Cubic Function Help Homework Statement

[JazzMasterKC]

Homework Statement

A table is given:
x | -2 | -1 | 0 | 1 | 2 | 3
h(x) | 0 | 9 | y | 3 | 0 | 6

Well actually the h(x) line was based on a graph that we had to look at, it should be right. I discovered y = 8 (the graph increments by 3 on y-axis and 1 on x axis, so I couldn't tell what y was)
so d = 8, b = -2
This graph is a cubic function, so I'm using ax^3+bx^2+cx+d formula
I got c=-8 and a=2

The first question that I got was to fill the h(x) part of table (thats all the direction I had)
Then the second part of the question was to figure h(3)-h(1).. ok yeah no problem easy enough..
However here is the catch.. using the found equation from my data.. well first this is what it is:
y=2x^3-2x^2-8x+8
.. so anyways I graphed in calculator and when I look at the table I noticed for x=-1 y=12.. which from viewing the graph its obvious that is false, The highest point on y-axis appears to be 10 at most from looking at graph... This has thrown me off and I'm not sure what to do now.. its clear I screwed up some where.

Homework Equations

my results for equation (wrong) y=2x^3-2x^2-8x+8
cubic form: ax^3+bx^2+cx+d
Here are the four equations I derived out of the data to get what I thought was my answer..
-8a+4b-2c+d=0
-a+b-c+0=9
a+b+c+d=3
8a+4b+2c+d=0

The Attempt at a Solution

I wish to evaluate h(3)-h(1) but I've noticed some errors with what I did in part 1 of the question.

 

[haruspex]

-a+b-c+0=9

Check that one again.

 

[Ray Vickson]

Homework Statement

A table is given:
x | -2 | -1 | 0 | 1 | 2 | 3
h(x) | 0 | 9 | y | 3 | 0 | 6

Well actually the h(x) line was based on a graph that we had to look at, it should be right. I discovered y = 8 (the graph increments by 3 on y-axis and 1 on x axis, so I couldn't tell what y was)
so d = 8, b = -2
This graph is a cubic function, so I'm using ax^3+bx^2+cx+d formula
I got c=-8 and a=2

The first question that I got was to fill the h(x) part of table (thats all the direction I had)
Then the second part of the question was to figure h(3)-h(1).. ok yeah no problem easy enough..
However here is the catch.. using the found equation from my data.. well first this is what it is:
y=2x^3-2x^2-8x+8
.. so anyways I graphed in calculator and when I look at the table I noticed for x=-1 y=12.. which from viewing the graph its obvious that is false, The highest point on y-axis appears to be 10 at most from looking at graph... This has thrown me off and I'm not sure what to do now.. its clear I screwed up some where.

Homework Equations

my results for equation (wrong) y=2x^3-2x^2-8x+8
cubic form: ax^3+bx^2+cx+d
Here are the four equations I derived out of the data to get what I thought was my answer..
-8a+4b-2c+d=0
-a+b-c+0=9
a+b+c+d=3
8a+4b+2c+d=0

The Attempt at a Solution

I wish to evaluate h(3)-h(1) but I've noticed some errors with what I did in part 1 of the question.

There is a bit of inconsistency in your data. You get three slightly different cubic fits if you use x=(-2,-1,1,2) vs. x = (-2,-1,2,3) vs. x = (-2,1,2,3). These will give three slightly different values for h(0); I get h(0) = 8, 7.8 and 8.4, respectively.

 

[ehild]

Your data do not fit exactly to a third order polynomial. If the data were read from a graph, try to read more accurately.

ehild

 

[JazzMasterKC]

thanks, though I can't stand these problems that are based from looking at a graph -_-
I ended up finding that 1x^3+2x^2-4x+8 (I think, just pullin from memory) fit exactly how it should with the exception x=3 y=5 looking a little sketchy from the graph, it looks like y=6 there to me... oh well this should suffice.

 

[haruspex]

I ended up finding that 1x^3+2x^2-4x+8

1x^3-2x^2-4x+8

 

[JazzMasterKC]

Yep that's what I had sorry.