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(ProQuest: ... denotes formulae omitted.)
INTRODUCTION
Just as two points determine a line, three (non-collinear) points determine a quadratic function and, in general, n +1 points determine a polynomial of degree n, presuming that they do not fall onto a polynomial of lower degree. The process of finding such a polynomial can use one of the most important approaches in curve fitting, the Lagrange interpolating formula. In this article, we show how the Lagrange interpolating polynomial can be introduced and developed at the precalculus level. This brings one of the most powerful and useful tools of numerical analysis to the attention of lower division students while simultaneously building on and reinforcing some of the fundamental ideas in precalculus mathematics.
INTERPOLATION AND PRECALCULUS MATHEMATICS
In college algebra and precalculus, we emphasize the connection between the real zeros of a polynomial and its linear factors. For example, if/(x) = x2 - 5x + 6 = (x - 2)(x - 3), then /(x) has two real zeros, x = 2 and x = 3, each corresponding to the linear factor x - 2 and x - 3 , respectively. Also, we connect a real zero with an x-intercept. If a parabola is given with two known x-intercepts x, and x2, then its quadratic function must be of the form /(x) = k(x - x, )(x - x2 ), where k is a constant that can be determined from additional information about the parabola.
For example, suppose that a parabola has two x-intercepts, x = -2 and x = 6, and passes through the point (3,20), as shown in Figure 1. Then we know f(x)-k(x + 2)(x-6). To determine k, we use the given point (3,20) to get /(3) = k(3 + 2)(3 - 6) = 20, and find that
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To see how the information about the parabola is used in the formula of the quadratic function, we express the quadratic function as
... 20
However, what if both x-intercepts are not given or if the parabola has no x-intercepts, so that it sits entirely above (or below) the xaxis? How do we find the quadratic polynomial if any three points on the parabola are given? If we are fortunate enough to have two of the three points...





