Question

how do I go about determing the polynomial equation that passes through these 4 points? (-3,10), (-1,0), (0,-5), (3,52)

Answer 1

Hmmm are you familiar with matrices?
What method should you be using?

Answer 2

it doesn't matter

Answer 3

Ok what grade r u in?

Answer 4

11

Answer 5

hmmm ok we will try matrices but I doubt u learnt that method. It may be too advanced

Answer 6

I know a little about matrices, mainly how to set them up using the coefficients but that's all

Answer 7

Ok for starters we have points so we are assuming that our polynomial is of degree 3 ok?

Answer 8

right

Answer 9

Now we know that a polynomial of degree 3 has the form of
\(f(x)=a+bx+cx^2+dx^3\)

Answer 10

ok-that's backwards from how I normally write it but ok

Answer 11

No we were given 4 diff points so we will plug them in
Our first point is (-3,10)
\(a+b(-3)+c(-3)^2+d(-3)^3=10\)
\(a-3b+9c-27d=10\)
Do you follow?

Answer 12

yes

Answer 13

do that with each set of points and then what

Answer 14

Ok so our next point is (-1,0)
\(a+(-1)b+(-1)^2c+(-1)^3d=0\)
\(a-b+c-d=0\)

Our next point is (0,-5)
\(a+(0)b+(0)^2c+(0)^3d=-5\)
\(a=-5\)

Our next point is (3,52)
\(a+(3)b+(3)^2c+(3)^3d=52\)
\(a+3b+9b+27d=52\)

Answer 15

Ok now that we know that a=-5 we can plug that into all the 3 equations

Answer 16

ok
which equation is the best one to use

Answer 17

No you gotta plug a=-5 into all the 3 equations

Answer 18

oh

Answer 19

i have done that with all 3 now what

Answer 20

Equation #1:
\((-5)-3b+9c-27d=10\)
\(-3b+9c-27d=15\)

Equation #2
\((-5)-b+c-d=0\)
\(-b+c-d=5\)

Equation #3
\((-5)+3b+9c+27d=52\)
\(3b+9c+27d=57\)

Answer 21

that's what i got

Answer 22

Ok next we will set up everything into a matrix
The first column will be our b's second will be c's and third will be d's and the last column will be our answer

Answer 23

ok

Answer 24

not sure what goes in the last column

Answer 25

\[ \left( \begin{array}{cccc}
-3 & 9 & -27 &15 \\
-1 & 1 & -1 &5\\
3 & 9 & 27 &57 \end{array} \right)\]

Answer 26

This is how our matrix shld look

Answer 27

good so far

Answer 28

Now we need to row reduce which is a pain
Do you know how to row reduce?

Answer 29

no

Answer 30

ughhhh Shi*

Answer 31

sorry

Answer 32

Well it will take me a few minutes to show you each step. Its really hard to show that matrix thing here. Ill try to explaing as much as possible

Answer 33

Ok first I will switch the rows
\(\left( \begin{array}{cccc}
-1 & 1 & -1 &5\\
-3 & 9 & -27 &15 \\
3 & 9 & 27 &57 \end{array} \right)\)

Answer 34

Then I multiplied the first row by -1
So -1*Row #1
\(\left( \begin{array}{cccc}
1 & -1 & 1 &-5\\
-3 & 9 & -27 &15 \\
3 & 9 & 27 &57 \end{array} \right)\)

Answer 35

Next I Multiply Row #1by 3 and add that to Row#2
Row#2=3*Row#1+Row#2
\(\left( \begin{array}{cccc}
1 & -1 & 1 &-5\\
0 & 6 & -24 &0 \\
3 & 9 & 27 &57 \end{array} \right)\)

Answer 36

why mult. by 3

Answer 37

Next you divide Row#2 by 6
Row#2=Row#2/6
\(\left( \begin{array}{cccc}
1 & -1 & 1 &-5\\
0 & 1 & -4 &0 \\
3 & 9 & 27 &57 \end{array} \right)\)

Answer 38

OK basically we are trying to get the matrix into this form
\(\left( \begin{array}{cccc}
1 & a & b &c\\
0 & 1 & d &e \\
0 & 0 & 1 &f \end{array} \right)\)
Where a,b,c,d,e,f are numbers
I just giving you an example. Just trying to give u an idea
So basically work row by row
We first get the first number in row 1 to be 1
Then we go to the next row and get the first number to be 0 and the second one to 1
And then we go to the third row and get the first 2 numbers to be 0 and the third one to be 1

Answer 39

not sure what to do to get equation

Answer 40

OK so basically lets look at the multiplying 3 one that you got confused by ok

Answer 41

In the second Row the first number is -3 correct?

Answer 42

how did u get -3

Answer 43

\(\left( \begin{array}{cccc}
1 & -1 & 1 &-5\\
-3 & 9 & -27 &15 \\
3 & 9 & 27 &57 \end{array} \right)\)
This our matrix we had in the beg

Answer 44

yes way back up at the top

Answer 45

Correct So u see that -3 ?
Its the first number in the second row

Answer 46

yes

Answer 47

Ok we need it to be 0 correct?

