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202011091423 Exercise 1 (Q2)

If f(x)=(x-1)(x+5), find the values of f(2), f(1), f(0), f(a+1), f\bigg(\displaystyle{\frac{1}{a}}\bigg), f(-5).

Solution.

\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}

On reflection.

Suppose you are given an unknown quadratic function f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2 such that

\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}

and you are asked to make polynomial interpolation.

\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}
\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}
\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}
\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}

And then is the interpolated polynomial f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1).

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