations and Inequalities

문제 19

6장, 문제 19

Find the quadratic function y = ax²+bx+c whose graph passes through the given points. (−1, 6), (1, 4), (2, 9)

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Start by writing the general form of the quadratic function: \(y = ax^{2} + bx + c\).

Substitute each given point into the quadratic equation to create a system of equations. For the point \((-1, 6)\), substitute \(x = -1\) and \(y = 6\) to get: \(6 = a(-1)^{2} + b(-1) + c\).

Similarly, substitute the point \((1, 4)\) into the equation: \(4 = a(1)^{2} + b(1) + c\).

Substitute the point \((2, 9)\) into the equation: \(9 = a(2)^{2} + b(2) + c\).

Solve the resulting system of three equations with three unknowns (\(a\), \(b\), and \(c\)) using substitution or elimination methods to find the values of \(a\), \(b\), and \(c\).

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주요 개념

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Quadratic Functions

A quadratic function is a polynomial of degree two, generally written as y = ax² + bx + c, where a, b, and c are constants and a ≠ 0. Its graph is a parabola, which can open upwards or downwards depending on the sign of a.

System of Equations from Points

Given points on the graph of a quadratic function, substituting their coordinates into y = ax² + bx + c creates a system of equations. Solving this system allows us to find the values of a, b, and c that define the specific quadratic.

Solving Linear Systems

To find the coefficients a, b, and c, we solve the system of linear equations formed by the points. Methods include substitution, elimination, or matrix operations, which yield the unique solution for the quadratic function passing through all given points.