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Integrate exponential question (1 Viewer)
- Thread starter enigma_1
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Assuming you got the primitive y = -e^[-xln(3)] + C, to find C; use the information given that a horizontal asymptote is y = 2
That is, an asymptote simply means a value for which the curve does not exist in - this case is y = 2.
So, let y = 2 into your primitive:
2 = -e^[-xln(3)] + C
What we can see is that there's 2 things on RHS, an exponential and a constant. We know that an exponential's asymptote (given no constants involved) is y = 0. So to make y = 0 (the LHS), let C = 2.
2 = -e^[-xln(3)] + 2
0 = -e^[-xln(3)] + 0
0 = e^[-xln(3)]
Looks like it all works out if C = 2. So yeah, pretty sure it's C = 2
That is, an asymptote simply means a value for which the curve does not exist in - this case is y = 2.
So, let y = 2 into your primitive:
2 = -e^[-xln(3)] + C
What we can see is that there's 2 things on RHS, an exponential and a constant. We know that an exponential's asymptote (given no constants involved) is y = 0. So to make y = 0 (the LHS), let C = 2.
2 = -e^[-xln(3)] + 2
0 = -e^[-xln(3)] + 0
0 = e^[-xln(3)]
Looks like it all works out if C = 2. So yeah, pretty sure it's C = 2