Why does the graph only touch the x axis at a zero of even multiplicity?

I've been thinking about this for a while and here's what I've come up with. Let's say, for example, that f(x) = ( x - 4 ) ( x - 1 )^2. ( x - 4 ) is a root of odd multiplicity. Notice that when x < 4, ( x - 4 ) is negative, but when x > 4, ( x - 4 ) is positive. So, depending on the value of x, the sign of ( x - 4 ) changes, which in turn changes the sign of f(x). But also notice that for roots of even multiplicity [ ( x - 1 ) in this example], it doesn't matter what value of x is chosen. Once raised to their EVEN power, they will always be positive, so will not be able to change the sign of f(x). So, if f(x) is negative as it approaches a zero of EVEN multiplicity, then f(x) will remain negative after it passes that zero (and likewise if f(x) was positive, it would remain positive). In other words, it would just touch the x-axis and then have to "bounce" away in the same (positive or negative) direction. But if f(x) is negative (or positive) as it approaches a zero of ODD multiplicity, then f(x) will change sign --- in other words, the graph will cross through the x-axis instead of bouncing back. I hope this has been helpful and hasn't ended up confusing you!

in the answer of the challenge question 8 how can there be 2 real roots . in total there are 3 roots as we see in the equation . but in the answer there are 2 real roots which will tell that there is only 1 imaginary root which does not exists. please help me . thanks in advance!!

There is no imaginary root. Sometimes, roots turn out to be the same (see discussion above on "Zeroes & Multiplicity"). That is what is happening in this equation. So, the equation degrades to having only 2 roots. If you factor the polynomial, you get factors of: -X (X - 2) (X - 2). You can see, 2 of the factors are identical. If you use these to solve for f(x) = 0, they create only 2 points: (0,0) and (2,0) because we have 2 identical factors that both create X=2. Hope this helps.

For problem Check Your Understanding 6), if its "6", then why is it odd, not even?

The question asks about the multiplicity of the root, not whether the root itself is odd or even. At a root of odd multiplicity, the graph will cross through the X-axis. At a root of even multiplicity, the graph will bounce off the X-axis and not go through it.

In challenge problem 8, I don't know understand how we get the general shape of the graph, as in how do we know when it continues in the positive or negative direction. So for example, from left to right, how do we know that the graph is going to be generally decreasing?

You don't have to know this to solve the problem. You can find the correct answer just by thinking about the zeros, and how the graph behaves around them (does it touch the x-axis or cross it). You can click on "I need help!" to see the solution. If you want to know how to determine the direction of the graph, check out the next tutorial: https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-end-behavior/a/end-behavior-of-polynomials

Why is Zeros of polynomials & their graphs important in the real world, when am i ever going to use this?

It depends on the job that you want to have when you are older. School is meant to prepare students for any career path, including those that have to do with math. You might think now that you don't want a career with math, but you never know if you might decide to change your aspirations. When my mother was a child she hated math and thought it had no use, though later in life she actually went into a career that required her to have taken high math classes. You might use it later on! I'm grateful enough that I even have the opportunity to have such a nice education compared to developing countries where most citizens never make it to college.

Using multiplity how can you find number of real zeros on a graph

So first you need the degree of the polynomial, or in other words the highest power a variable has. So if the leading term has an x^4 that means at most there can be 4 0s. There can be less as well, which is what multiplicity helps us determine. If a term has multiplicity more than one, it "takes away" for lack of a better term, one or more of the 0s. So for instance (x-1)(x+1)(x-2)(x+2) will have four zeros and each binomial term has a multiplicity of 1 Now, if you make one of them have a multiplicity of 2 that takes away one of the zeroes. so (x-1)(x-1)(x+2)(x-2), here there are two (x-1) terms so it has multiplicity 2, this means there is one less zero. So now there are only three zeroes at 1, 2 and -2. ALSO if a term has an even multiplicity it means it touches the x axis rather than crosses it. Let me know if that didn't help.

What if there is a problem like (x-1)^3 (x+2)^2 will the multiplicity be the addition of 3 and 2 or the highest exponent will be the multiplicity?

A polynomial doesn't have a multiplicity, only its roots do. The roots of your polynomial are 1 and -2. 1 has multiplicity 3, and -2 has multiplicity 2.

why the power of a polynomial can not be negative or in fraction?

If you found the zeros for a factor of a polynomial function that contains a factor to a negative exponent, you’d find an asymptote for that factor with the negative power. For example: f(x)=(x+3)^2+(x-5)(x-3)^-1 That (x-3)^-1 is equal to 1/(x-3), so it would be undefined at x=3 because that would make the denominator equal to 0, so a vertical asymptote is created at x=3. It doesn’t necessarily mean you can’t have negative powers in polynomial fractions, but it does mean that for negative exponents you will find an asymptote where the zero would be in the case of a positive exponent.

Why's it called a 'linear' factor?

Linear equations are degree 1 (the exponent on the variable = 1). This same terminology is being used for the factor. It is a linear factor because it is degree = 1. Hope this helps.

how to find weather the graph is (odd or even)

A function is even when it's graph is symmetric about the y-axis. A function is odd when rotating it 180º about the origin leaves it unchanged.

Main content

Algebra 2

Course: Algebra 2 > Unit 5

Lesson 2: Positive and negative intervals of polynomials

Zeros of polynomials & their graphs

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Learn about the relationship between the zeros, roots, and x-intercepts of polynomials. Learn about zeros multiplicities.

