Another graph question (1 Viewer)
- Thread starter BlueGas
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davidgoes4wce
Well-Known Member
Well its being shift horizontally to the left by 3, its been vertically shifted up by 2. So b=-3, c=2. So that rules out A and D.
I think there is something wrong with this question because if you substitute the coordinate (2,3) back into this equation you get a value of a=-1
Here is my working out:
I think there is something wrong with this question because if you substitute the coordinate (2,3) back into this equation you get a value of a=-1
Here is my working out:
There's no option for a=-1. And you don't really have to do any working out for this one. So here is my method
You see the hyperbola has a negative a because it is in the 2nd and 4th "quadrants" formed by it's asymptotes. B will be -3, as x cannot be 3, and similarly c will be 2 as y cannot be 2. So the only choice that fits all of these is C.
You see the hyperbola has a negative a because it is in the 2nd and 4th "quadrants" formed by it's asymptotes. B will be -3, as x cannot be 3, and similarly c will be 2 as y cannot be 2. So the only choice that fits all of these is C.
It isn't given that the coordinate is (2,3), so you can't use it. Just look at it and C will be the only one that makes sense. But I agree the graph is off because subbing a, b and c for x=2 will get (2,4) which is clearly not right.Well its being shift horizontally to the left by 3, its been vertically shifted up by 2. So b=-3, c=2. So that rules out A and D.
I think there is something wrong with this question because if you substitute the coordinate (2,3) back into this equation you get a value of a=-1
Here is my working out:
davidgoes4wce
Well-Known Member
I guess I am making an assuming that the coordinate which is pretty much directly on the line is (2,3)
It clearly has to be a negative 'a' value as the curve is a negative hyperbola function.
Yeah, thats why I said C.