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Q10 James Ruse Ext 1 2022 (1 Viewer)

jimmysmith560

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Would the following information help?

For to have only one asymptote, the discriminant of must be a negative value.





Since , satisfies the requirement .

The answer is therefore C.
 

carrotsss

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Because it’s a reciprocal function, there’s by default gonna be an asymptote where y=0. Therefore, there must be no other asymptotes, which means that there are no real x values for which the polynomial on the denominator is =0. Hence, the discriminant for the denominator must be less than (not equal to) 0. Rearranging,
0>b^2-4ac
0>b^2-16a^2
b^2<16a^2
therefore -4a<b<4a

So therefore, A, B and D don't work. But, because a>1, technically we can also say that
-4<=b<=4
and -1<=b<=4 is within that domain, so it is also true.
 

5uckerberg

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Anyone know how to do this?
View attachment 38456
The answer should be C
For the denominator we use the quadratic formula where . For only one asymptote in this case we would like to have to not exist. Here, we know that a solution does not exist when which in this case is which in reality is simply and . In the question , so therefore, . On the negative side of things . As such your answer will be C.
But if it is a post MC question the complete answer is or .
 

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