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How are you able to find a second degree polynomial based on only knowing three values of the polynomial like in the title? P(-1) = 132, P(0) = 98, P(1) = 88

Lukas
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  • Write down a general polynomial of second degree, insert the values, and solve the resulting system of equations. – celtschk Nov 12 '17 at 16:38
  • Consider the general quadratic: $ax^2 + bx + c$. You get three equations in three unknowns. – jonsno Nov 12 '17 at 16:39

2 Answers2

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using that $P(x)$ has the form $$P(x)=ax^2+bx+c$$ then we get $$P(-1)=a-b+c=132$$ $$P(0)=c=98$$ $$P(1)=a+b+c=88$$ for your Control: $$a=12,b=-22,c=98$$

Dr. Sonnhard Graubner
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All you have to do is to solve the system:

$$\begin{cases} a-b+c=132 \\ c=98 \\ a+b+c=88 \end{cases} \Rightarrow \begin{cases} a=12\\ b=-22 \\ c=98 \end{cases}$$

So the polynomial is $12x^2-22x+98$

Karagum
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