Answer 48

why

Answer 49

Cuz we need it to be in that form
\(\left( \begin{array}{cccc}
1 & a & b &c\\
0 & 1 & d &e \\
0 & 0 & 1 &f \end{array} \right)\)

Answer 50

k

Answer 51

Ok so how can we get the -3 to become a 0?

Answer 52

add 3

Answer 53

Well we can't do that
Cuz remember the whole row in one equation remember?
And there are rules to matrices
Firstly you can only divide and multiply a single row
And secondly you can add and subtract rows

Answer 54

You cant just add -3 to a whole row
So what I did was multiply Row#1 by 3 and then add that to Row #2
So our new Row#2 =3*Row#1+Row#2

Answer 55

Like I think this method is too advanced for you

Answer 56

i agree

Answer 57

hmmmm thinking how else to solve this give me a sec

Answer 58

hahahha ok So I have another way to solve Using elimination
Are you familiar with that?
I remember using that method back in HS so i bet u do

Answer 59

I have used it before

Answer 60

Ok so lets write out the 3 equations again so it will be easier

Answer 61

-3b+9c-27d=15
-b+c-d=5
3b+9c+27d=57

Answer 62

I have c+4

Answer 63

=4

Answer 64

Ok so we will set subtract equation #1 and equation #2
And we will subtract Equation #1 and equation #3
We will eliminate the "b" ok
And then we will have 2 brand new equations
\(-3b+9c-27d=15\)
\(-3( -b+c-d=5))\)
We can only eliminate the b by multiplying the second equation by -3
\(-3b+9c-27d=15\)
\(3b-3b+3b=-15\)
---------------------
0b+6c-24b=0 ----> Equation #4


\(-3b+9c-27d=15\)
\(3b+9c+27d=57\)
---------------------
0b+18c+0d= 72 -----> Equation #5

So 18c=72
\(\large {18c \over 18} ={72 \over 18}\)
c=18

Ok so now lets plug in c=18 into equation #4
6(18)-24b=0
108-24b=0
-24b=-108
b=4.5

Now lets plug c=4 and b=4.5 into Equation #2
(-4.5)+4-d=0
-.5-d=0
-.5=d

Answer 65

my choicces are x^3+4x^3+2x-5 or x^3-4x^2-2x-5

Answer 66

hmmm we got the wrong answer let me recheck my work

Answer 67

and we came up with a positive fouand a -5 but not a 1 or a 2

Answer 68

positive four

Answer 69

ohh I see where i went wrong ugghhh

Answer 70

u used 18 instead of 4 for c

Answer 71

yesssssssssssssssss

Answer 72

Ok so now lets plug in c=4 into equation #4
6(4)-24b=0
24-24b=0
-24b=-24
b=1
Now lets plug c=4 and b=1 into Equation #2
-(1)+4-d=0
3-d=0
3=d

Answer 73

Still getting a diff answer

Answer 74

hmmmmm

Answer 75

this might be a problem that needs to be done on a calculator. I think I will just mark the first answer listed with the positive four. I didn't expect all of this help from you. thanks so m;uch

Answer 76

Ill brb I just need to eat dinner. Everything is cold by now

Answer 77

I see, First off, I'd graph it.

Answer 78

but how do I get an equation from a graph

Answer 79

Nah u r not suppossed to graph this stuff. Its 3 variable so it will be annoying

Answer 80

ok Let me go over my work. and see where i messed up

Answer 81

I have a few ways although I'm just checking if they work first.

Answer 82

Ohhh I see where i messed up lol

Answer 83

hahah I totally mixed up variables all over

Answer 84

Today isnt my day, by accident i wrote out a check for $300 instead of $800
I think its the lack of food
OK let me start form way back there

Answer 85

f(x) =af(k(x-d))+c

It was -5 down from the y-axis axis so the c is is -5

Answer 86

Lagging so much it's unbelievable.

Answer 87

the d is 0 since the x-axis is the same.

Answer 88

you mean d is -5 ax^4+bx^3+cx^2+x

Answer 89

−3b+9c−27d=15
3b−3c+3d=−15
---------------------
0b+6c-24d=0 ----> Equation #4

−3b+9c−27d=15
3b+9c+27d=57
---------------------
0b+18c+0d= 72 -----> Equation #5

18c=72
c=4
Now we plug that c=4 into equation #4
6(4)-24d=0
24-24d=0
-24d=-24
d=1

Answer 90

my choices listed all have d as -5

Answer 91

Oh wow Im lagging wayy too much . Sorry.

Answer 92

right swiss, your d shoulld be one

Answer 93

No ure d=1 in all cases
Remember the formula
f(x)=a+bx+cx^2+dx^3

Answer 94

d - 1 in all my choices if written your way

Answer 95

a= -5 in all my choices

Answer 96

c is either 4 or -4 and b is either 2 or -2

Answer 97

a=-5 as we saw in the beg
c=4
d=1
and now we gotta show what b is

Answer 98

-2