The video of lesson 3.5 reviews this material.

Sometimes students are asked to find a quadratic formula \(y=ax^2+bx+c\) to fit three data points. Lagrange’s Interpolation formula works very well for this. The task of simplifying the formula one gets (if one is required to simplify answers) is not too onerous.

Practice 1: a) Show that \(y=7\dfrac{\left(x-4\right)\left(x-5\right)}{\left(-5\right)\left(-6\right)} +  \dfrac{\left(x+1\right)\left(x-5\right)}{\left(5\right)\left(-1\right)} + 10\dfrac{\left(x+1\right)\left(x-4\right)}{\left(6\right)\left(1\right)}\) is a quadratic equation in disguise. Show that the graph of this equation passes through the data points \((-1,7)\), \((4,1)\), and \((5,10)\).   

b) Find a quadratic formula that fits the data \((2,5)\), \((-1,6)\), and\((5,46)\) and make your answer look as friendly as possible.

c) Find a quadratic formula that fits the data \((2,5)\), \((3,8)\), and \((5,14)\) and make your answer look as friendly as possible. Explain what happens!

d) There is no quadratic formula \(y=ax^2+bx+c\) that fits the data \((2,5)\), \(10,6)\), and\((10,100)\). (The same input value of \(x=10\) cannot give two different output values.) So then, how does Lagrange’s Interpolation method fail when you try to use it?

Of course, from the previous lesson we know how to find a polynomial that fits any number of data points, not just three.

Practice 2: Use Lagrange’s Interpolation Formula to find the equation of the line between two points \((p,m)\) and \((q,n)\) with \(p \neq q\). Is your equation equivalent to “\(y=mx+b\)” where \(m\) is the slope of the line segment between the two points?      

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