Help please?

Which polynomial has a graph that passes through the given points?

(–3, 10), (–1, 0), (0, –5), (3, 52)

Question

Help please?

Which polynomial has a graph that passes through the given points?

(–3, 10), (–1, 0), (0, –5), (3, 52)

anonymous · Answer

any clue as to the degree?

anonymous · Answer

unfortunatly no :/ thats all the information the question gives

anonymous · Answer

ok you have 4 points, so it can determine a polynomial of degree 3 since that has 4 coefficients

anonymous · Answer

since $p(0)=-5$ youi know that the constant is $-5$

anonymous · Answer

y = x3 – 4x2 + 2x – 5
y = x3 + 4x2 – 2x – 5
y = x3 + 4x2 + 2x – 5
y = x4 + 4x3 – 2x2 – 5x

here are the possible answers though

anonymous · Answer

ok since you have answers, you have a rather easier job
check which numbers work

anonymous · Answer

last one is wrong because if you replace $x$ by 0 you are supposed to get -5 
you can cross that off the list

anonymous · Answer

@amistre64  has a nice way to do this

amistre64 · Answer

xint     yint
(–3, 10), (–1, 0), (0, –5), (3, 52)
high         low      lower    high

amistre64 · Answer

with multiple points its doable, but labor intensive

amistre64 · Answer

it not the last one, since that y int is zero

anonymous · Answer

and you can replace $x$ by $-1$ and see which of the top three gives 0

amistre64 · Answer

(–3, 10), (–1, 0), (0, –5), (3, 52)

y = a + b(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x

anonymous · Answer

actually i think that gives it right away

amistre64 · Answer

yeah, it simpler to just fill in for x and do the math to narrow the options for this one

anonymous · Answer

i still would like to see your method
what are $a,b,c,d$?

anonymous · Answer

looks like $a=10$ from what you wrote

anonymous · Answer

which confuses me because the constant is $-5$

anonymous · Answer

oops i see $a$ is not the constant
i will shut up and learn

amistre64 · Answer

lol, fine

(–3, 10)

 y   = a + b(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x
10 = 10                0             0                      0


 y  = 10 + b(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x


(–1, 0)

 y -10  = b(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x
0                2b                    0                      0
                   -5



(0, –5)

   y  = 10 -5(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x

y-10  = -5(x+3) + c(x+3)(x+1) + d(x+3)(x+1) x
-5-10        -15             3c                               0
                                      0



(3, 52)

 y  = 10 -5(x+3) + d(x+3)(x+1) x
52 = 10   -  30          d 6     4    3
72 =                          72 d
                                         1



 y  = 10 -5(x+3) + (x+3)(x+1) x

anonymous · Answer

oh!! replace $x$ by -3 set result equal to 10 
then move on down the line!

amistre64 · Answer

correct, i learned later that this is a newton method, so that when more information is added, you dont have to redo all the stuff prior to it

anonymous · Answer

that is great!! also rather easier than solving a 4 by 4 system of equations. 
does this have a name?

anonymous · Answer

not "newton's method" as in "newton ralphson" right?  must be something else

amistre64 · Answer

legendre polynomial or some such is what i think we determined; i just call it the amistre64 chain of pain lol

anonymous · Answer

reminds me of pfd

amistre64 · Answer

y  = 10 -5(x+3) + (x+3)(x+1) x

y = 10 - 5x - 15 + x^3 +x^2 +3x^2 +3x

y = x^3 +4x^2 -2x -5

anonymous · Answer

damn i can't get a trademark symbol in latex
any ideas?

amistre64 · Answer

... dont use latex ;)

anonymous · Answer

i don't do windows

amistre64 · Answer

^{tm}  maybe?

amistre64 · Answer

$$amistre64^{tm}$$

amistre64 · Answer

$$\textsuperscript{\texttrademark}$$ doesnt seem to be coded

anonymous · Answer

Thanks guys! Great explanations :D