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September 23, 2025, 09:11:57 am

Author Topic: 2009 Exam 2 MCQ  (Read 1330 times)  Share 

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sajib_mostofa

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2009 Exam 2 MCQ
« on: October 31, 2010, 12:54:21 pm »
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Can anyone help with me with Q2 of the multiple choice. They say there's three pts of intersection but I'm only getting two with the calculator.

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Re: 2009 Exam 2 MCQ
« Reply #1 on: October 31, 2010, 12:57:57 pm »
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You're using the solve function aren't you? Graph the two and there's clearly 3.
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sajib_mostofa

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Re: 2009 Exam 2 MCQ
« Reply #2 on: October 31, 2010, 12:58:45 pm »
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Any reason as to why the solve function gives me two?

cipherpol

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Re: 2009 Exam 2 MCQ
« Reply #3 on: October 31, 2010, 12:59:55 pm »
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o_O i used the solve function and got 3
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azn_dj

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Re: 2009 Exam 2 MCQ
« Reply #4 on: October 31, 2010, 01:00:51 pm »
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How do you graph it???
Using ti-89
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kyzoo

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Re: 2009 Exam 2 MCQ
« Reply #5 on: October 31, 2010, 01:04:28 pm »
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Any reason as to why the solve function gives me two?

Because two of the intersections are on top of each other (in the same vertical line) so if you solve for "x" you only get 2 values of "x". I think there are three solutions for "y"? I dunno I made the same mistake.
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98.40_for_sure

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Re: 2009 Exam 2 MCQ
« Reply #6 on: October 31, 2010, 01:05:00 pm »
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Any reason as to why the solve function gives me two?

Because you're solving for the variable x. At the positive X value it comes up, there is TWO y values. So there is TWO intersection points for x, but 3 for y.
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sajib_mostofa

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Re: 2009 Exam 2 MCQ
« Reply #7 on: October 31, 2010, 01:07:54 pm »
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Ahh yep thanks for that. Quite cheeky..

sajib_mostofa

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Re: 2009 Exam 2 MCQ
« Reply #8 on: October 31, 2010, 01:15:49 pm »
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Also for Q7, it asks for the minimum degree but isnt 3 the highest degree?

azn_dj

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Re: 2009 Exam 2 MCQ
« Reply #9 on: October 31, 2010, 01:18:18 pm »
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Nope. For Question seven, it says real coefficients, and you have a complex number and a real number.
so instantly you know there must be a complex conjugate to have real coefficients.
Hence the minimum is 3. For more you can just multiply by (z-1) or (z-2) or anything random. So the maximum is well, anything =D
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sajib_mostofa

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Re: 2009 Exam 2 MCQ
« Reply #10 on: October 31, 2010, 01:20:52 pm »
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Nope. For Question seven, it says real coefficients, and you have a complex number and a real number.
so instantly you know there must be a complex conjugate to have real coefficients.
Hence the minimum is 3. For more you can just multiply by (z-1) or (z-2) or anything random. So the maximum is well, anything =D

Thanks for that. I originally assumed that there were ONLY those three factors that you mentioned.

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Re: 2009 Exam 2 MCQ
« Reply #11 on: October 31, 2010, 02:09:26 pm »
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How do you draw more than one conics equation simultaneously (for comparison) in Classpad 330? :O

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Re: 2009 Exam 2 MCQ
« Reply #12 on: October 31, 2010, 02:14:43 pm »
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How do you draw more than one conics equation simultaneously (for comparison) in Classpad 330? :O

Sketch the graph in parametric form
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Re: 2009 Exam 2 MCQ
« Reply #13 on: October 31, 2010, 02:18:13 pm »
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How do you draw more than one conics equation simultaneously (for comparison) in Classpad 330? :O
eg for an ellipsis Form -> Insert, then whatever
Then you want to see a hyperbola on top of it:
Form -> Insert, then type out the equation
instead of clicking on the graph button, highlight the equation and drag it onto the graph (with an ellipsis already present)
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Re: 2009 Exam 2 MCQ
« Reply #14 on: October 31, 2010, 02:26:07 pm »
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Sweet as! :D
Cheers