Question

find a second - degree polynomial $p$ such that $p(2)=19, p(2)=17$, and $p(2)=10$.
$p(x)=$

Explanation:

Step1: Define the polynomial

Let $P(x)=ax^{2}+bx + c$. Then $P'(x)=2ax + b$ and $P''(x)=2a$.

Step2: Find the value of $a$

Since $P''(2) = 10$ and $P''(x)=2a$, substituting $x = 2$ into $P''(x)$ gives $2a=10$, so $a = 5$.

Step3: Find the value of $b$

Since $a = 5$, then $P'(x)=10x + b$. Substituting $x = 2$ and $P'(2)=17$ into $P'(x)$, we have $10\times2 + b=17$, which simplifies to $20 + b=17$, so $b=- 3$.

Step4: Find the value of $c$

Since $a = 5$ and $b=-3$, then $P(x)=5x^{2}-3x + c$. Substituting $x = 2$ and $P(2)=19$ into $P(x)$, we get $5\times2^{2}-3\times2 + c=19$, which is $20 - 6 + c=19$, so $c = 5$.

Answer:

$P(x)=5x^{2}-3x + 5$