
quadratic equations
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^24x+1 by the way

Replace the values of x in the equation given to you!
For example, under 3, you put:
$\displaystyle (3)^2  4(3) + 1 = 9 + 12 + 1 = 22$
Hence:
$\displaystyle \begin{array}{cccccccc}\hline
x & 3 & 2 & 1&0&1&2&3 \\ \hline
y & 22 &&&&&&\\ \hline \end{array}$

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

$\displaystyle (2)^2  4(2) + 1 = 4 + 8 + 1$
Does this equal 11?

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

Quote:
Originally Posted by
andyboy179 would 1=3?
I get 6

Think of it again:
$\displaystyle (1)^2  4(1) + 1 = 1 + 4 + 1$
Does this make 1?
remember that when you multiply two negative numbers, they become a positive number.

would 0 =1
1=0
2=1
3=1???

for 0: yes.
For 1: $\displaystyle (1)^2  4(1) + 1 = 1  4 + 1$
For 2: $\displaystyle (2)^2  4(2) + 1 = 4  8 + 1$
For 3: $\displaystyle (3)^2  4(3) + 1 = 9  12 + 1$
Hm.. you need to work your basic calculations more...

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

Are you sure you posted the right equation?

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

I still have no idea what the equation actually is... (Speechless)