The function f(x)=ax^2 satisfies f(2)=1 and f(32)=4. Find r.
Rachel, Rachel !
Since both Damon and I went farther than the question asked for and also found the value of a ....
all you needed is r = 1/2
(the value of a indeed has √2 in it, but your teacher was not referring to a but rather to r .
Note we both had that √2 in the answer to a )
where does r come in ?
did you mean , find a ?
You either have a typo or your text was written in some other universe.
r ? what r?
from your first constraint a = 1/4
that does not work for your second constraint
I am so sorry the right problem:
The function f(x)=ax^r satisfies f(2)=1 and f(32)=4. Find r.
ahh, so we have two ordered pairs (2,1) and (32, 4) that satisfy our equation
----> 1 = a(2)^r
----> 4 = a(32^r
a(32)^r
= a(2)5r
so a(2)5r = 4
and a(2)^r = 1
divide them
2^(5r-r) = 4
2^4r = 2^2
then 4r = 2
r = 1/2
back into the first:
a(2)^(1/2) = 1
a√2 = 1
a = 1/√2 or √2/2
a (2)^r = 1 so a = 1/2^r
a (32)^r = 4
but 32 = 2^5
a (2)^5r = 4
2^5r / 2^r = 4
2^4r = 4
so 4 r = 2
r = 1/2
check
a = 1/sqrt 2
a sqrt(32) = sqrt 32/sqrt 2 = 4 sqrt 2/sqrt 2
= 4 sure enough
Hi Reiny,
Thank you for answering my question, but I don't think there is �ã2 at the top because, my teacher said it is just supposed to be whole numbers in the denominator and numerator.