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[BudgieLover5000]

I HAVE A QUESTION!
TYPE OF QUESTION: Algebra 2
YOUR QUESTION:
How do you graph (x-2)(x+2)(x-1)?
I know how to get the x intercepts (I got -2, 1, and 2) but how do you get the y-intercepts (I think that's what it's called?)

to get the y-intercept, put 0 in for x, so y=(0-2)(0+2)(0+1)

Thanks!
I have a few questions though:
How come you substitute with a 0?
How come (x - 1) changes into (x + 1)?
And how do you get the other y-intercept? I thought there was supposed to be 2?

The way I was taught to do it was to substitute with the missing x, so if you put -2, 1, and 2 on a graph then -1 and .5 would be missing in between those points. So you would substitute like so:
Example of what I'm talking about: (-1 -2)(-1+2)(-1-1)
(.5 - 2)(.5 +2)(.5 - 1)

So that is why I'm really confused, sorry about that :C

[Kaidaft]

I HAVE A QUESTION!
TYPE OF QUESTION: Algebra 2
YOUR QUESTION:
How do you graph (x-2)(x+2)(x-1)?
I know how to get the x intercepts (I got -2, 1, and 2) but how do you get the y-intercepts (I think that's what it's called?)

to get the y-intercept, put 0 in for x, so y=(0-2)(0+2)(0+1)

Thanks!
I have a few questions though:
How come you substitute with a 0?
How come (x - 1) changes into (x + 1)?
And how do you get the other y-intercept? I thought there was supposed to be 2?

The way I was taught to do it was to substitute with the missing x, so if you put -2, 1, and 2 on a graph then -1 and .5 would be missing in between those points. So you would substitute like so:
Example of what I'm talking about: (-1 -2)(-1+2)(-1-1)
(.5 - 2)(.5 +2)(.5 - 1)

So that is why I'm really confused, sorry about that :C

ah the (x+1) was a typo
but you substitute it as 0 because all y-intercept values are (0,y) so, to find the y-intercept your x value would have to be 0, or it just turns into a coordinate. because, a y intercept lies on the y axis.


so, when putting in 0, (0-2)(0+2)(0-1), y=4, or (0,4)
there also wouldn't be more than 1 y intercept

[BudgieLover5000]

ah the (x+1) was a typo
but you substitute it as 0 because all y-intercept values are (0,y) so, to find the y-intercept your x value would have to be 0, or it just turns into a coordinate. because, a y intercept lies on the y axis.


so, when putting in 0, (0-2)(0+2)(0-1), y=-1, or (0,-1)
there also wouldn't be more than 1 y intercept

Ohh!!!! That makes a lot of sense. The problem was that I was confusing y-intercept for how to find hills and valleys, but I get it now. Thank you so much!!

[nev]

hh so i'm doing a unit test for math and im stuck on this one thing
everything else is easy but

here

if someone has 4/5 of a tank of fuel, and they use 1/10 of it a day, whats the answer?

thanks

[cuervos]

1/10 of 4/5 is 8/100, simplified as 2/25. So, every day 2/25 of fuel would be subtracted from the initial value. So, after one day, there would be 18/25 of fuel left.

[BudgieLover5000]

I HAVE A QUESTION!
TYPE OF QUESTION: Algebra 2
YOUR QUESTION: Sorry, this is my second time today, but my last one hopefully!!

Could I get a simple rundown on how to find the zeros a function and state the multiplicity of multiple zeroes?
You can use these as examples (I learn better when I see it done on a math problem, but it's not required)
1) x = (x + 3)^3

2) y = 3x^3 − 3x

Thank you for any help!

[nev]

Hey, I was wondering how you do fractions on a calculator? I normally use a scientific calculator.

[sparrow,,]

will be using for tomorrow whooooop

[CᴀɴᴅʏNᴜᴛᴍᴇɢ]

I HAVE A QUESTION!
TYPE OF QUESTION: Algebra 2
YOUR QUESTION: Sorry, this is my second time today, but my last one hopefully!!

Could I get a simple rundown on how to find the zeros a function and state the multiplicity of multiple zeroes?
You can use these as examples (I learn better when I see it done on a math problem, but it's not required)
1) x = (x + 3)^3

2) y = 3x^3 − 3x

Thank you for any help!

First, I'm going to assume that was a typo up there and you meant "y = (x+3)^3".

Let's go through them one at a time.
1. To find the zeros, or x-intercepts, you need to equate the entire equation to 0.
(Why? Because your entire equation is y, and y has to be zero if you want an intercept on the x-axis.)
This example's pretty straightforward.
y = (x+3)^3
0 = (x+3)^3
Now, solve for x by reversing each operation, starting from the outside.
cube root(0) = (x+3)
0 = (x+3)
-3 = x
Note that if you have any factored form, for example, a(x-2)(x-1)(x+3), where a is any random constant, your zeros will be the numbers in the brackets, but with the signs flipped. (Why flipped signs? Because isolating for x requires the sign to be flipped if the number's moved across the equals sign.)
What if you have the general form a(bx-c), where x is a variable and a, b, and c are all constants? Feel free to check this by isolating for x, but your zero will be c/b. (This is the same as the process in my example above, but if the coefficient of x is 1, it doesn't change the value, so it can be ignored.)

For the multiplicity, it's just how many times the factor repeats. If the equation is (x+3)^3, x=-3 repeats 3 times (as seen from the exponent). Therefore multiplicity = 3.

2. y = 3x^3 - 3x
Here you need to take the common element out of each term.
In 3x^3 and 3x, we have the common "3x", so move this out of the brackets by dividing each term by it.
y = 3x(x^2 - 1)
I don't know if you've learned this yet, but note that x^2 and 1 are both squares, i.e. y = 3x((x)^2 - (1)^2)
We can turn this into a difference of squares, by sort of splitting it into halves; y = 3x(x-1)(x+1).
(Compare this with the original by expanding the brackets and multiplying.)
Now we've got a fully simplified function.
Equate to 0 and solve for the brackets as in example 1. x = -1, x = 1.
However, there's also another factor that can make the entire equation equal 0; x = 0. (Because 3(0) = 0, and anything multiplied by 0 is 0, which negates the numbers in the brackets.)
Each factor repeats once in the equation, therefore the multiplicity of each factor is 1.

Hey, I was wondering how you do fractions on a calculator? I normally use a scientific calculator.

Some calculators have the ability to flip through various forms of the answer (fraction, rounded decimal, repeating decimal with bar notation, whatever). Look through your manual, yours might have that as a hidden option.

Alternatively, you could use fractions the whole way through your solution, which might require changing some denominators and extra mental math.

Alternatively #2, if your result is a good (non-repeating, finite) decimal, you could take the decimal, multiply by 100, then put it over 100. (That sounded vague, so - let's say you get 0.25 as your answer. 0.25 = 25/100. This is the kind of thing I mean.) Then, simplify, (25/25)/(100/25) = 1/4.

(If it's a repeating decimal, you memorize the patterns for the more common ones if you work with them enough,
1/7 = 0.142857...
2/7 = 0.285714...
3/7 = 0.428571...
Notice how it's the same 6 numbers in the same order, but starting from different places every time in an increasing pattern.
Anything in the form 0.xxxxxxxx is x/9. (1/3 and 2/3 are simplifications of this.)
Anything in the form 0.xoxoxoxo is xo/99.
etc.)

[protect]

is anyone really good at geometry? I'm having trouble and really need help for my test tomorrow
if u can help please pm me