SOLUTION: State the equation of the line asymptotic to the graph (if any) y=(1/4)^x-3. Could you help me with this please?

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Question 270116: State the equation of the line asymptotic to the graph (if any)
y=(1/4)^x-3. Could you help me with this please?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
It is not possible to know whether your equation is
y+=+%281%2F4%29%5E%28x-3%29
or
y+=+%281%2F4%29%5Ex-3
Please use parentheses generously to make your problem clear. The first equation should be typed in as
y = (1/4)^(x-3)

I will answer the question for both equations. The key to both solutions is to understand what values a positive number raised to a power can be. Your equations both have a positive number, 1/4, raised to some power. No matter what exponent you put on 1/4, the result will NEVER be:
  • Zero. (Remember %281%2F4%29%5E0+=+1 not zero!)
  • Negative. (Remember 1/4 to a negative power is not a negative number. Negative exponents just mean reciprocals. And any reciprocal of 1/4 is still positive.)

While 1/4 to some power can never be zero, it can, however, be an extremely small positive fraction (i.e. a very tiny fraction just above zero). 1/4 to very large positive powers will be very small fractions near zero. The larger the exponent, the closer we get to zero. This is how an asymptote works.

So for
y+=+%281%2F4%29%5E%28x-3%29
where y equals a power of 1/4, y = 0 is a horizontal asymptote.
and for
y+=+%281%2F4%29%5Ex-3
where y equals a power of 1/4 minus 3, then y = 0-3 = -3 is a horizontal asymptote.