SOLUTION: find all positive real number m so that the graphs of x+my=0 and x=y^2 have respectively:
a. exactly one point of intersection
b. no point of intersection
c. exactly 2 points of
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-> SOLUTION: find all positive real number m so that the graphs of x+my=0 and x=y^2 have respectively:
a. exactly one point of intersection
b. no point of intersection
c. exactly 2 points of
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Question 729832: find all positive real number m so that the graphs of x+my=0 and x=y^2 have respectively:
a. exactly one point of intersection
b. no point of intersection
c. exactly 2 points of intersection. Answer by hoomanc(14) (Show Source):
You can put this solution on YOUR website! first re-arrange the equation base on : and or by writing the equilibrium for both sides base on "y" we would have: then then then then that means: or so, base on this answer it appears that regardless of what value taken we always would have an intersection in the origin of & .
In addition, if we try to use a very small value for , lets say for example the second answer of would be very close value to zero , In other words, as much as we take a smaller value for we are more closer to a unique intersection value of zero, otherwise in bigger cases we can't neglect bigger values of and the second intersection would be at .
In above graph the two curve line are and
As much as the take a smaller value for the straight line is going to get more vertical {(red m=1) (green m=0.2) (blue m= 0.1)}. so the second intersection point get closer more and more to zero.
In conclusion >>>
1)for case b: we can't guarantee that for sure cause there will always be at least one intersection at
2) for case a if we try to minimize the value of we are close to unique answer of one intersect at but it's never gonna happen (try to draw both graphs to understand why)
3)for case c: it's what is always going to happen for the rest of non-small values of .
hope that helps.