Write an equation for the nth term of the arithmetic sequence 6,3,0,-3,...?
2 answers:
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Answer:
Step-by-step explanation:
1. x^1/3(x^1/2) =
2. x^1/3(2x^2) =
3. = +
The asymptote cannot be x= because x can be any number. If you think about it, you can take a number to any exponent.
If x is a positive exponent, y is positive.
If x is a nevative exponent, y decreases, but is still positive. This is because a number to a negative exponent equals 1 over the number to the positive exponent. Thus, it is smaller, but still positive.
If x is 0, y is positive again because anything to the zero is positive 1.
There is no way y could be less than or equal to zero. So, there is an asymptote at y=0.
Also, set the equation equal to 0 and solve. You should end up with 4^x=0. Since no exponenent can make a number zero, this isn't possible, so y cannot equal zero.
Here is the graph for a visual:
If x is a positive exponent, y is positive.
If x is a nevative exponent, y decreases, but is still positive. This is because a number to a negative exponent equals 1 over the number to the positive exponent. Thus, it is smaller, but still positive.
If x is 0, y is positive again because anything to the zero is positive 1.
There is no way y could be less than or equal to zero. So, there is an asymptote at y=0.
Also, set the equation equal to 0 and solve. You should end up with 4^x=0. Since no exponenent can make a number zero, this isn't possible, so y cannot equal zero.
Here is the graph for a visual: