Find the zeros of the function and state their multiplicities
2 answers:
I'll explain.
First of all, when people say 'find the zeros' they basically mean 'find the solutions.' The reason for calling them 'zeroes' is ACTUALLY important. It tells you WHAT a solution is.
Many people, even people who are all the way up in pre-calculus don't know this...
The solutions of a function are the points at which it crosses the x-axis.
With that being said, let's find the zeros of this function:
y=x^4-8x^2+16
First of all, we must factor it.
y=x^4-8x^2+16
Becomes...
y=(x+2)(x-2)(x+2)(x-2)
We can find the solutions by setting these equal to 0!
x+2=0, x=-2
x-2=0, x=2
x+2=0, x=-2
x-2=0, x=2
Now, those are your answers. Continue if you want to learn something interesting.
Look at the above picture of the graph. I want you to look at where the graph touches the x-axis. These are the solutions.
Now I will point out something interesting.
There are 3 ways in which a polynomial function touches the x-axis. I like to call them the...
Cross through: this is when the graph goes strait through the x-axis like a line.
Bounce: this is when the graph touches the x-axis at one point and then bounces back.
Wiggle through: this is when the graph makes a little 'wiggle' as it crosses the x-axis.
WHY am I pointing this out?
Well, the way in which the graph crosses the x-axis will tell you how many of the same factor the equation has.
Cross through: This indicates that the graph only has one factor that equals the value of the point it crosses through
Example: y=(x-2)(x-2)(x+1) this means at -1, the graph crosses strait through the x-axis.
Bounce: like in our example, a bounce indicates that there are two factors that equal the value of the point it touches.
Example: y=(x+2)(x-2)(x+2)(x-2) this means that at -2 and 2, the graph bounces off the x-axis
Wiggle: this indicates that 3 factors that equal the value of the point it crosses through
Example: y=(x+1)(x+1)(x+1)(x-2) the graph wiggles through the x-axis at -1
I hope this helps! Ty for reading:)
First of all, when people say 'find the zeros' they basically mean 'find the solutions.' The reason for calling them 'zeroes' is ACTUALLY important. It tells you WHAT a solution is.
Many people, even people who are all the way up in pre-calculus don't know this...
The solutions of a function are the points at which it crosses the x-axis.
With that being said, let's find the zeros of this function:
y=x^4-8x^2+16
First of all, we must factor it.
y=x^4-8x^2+16
Becomes...
y=(x+2)(x-2)(x+2)(x-2)
We can find the solutions by setting these equal to 0!
x+2=0, x=-2
x-2=0, x=2
x+2=0, x=-2
x-2=0, x=2
Now, those are your answers. Continue if you want to learn something interesting.
Look at the above picture of the graph. I want you to look at where the graph touches the x-axis. These are the solutions.
Now I will point out something interesting.
There are 3 ways in which a polynomial function touches the x-axis. I like to call them the...
Cross through: this is when the graph goes strait through the x-axis like a line.
Bounce: this is when the graph touches the x-axis at one point and then bounces back.
Wiggle through: this is when the graph makes a little 'wiggle' as it crosses the x-axis.
WHY am I pointing this out?
Well, the way in which the graph crosses the x-axis will tell you how many of the same factor the equation has.
Cross through: This indicates that the graph only has one factor that equals the value of the point it crosses through
Example: y=(x-2)(x-2)(x+1) this means at -1, the graph crosses strait through the x-axis.
Bounce: like in our example, a bounce indicates that there are two factors that equal the value of the point it touches.
Example: y=(x+2)(x-2)(x+2)(x-2) this means that at -2 and 2, the graph bounces off the x-axis
Wiggle: this indicates that 3 factors that equal the value of the point it crosses through
Example: y=(x+1)(x+1)(x+1)(x-2) the graph wiggles through the x-axis at -1
I hope this helps! Ty for reading:)
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9514 1404 393
Answer:
53/99
Step-by-step explanation:
When an n-digit repeat begins at the decimal point, the repeating digits can be placed over an equal number of 9s to make a fraction. Sometimes, that fraction can be reduced to lower terms.
__
If the repeat does not start at the decimal point, the conversion can be done as follows.
For a number x with a decimal fraction that has n digits repeating, multiply by 10^n, subtract x and then divide by (10^n -1).
The fraction will be ...
(x·10^n -x)/(10^n -1) . . . where x has n repeating digits in the decimal fraction
Here, that looks like ...
No need to fear, thehotdogman93 is here!
The first step is to get rid of those very large numbers. It's going to be very difficult to factor unless we can bring those high numbers down. So lets see if we can factor each term.
So after dividing 49 with every single digit. The only number that divides evenly is 7 and one, and 16 isnt divisible evenly by 7 so that didn't work. Looks like we're gonna have to work with these big numbers.
There is something interesting though about these numbers. 16 and 49 are both perfect squares. 16 is the same as 4^2 and 49 is the same as 7^2. So we can factor the whole trinomial as:
If we were to expand this out as:
and multiply it back into the original form. It would match with the expression we started with. The 4's would multiply back into 16x^2 and the 7's would multiply back into 49.
Additionally 4 * -7 is -28, so you can combine two -28x's into the -56x term in the original trinomial.
Thus, the answer is yes you can, and the answer is: