Which of the following is the function representing the graph below? f(x) = (1/3)^x + 1 f(x) = (1/3)^x – 1 f(x) = 3x – 1 f(x) = 3x + 1
well, you have a noticeable increament/decrement on the x, y relationship that means as one changes a bit, the one changes a lot
what does that mean? is an exponential equation
Im confused.
now at 1/3, \(1^3 = 1, 1^{100000} = 1\)
so as the exponent for 1/3 changes, 1 remains 1, and changes dramatically from 3 to 9 to 81 to 6561
from 1/3 to 1/6561, the latter number is much smaller of a fraction, and thus closer to 0
linear equations - straight lines - follow the formet y = mx +c. m is the gradient, c is where the line hits the x axis. the line you show is clearly not linear - it is not a straight line. that eliminates c) and d). looking at the other options, why not start by setting x to 0 and seeing what y becomes. the solutions are: A) +1 B) -1 anything to the power of zero = 1. so if your curve intercepts x = 0 at +2, the answer is........?!?!
so, if you were to use say 1/6561 - 1 as choice b) suggests you'd get -6560/6561 a negative number but "y" in the graphic shows no negative numbers so, which one do you think it's?
a ? idk im confused.:(
well, it's an exponential function so, which choices give you an exponential function?
look, ignore jdoe and read what i wrote. the answer is in there.
b ?
i got lost when you said so if your curve intercepts x = 0 at +2.
oh no nvm i understand that part lol sorry.,
(1/3)^0 + 1 = 2 (1/3)^0 - 1 = 0 does that make sense?
yess.
so how do i find the answer from that .
out of (1/3)^1000 + 1 = ? (1/3)^1000 - 1 = ? which is positive?
the first one ?
yes
so, the 1st one is positive, the graphic shows only POSITIVE for "y" it shows no negative ones (1/3)^1000 - 1 means ^ smaller -1 than 1 so a number smaller than 1 -1, will give a negative value but in the graphic "y" has no negative ones so, there's only 1 option :)
so it is a ? the positive 1 ?
yes Lesha, and I hope you can see why! Well done you.
Thank you. (: . Veryy much !
yw
ii have one more graph question. can you help ?
sure, post in channel so we all can see it and help, and revise each other :)
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