Answer to Question #142804 in Differential Equations for liam donohue

Question

Determine coefficients a and b such that p(x)=x2+ax+b satisfies p(1)=−6 and p′(1)=3.

a=


b=

Answer 1

First of all, let us start with the condition $p'(1)=3$ and we will see why in a second :

Let's calculate $p'(x) = (x^2+ax+b)' = 2x+a$ (because $(x^2)'=2x, (ax)'=a, b'=0$ ). The condition $p'(1)=3$ gives us $2\times 1 + a = 3$ , equation where b vanishes and so we have a simple solution $a=1$ .

Now we can proceed to the condition $p(1)=-6$ , it gives us $1^2+a\times1+b=-6$ , knowing that a=1, so we have $b= -8$ .

So our final answer is: $p(x) = x^2+x-8$ .