hi, i need to make this into a quadrattic equation but there is a pert on the table which i can't remember what to do!

when you have x then -3,-2,-1,0,1,2,3
under that you puy Y and i can remember how to work that out! if you can help i will be very thankful!

my question is Y= x^2-4x+1 by the way

2. Replace the values of x in the equation given to you!

For example, under -3, you put:

$(-3)^2 - 4(-3) + 1 = 9 + 12 + 1 = 22$

Hence:

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline
x & -3 & -2 & -1&0&1&2&3 \\ \hline
y & 22 &&&&&&\\ \hline \end{array}$

3. thanks for answering my question! under -2 it would be 11?

4. $(-2)^2 - 4(-2) + 1 = 4 + 8 + 1$

Does this equal 11?

5. no its 13, i think i got it now!! would -1=3?

6. Originally Posted by andyboy179
would -1=3?
I get 6

7. Think of it again:

$(-1)^2 - 4(-1) + 1 = 1 + 4 + 1$

Does this make 1?

remember that when you multiply two negative numbers, they become a positive number.

8. would 0 =1
1=0
2=1
3=1???

9. for 0: yes.

For 1: $(1)^2 - 4(1) + 1 = 1 - 4 + 1$

For 2: $(2)^2 - 4(2) + 1 = 4 - 8 + 1$

For 3: $(3)^2 - 4(3) + 1 = 9 - 12 + 1$

Hm.. you need to work your basic calculations more...

10. actually would this be it:
-3=19
-2=11
-1=5
0=1
1=-1
2=-1
3=1

11. Are you sure you posted the right equation?

12. cr@p, no i didn't its -4X sorry!!!!!!!!

13. I still have no idea what the equation actually is...