Question

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

Answer 1

dude let the polynomial be ax^2+bx+c
find P(2) P'(2) and P"(2)

Answer 2

I am not sure what that means.. I sort of need it explained...

Answer 3

if P(x)=ax^2+bx+c
then whats P(2)

Answer 4

a2x^2+b2+c

Answer 5

nope
it should be a(2)^2+b(2)+c=4a+2b+c=5(given)

Answer 6

2 things.. how does that = 5, and where did you get " P(x)=ax^2+bx+c"

Answer 7

we assumed the equation to be that
and its given that P(2)=5

Answer 8

ok, but where did you get " P(x)=ax^2+bx+c" Is that some standard equation?

Answer 9

yes that is standard quadratic (2nd degree) equation

Answer 10

its a general form of the quadratic equation

Answer 11

and their answer to P(2)=5 was x^2-x+3

Answer 12

did you find \(P'\) and \(P''\)?

Answer 13

P(x) = ax^2 +bx +c

P'(x) = 2ax +b

P''(x) = 2a

Answer 14

then work backwards to solve for a,b,c

Answer 15

what one would i use again?

Answer 16

and as per dumbcow's answer since \(P''(x)=2a\) and \(P''(x)=2\) you have
\[2a=2\implies a=1\]

Answer 17

wait, how does "P′′(x)=2a and P′′(x)=2 " happen?

Answer 18

wait is that P''(x) = 2 or P''(2) = ?

Answer 19

what?

Answer 20

in your post what does P''(2) equal ?

Answer 21

no, how is "P′′(x)=2a" and "P′′(x)=2 " the same..

Answer 22

if its given that P''(x) = 2, then you can set 2a = 2, --> a=1

P''(x) = 2a because it is 2nd derivative of P(x)
do you know how to take derivative of polynomial?

Answer 23

I am pretty sure i know, let me try to make sure

Answer 24

anyway there is a typo in your question....
you have P(2) =5 , P'(2) = 3, P''(2) = what??

Answer 25

2

Answer 26

then satellite was correct,
set 2a = 2
--> a=1

Answer 27

ok

Answer 28

now set
2a(2) +b = 3

plug in a value, solve for b

Answer 29

a(2)^2 +b(2) +c = 5

plug in "a" and "b", solve for c

Answer 30

ahhh, ok, now i get it!

Answer 31

Thanks a million!

Answer 32

no problem