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Let $f(x) = x^3 + 2x^2 + x + 1$. Find the polynomial of degree $2$ that interpolates the values of $f$ at $x = -1,0,1$.

I was able to do the an initial part of this problem (not written), but I don't really know how to meet the restriction that the polynomial has to be degree $2$. I've found the values of $f$ at the indicated points, but from there i'm not sure what to do.

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2 Answers 2

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You need to find $p(x)=ax^2+bx+c$ such that $p(-1,0,1)=f(-1,0,1)=(1,1,5)$. The system is therefore $$a-b+c=1$$ $$c=1$$ $$a+b+c=5$$ So $a=2,b=2,c=1$ and therefore $$p(x)=2x^2+2x+1$$

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  • $\begingroup$ Ah, that makes a lot of sense. Thanks! $\endgroup$
    – Ozera
    Commented Apr 14, 2014 at 18:09
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A polynomial of degree 2 is a function of the following form:

$P_2(x) = ax^2 + bx + c$

(degree 2 corresponds to the exponent of the highest monomial)

Wanting this polynomial to interpolate your function at $x_0$, means that $P_2(x_0) = f(x_0)$. This means you need to solve the following three equations for a,b and c:

$P_2(-1,0,1) = f(-1,0,1)$ which gives you the coefficients of the polynomial $P_2$ which interpoaltes the function $f(x)$ at the points $ x = -1,0,1$.

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