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Updated on: 09 Dec 2018, 01:48
Originally posted by vanam52923 on 09 Dec 2018, 01:42.
Last edited by vanam52923 on 09 Dec 2018, 01:48, edited 1 time in total.
Last edited by vanam52923 on 09 Dec 2018, 01:48, edited 1 time in total.
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ruchik
JeffYin
ruchik
thankyou so much for clearing it.
I am very much clear about statement 1 now.
But i am not clear on statement 2.
It is given x>0 so we have to take positive x so quad 1 or 4th
but why cannot i take parabola as shown in my diagram case 2:
second parabola
its x intercept is positive and y is also positive.but it lies in 4th quad too(as shown in diagram)
Is it because then it wont b symmetric around y axis?
this is my confusion.
here
i am getting parabola should be symmetric to y and it should have +ve x intercept with positive y intercept so Case 2 second diagram fails but in case 2 diagram 1 ,i am not intercepting parabola any where on x axis.so how is this possible.
I read somewhere on net that
here for x intercept
y=0 if i put
i get x^2=-k
so x will have imaginary coordinates which is not possible so this parabola cannot have x intercept.Is this true.If yes then only Case 2 diagram 1 holds.Is it a thumb rule?
Thanking you in anticipation
09 Dec 2018, 01:45
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ruchik
ruchik
thankyou so much for clearing it.
I am very much clear about statement 1 now.
But i am not clear on statement 2.
It is given x>0 so we have to take positive x so quad 1 or 4th
but why cannot i take parabola as shown in my diagram case 2:
second parabola
its x intercept is positive and y is also positive.but it lies in 4th quad too(as shown in diagram)
Is it because then it wont b symmetric around y axis?
this is my confusion.
here
i am getting parabola should be symmetric to y and it should have +ve x intercept with positive y intercept so Case 2 second diagram fails but in case 2 diagram 1 ,i am not intercepting parabola any where on x axis.so how is this possible.
I read somewhere on net that
here for x intercept
y=0 if i put
i get x^2=-k
so x will have imaginary coordinates which is not possible so this parabola cannot have x intercept.Is this true.If yes then only Case 2 diagram 1 holds.Is it a thumb rule?
Thanking you in anticipation
09 Dec 2018, 12:38
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vanam52923
In a sense, you are correct that your case 2 diagram 2 is not allowed because it isn't symmetric about the y axis. This is because the question itself tells us that y = x^2 + k, which means that the parabola is symmetric about the y axis, and no scenario is allowed that contradicts the information in the question itself.
There is no "rule of thumb" involved here. Instead, what you did points to a couple of issues to pay attention to:
1) First, as mentioned above, we can't contradict any information provided in the question itself. If you find yourself considering a scenario that is not consistent with the information provided in the question, stop yourself once you notice this, because this scenario is not allowed.
2) Second, we need to read carefully. Notice that statement 2 says "x is greater than 0". You interpreted this as "the x-intercept is greater than 0". In this case, you added some information (x-intercept) that was not mentioned in the statement. So, ideally, you don't want to add any assumptions beyond what is provided in the question or the statement. If you are not sure, and you start going down the road that you went down in this question, you can stop yourself as soon as you realize that the scenario you are considering isn't consistent with the information provided in the question.
Please let me know if you have any more questions!
