anonymous
  • anonymous
math homework helppppppp
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1377624206768:dw|
DebbieG
  • DebbieG
omg no, that's so wrong. You CANNOT cancel like that!!
DebbieG
  • DebbieG
@MorganMorganMorgan , you need to factor the denominator first. Do you know how to do that?

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anonymous
  • anonymous
thats what ive been having a hard time doing
DebbieG
  • DebbieG
OK, let's work on that then. You have \(\Large 3x^2-5x-2\) Depending on what methods you've learned for factoring, you might want to proceed with some trial and error, or use the a*c method? but you're looking for \(\Large 3x^2-5x-2=(....?...)(....?....)\) right?
anonymous
  • anonymous
yes
DebbieG
  • DebbieG
OK, well, trial and error actually probably isn't too bad here. Since the coefficient on the squared term is 3, you KNOW that the FIRST terms in the ( )'s have to be: 3x in one x in the other Do you see why that must be so?
anonymous
  • anonymous
yes
DebbieG
  • DebbieG
Good! Similarly, since the constant term (the 3rd term) is -2, the LAST terms in the ( )'s have to be: \(\pm2\) in one \(\pm1\) in the other and the signs MUST be opposite. Do you see why that is true?
DebbieG
  • DebbieG
So your factoring has to be one of the following: \(\Large (3x\pm1)(x\pm2)\) \(\Large (3x\pm2)(x\pm1)\) You just have to figure out where the 2 and the 1 go, and which gets the + and which the - Now, a HINT here is found in the fact that you need the middle term to end up being -5x And the COEFFICIENT on that middle term, the -5, will come from the inner and outer products of the FOIL.
anonymous
  • anonymous
okay im going to try it right now
anonymous
  • anonymous
(3x+1)(x-2) worked for me
DebbieG
  • DebbieG
Yes, that's perfect! Good job. So now we have:
DebbieG
  • DebbieG
\[\Large \frac{ 2-x}{ (3x+1)(x-2) }\] right?
DebbieG
  • DebbieG
Now, notice that the 2-x in the num'r is ALMOST the same as one of your terms in the den'r, right? If only it was x-2 instead of 2-x, they would cancel, right?
DebbieG
  • DebbieG
So here is a handy little algebraic "trick"..... always keep in mind that a - b = (-1)(b - a) Do you see why?
DebbieG
  • DebbieG
If I distribute the -1 through on the right, I get: (-b) + a And since I can switch the order of the terms in a SUM (not in a difference, but this is now a sum) (-b) + a = a + (-b) = a - b
DebbieG
  • DebbieG
Sooooooooo...... you can re-write your numerator 2 - x as (-1)(.....?.......) ?
anonymous
  • anonymous
so i multiply the top and bottom by -1 ?
anonymous
  • anonymous
-2+x
DebbieG
  • DebbieG
Ok, and then factor out a -1 That is, write it so that the -1 is out in front of the ( ) x-2 = -2+x = -1(2-x)
DebbieG
  • DebbieG
oh wait, I did that backwards, lol
DebbieG
  • DebbieG
2 - x = -x + 2 = -1(x - 2) You want to re-write your num'r of 2-x so that it is a product that has a factor of x-2. That will allow you to cancel it with the den'r factor of x-2
DebbieG
  • DebbieG
\[\Large \frac{ 2-x}{ (3x+1)(x-2) }=\frac{ -1\cancel{(x-2)}}{ (3x+1)\cancel{(x-2)} }=?\]
anonymous
  • anonymous
-1/3x+1
anonymous
  • anonymous
thank you
DebbieG
  • DebbieG
You're welcome. :)
anonymous
  • anonymous
the next step of this question is finding the x and y intercept. can you help me with that also
DebbieG
  • DebbieG
So it's an equation, \[\Large y=\frac{ 2-x}{ (3x+1)(x-2) }\] right?
anonymous
  • anonymous
yes
DebbieG
  • DebbieG
Which you've now simplified to \[\Large y=\frac{-1}{ (3x+1) }\] right?
DebbieG
  • DebbieG
So let's start with the y-intercept, because that's the easiest. :) The y-intercept is the point on the y-axis where the graph will intersect the axis. Every point on the y-axis has the same x-coordinate. What is that?
DebbieG
  • DebbieG
In other words, for every point that is on the y axis, the x-coordinate of that point is x=?
anonymous
  • anonymous
it equal to the same value of y? so if y is equal to 3 x is equal to 3 as well?
DebbieG
  • DebbieG
Noooo. That's not true on the y-axis. In fact, there is only ONE point on the y-axis where the x-coordinate is equal to the y-coordinate (that's at the origin).
DebbieG
  • DebbieG
Look at a coordinate plane. Look at a few points on the y-axis. Think about their coordinates. what is true of the x-coordinate for all of them?|dw:1377627029834:dw|
anonymous
  • anonymous
theyre all zero
DebbieG
  • DebbieG
RIGHT! If a point is on the y-axis, then it has x=0 So to find the y-intercept of this equation (or for any equation), you simply set x=0 and see what you get for y.
DebbieG
  • DebbieG
\(\Large y=\dfrac{-1}{ (3(0)+1) }=?\)
anonymous
  • anonymous
-1
DebbieG
  • DebbieG
GOOD! That's right!
DebbieG
  • DebbieG
Now the x-intercept is the same idea, but on the x-axis. So you want to find any values of x for which y=0, because if y=0, then that point is ON the x-axis, and so it's an x-intercept. Now that means setting y=0 and solving the equation to find the x that makes that true. But keep in mind, there isn't necessarily ANY x-intercepts. (IN fact, there isn't necessarily any y-intercepts either, although we see that here there is). So you need to know whether: \(\Large 0=\dfrac{-1}{ (3x+1) }\) has any solutions That is, is there ANY x value that will make that a true equality?
DebbieG
  • DebbieG
And the trick to remember with a function like this, called a RATIONAL FUNCTION (which really just means a big ol' fraction, at least for our purposes here that's good enough), is this: The ONLY WAY a fraction is ever =0 is if its NUMERATOR (the top) is = 0, right?
DebbieG
  • DebbieG
So for example, something like \(\Large \dfrac{2x}{ 5 }=0\) DOES have a solution, since if x=0 then we have \(\Large \dfrac{0}{ 5 }=0\) and that's true. But something like \(\Large \dfrac{2}{ 5x }=0\) can NEVER be true, since the only thing in the numerator is a "2" - nothing involving x - so no matter WHAT we plug in for x, that numerator is never, ever, going to =0.
anonymous
  • anonymous
its not equal to zero so there is no x intercept
DebbieG
  • DebbieG
So what are your thoughts about \[\Large 0=\dfrac{-1}{ (3x+1) }\]
DebbieG
  • DebbieG
Right!! Good job! :)
anonymous
  • anonymous
thank you so much you helped me out a lot
DebbieG
  • DebbieG
You're very welcome. Happy to help, and you did good work yourself. Stick with it, you are obviously getting it!! :)

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