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Numerade Educator



Problem 67 Hard Difficulty

Find a second-degree polynomial & P & such that $ P(2) = 5, P'(2) = 3, $ and $ P"(2) = 2. $


Let $P(x)=a x^{2}+b x+c$. Then $P^{\prime}(x)=2 a x+b$ and $P^{\prime \prime}(x)=2 a . P^{\prime \prime}(2)=2 \Rightarrow 2 a=2 \Rightarrow a=1$
$P^{\prime}(2)=3 \Rightarrow 2(1)(2)+b=3 \Rightarrow 4+b=3 \Rightarrow b=-1$
$P(2)=5 \Rightarrow 1(2)^{2}+(-1)(2)+c=5 \Rightarrow 2+c=5 \Rightarrow c=3 .$ So $P(x)=x^{2}-x+3$

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Video Transcript

it's clear. So when you read here so we have a P of X is equal to a X square plus B x plus c. So our first derivative, it has to be to a X plus be in our second derivative. Must to be too, eh? We're gonna plug in to what's to pay Becomes a is equal to one. So our first derivative is equal to two x plus being the first orbit of to a sequel to two comes too close. Be just equal to three. He is equal to negative one. So P f X is equal to x square minus X plus c. You have PF two is equal to two square minus two plus C, which is equal to five. C is equal to three, so you get your backs is equal to X squared minus X plus three.