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How do you graph y= (1/2)X ? Not sure what exponential rule to apply to plot the points. please help?
its a exponential function where x is the exponent.
y = (1/2)^x
is an exponential function where x is the exponent
1° case when x approaches negative infinity
Note that y = (1/2)^x = 2^(-x) so when x -> (-inf.), function approaches y = 2^[-(-inf)] = 2^(+inf)
and lim(y) {x -> -inf} = +inf
When x decreases to -inf., y grows to +inf
2° case when x approaches positive infinity
We can write
y = (1/2)^x = f(x)/g(x)
where f(x)=1 and g(x)=2^x
and then use following rule
lim[f(x)/g(x)] {x -> inf} = lim f(x) {x -> inf} / lim g(x) {x -> inf}
therefore
lim[(1/2)^x] {x -> inf} = lim[1/(2^x)] {x -> inf} = lim (1){x -> inf} / lim (2^x){x -> inf}
lim (1){x -> inf} = 1
lim (2^x){x -> inf} = inf
lim[(1/2)^x] {x -> inf} = 1 / inf. = 0
The limit of y as x approaches +infinity is 0.
That means horizontal line y=0 is horizontal asymptote of function y = (1/2)^x.
3° intercept with y-axis
For x=0,
y = (1/2)^0 = 1
Graph of this function passes through (0,1).
Black Butler – 11 – His Butler, However You Please (SUB)
