math homework helppppppp

- anonymous

math homework helppppppp

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

|dw:1377624206768:dw|

- DebbieG

omg no, that's so wrong. You CANNOT cancel like that!!

- DebbieG

@MorganMorganMorgan , you need to factor the denominator first. Do you know how to do that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

thats what ive been having a hard time doing

- 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

yes

- 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

yes

- 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

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

okay im going to try it right now

- anonymous

(3x+1)(x-2) worked for me

- DebbieG

Yes, that's perfect! Good job. So now we have:

- DebbieG

\[\Large \frac{ 2-x}{ (3x+1)(x-2) }\]
right?

- 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

So here is a handy little algebraic "trick"..... always keep in mind that
a - b = (-1)(b - a)
Do you see why?

- 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

Sooooooooo...... you can re-write your numerator
2 - x
as (-1)(.....?.......) ?

- anonymous

so i multiply the top and bottom by -1 ?

- anonymous

-2+x

- 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

oh wait, I did that backwards, lol

- 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

\[\Large \frac{ 2-x}{ (3x+1)(x-2) }=\frac{ -1\cancel{(x-2)}}{ (3x+1)\cancel{(x-2)} }=?\]

- anonymous

-1/3x+1

- anonymous

thank you

- DebbieG

You're welcome. :)

- anonymous

the next step of this question is finding the x and y intercept. can you help me with that also

- DebbieG

So it's an equation, \[\Large y=\frac{ 2-x}{ (3x+1)(x-2) }\]
right?

- anonymous

yes

- DebbieG

Which you've now simplified to
\[\Large y=\frac{-1}{ (3x+1) }\]
right?

- 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

In other words, for every point that is on the y axis, the x-coordinate of that point is x=?

- anonymous

it equal to the same value of y? so if y is equal to 3 x is equal to 3 as well?

- 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

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

theyre all zero

- 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

\(\Large y=\dfrac{-1}{ (3(0)+1) }=?\)

- anonymous

-1

- DebbieG

GOOD! That's right!

- 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

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

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

its not equal to zero so there is no x intercept

- DebbieG

So what are your thoughts about \[\Large 0=\dfrac{-1}{ (3x+1) }\]

- DebbieG

Right!! Good job! :)

- anonymous

thank you so much you helped me out a lot

- DebbieG

You're very welcome. Happy to help, and you did good work yourself. Stick with it, you are obviously getting it!! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.