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September 23, 2025, 04:45:44 pm
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Author Topic: TyErd's questions  (Read 42330 times)  Share 

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TyErd

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TyErd's questions
« on: April 24, 2010, 12:10:13 pm »
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solve:
« Last Edit: April 27, 2010, 10:09:47 pm by TyErd »
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Damo17

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Re: TyErd's questions
« Reply #1 on: April 24, 2010, 12:17:58 pm »
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TyErd

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Re: TyErd's questions
« Reply #2 on: April 24, 2010, 12:24:19 pm »
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oook thanks I get it!
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TyErd

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Re: TyErd's questions
« Reply #3 on: April 24, 2010, 12:31:53 pm »
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How do you sketch the graph of
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superflya

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Re: TyErd's questions
« Reply #4 on: April 24, 2010, 12:45:27 pm »
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as f(-x)=f(x) it is an even function. so reflect in y.
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TyErd

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Re: TyErd's questions
« Reply #5 on: April 24, 2010, 12:58:51 pm »
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what graph do you reflect though?
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kyzoo

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Re: TyErd's questions
« Reply #6 on: April 24, 2010, 01:37:14 pm »
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As the numerator of the index (2/3) is an even number, therefore the function will always be a positive number; it will always be above the x-axis. The denominator being an odd number means that the function exists for all values of x, rather than just positive values. So you have a function that exists in the 1st and 2nd quadrants.

It will have a similar appearance to y = x^0.5, just that it exists in both the 1st and 2nd quadrants. So you have a graph that looks like y = x^0.5 combined with the graph of y = x^0.5 reflected about the y-axis.
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TyErd

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Re: TyErd's questions
« Reply #7 on: April 24, 2010, 01:46:30 pm »
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so you have to pretty much memorise the shapes of those type of functions?
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TyErd

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Re: TyErd's questions
« Reply #8 on: April 24, 2010, 01:58:07 pm »
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When , Find the real values of h for which only one of the solutions of the equation is positive.
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Blakhitman

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Re: TyErd's questions
« Reply #9 on: April 24, 2010, 03:20:04 pm »
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This is how I'd go about this question, should be right!

factorised:

basic shape will be,



then we need to solve both solutions when they are >0





And





and the question asks for when only one is positive, so we'll use the more positive one, cause it'll always be more positive than the other one!

so the answer is:
« Last Edit: April 24, 2010, 03:41:41 pm by Blakhitman »
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TyErd

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Re: TyErd's questions
« Reply #10 on: April 24, 2010, 03:38:40 pm »
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okay i get it up to the point you drew thegraph lol, but i think its the question that im finding it difficult to understand. Whats the questions even asking. Can someone rephrase it for me?
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Blakhitman

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Re: TyErd's questions
« Reply #11 on: April 24, 2010, 03:41:05 pm »
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It's asking when is only one x intercept positive.
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TyErd

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Re: TyErd's questions
« Reply #12 on: April 24, 2010, 03:43:59 pm »
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OHHHHHHHHHH! i get it! so pretty much were shifting the graph left 'h' units until we have 1 solution, right?
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Blakhitman

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Re: TyErd's questions
« Reply #13 on: April 24, 2010, 04:01:18 pm »
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Well, we're trying to find the values of 'h' so that only one of the x-int. is positive.

and the x-ints, are given by and

So what we do is solve for when these two intercepts are .

we get and .

and since the question is asking when only one is positive, we need to eliminate when one of them is positive, so h will not be <1 so we are left with .

Hope this makes sense!
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TyErd

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Re: TyErd's questions
« Reply #14 on: April 24, 2010, 04:07:32 pm »
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Yeaah i get it thankyou! a Graph really does help! thnx
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