Question

help with polynomials

Answer 1

the graphs show that f(1/2t) = g(t), how do I find the ratio a1/a2 ?
http://img38.imageshack.us/img38/3199/v2p9.jpg

Answer 2

hi

Answer 3

I guess this has to do with ax^2 + bx + c = mx^2 + nx + p, a=m, b=n, c=p

Answer 4

First interpret the graph and express them in two functions, S1 and S2.

S1(t)=a1 t^2 + b1 t
S2(t)=a2 t^2 + b2 t

Since both S1(t1/2) and S2(t1) equal h (check from graph), we write
S1(t1/2)=S2(t1)
which expands to
(a1/4) t1^2) + (b1/2) t1 = a2 t1^2 + b2 t1

For the equality to be true for any value of t1, the coefficients of each power of the polynomials on each side must equal, in other words,

a1/4 = a2
and
b1/2 = b2

Answer 5

I get now, I wasn't understanding the explanation

Answer 6

An easier solution is to first observe that both parabolae have an x-intercept at 0. The first one has an intercept at t1 and the other one at t2. Hence the factored form of both can be written as:\[\bf S_1(t)=t(t-t_1)\]\[\bf S_2(t)=t(t-2t_1)\]Expanding them we get:\[\bf S_1(t)=t(t-t_1)=t^2-(t_1)t\]\[\bf S_2(t)=t^2-(2t_1)t\]We also know that \(\bf a_1\) and \(\bf a_2\) are the leading coefficients in each quadratic, i.e. they are the coefficients of \(\bf t^2\) in each equation. Both equations have a 1 in front of \(\bf t^2\) which means that the leading coefficient is 1 for both of them, which implies \(\bf a_1=a_2=1\). Since they both equal 1, their ratio must also be 1.

@lucaz

Answer 7

1 or 4 ?

Answer 8

@genius12
If we consider that S1(t1/2)=S2(t1)=h, then
we need a different leading coefficient for S1 and S2

Answer 9

@mathmate I see now. But then it should be 1/4 not 4.

Answer 10

@mathmate Do you agree?

\(\bf a_1/a_2=1/4\)

Answer 11

@lucaz

Answer 12

I see 4 not 1/4

Answer 13

It depends on whether we are calculating a1/a2 or a2/a1.

Try
T1(t)=t(t-2)
T2(t)=(t/4)(t-4)
T1(1)=T2(2)=-1
so
a1/a2=1/(1/4)=4
@genius12

Answer 14

it's a1/a2

Answer 15

yes I know @mathmate He asked for a1/a2 which is 1/4. if he had asked a2/a1, it would be 4.

Answer 16

Did you have a chance to look at the counter-example?

Answer 17

Let me check some stuff over lol.

Answer 18

@mathmate K makes sense. I was doing something unnecessary.

Answer 19

Great! Thank you for keeping me on my toes!