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Rutwik Pasani
Posted 7 years ago. Direct link to Rutwik Pasani's post “Why does the graph only t...”
Why does the graph only touch the x axis at a zero of even multiplicity?
Button navigates to signup pageComment on Rutwik Pasani's post “Why does the graph only t...”
(65 votes)
Answer
- Judith Gibson
  Posted 7 years ago. Direct link to Judith Gibson's post “I've been thinking about ...”
  I've been thinking about this for a while and here's what I've come up with.
  Let's say, for example, that f(x) = ( x - 4 ) ( x - 1 )^2.
  ( x - 4 ) is a root of odd multiplicity.
  Notice that when x < 4, ( x - 4 ) is negative,
  but when x > 4, ( x - 4 ) is positive.
  So, depending on the value of x, the sign of ( x - 4 ) changes, which in turn changes the sign of f(x).
  But also notice that for roots of even multiplicity [ ( x - 1 ) in this example], it doesn't matter what value of x is chosen. Once raised to their EVEN power, they will always be positive, so will not be able to change the sign of f(x).
  So, if f(x) is negative as it approaches a zero of EVEN multiplicity, then f(x) will remain negative after it passes that zero (and likewise if f(x) was positive, it would remain positive). In other words, it would just touch the x-axis and then have to "bounce" away in the same (positive or negative) direction.
  But if f(x) is negative (or positive) as it approaches a zero of ODD multiplicity, then f(x) will change sign --- in other words, the graph will cross through the x-axis instead of bouncing back.
  I hope this has been helpful and hasn't ended up confusing you!
  Comment on Judith Gibson's post “I've been thinking about ...”
  (232 votes)
Harsh Agrawal
Posted 7 years ago. Direct link to Harsh Agrawal's post “in the answer of the chal...”
in the answer of the challenge question 8 how can there be 2 real roots . in total there are 3 roots as we see in the equation . but in the answer there are 2 real roots which will tell that there is only 1 imaginary root which does not exists. please help me . thanks in advance!!
Button navigates to signup pageComment on Harsh Agrawal's post “in the answer of the chal...”
(16 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “There is no imaginary roo...”
  There is no imaginary root. Sometimes, roots turn out to be the same (see discussion above on "Zeroes & Multiplicity"). That is what is happening in this equation. So, the equation degrades to having only 2 roots.
  If you factor the polynomial, you get factors of: -X (X - 2) (X - 2). You can see, 2 of the factors are identical.
  If you use these to solve for f(x) = 0, they create only 2 points: (0,0) and (2,0) because we have 2 identical factors that both create X=2.
  Hope this helps.
  Button navigates to signup page
  (36 votes)
Timothy (Tikki) Cui
Posted 6 years ago. Direct link to Timothy (Tikki) Cui's post “For problem Check Your Un...”
For problem Check Your Understanding 6), if its "6", then why is it odd, not even?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Judith Gibson
  Posted 6 years ago. Direct link to Judith Gibson's post “The question asks about t...”
  The question asks about the multiplicity of the root, not whether the root itself is odd or even.
  At a root of odd multiplicity, the graph will cross through the X-axis.
  At a root of even multiplicity, the graph will bounce off the X-axis and not go through it.
  Comment on Judith Gibson's post “The question asks about t...”
  (13 votes)
Kevin
Posted 6 years ago. Direct link to Kevin's post “Why is Zeros of polynomia...”
Why is Zeros of polynomials & their graphs important in the real world, when am i ever going to use this?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- QUINN767
  Posted 4 years ago. Direct link to QUINN767's post “It depends on the job tha...”
  It depends on the job that you want to have when you are older. School is meant to prepare students for any career path, including those that have to do with math. You might think now that you don't want a career with math, but you never know if you might decide to change your aspirations. When my mother was a child she hated math and thought it had no use, though later in life she actually went into a career that required her to have taken high math classes. You might use it later on! I'm grateful enough that I even have the opportunity to have such a nice education compared to developing countries where most citizens never make it to college.
  Button navigates to signup page
  (14 votes)
Michael Gomez
Posted 7 years ago. Direct link to Michael Gomez's post “In challenge problem 8, I...”
In challenge problem 8, I don't know understand how we get the general shape of the graph, as in how do we know when it continues in the positive or negative direction. So for example, from left to right, how do we know that the graph is going to be generally decreasing?
Button navigates to signup pageComment on Michael Gomez's post “In challenge problem 8, I...”
(7 votes)
Answer
- Tomer Gal
  Posted 7 years ago. Direct link to Tomer Gal's post “You don't have to know th...”
  You don't have to know this to solve the problem. You can find the correct answer just by thinking about the zeros, and how the graph behaves around them (does it touch the x-axis or cross it). You can click on "I need help!" to see the solution.
  
  If you want to know how to determine the direction of the graph, check out the next tutorial:
  
  https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-end-behavior/a/end-behavior-of-polynomials
  Comment on Tomer Gal's post “You don't have to know th...”
  (8 votes)
shub112
Posted 4 years ago. Direct link to shub112's post “Using multiplity how can ...”
Using multiplity how can you find number of real zeros on a graph
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “So first you need the deg...”
  So first you need the degree of the polynomial, or in other words the highest power a variable has. So if the leading term has an x^4 that means at most there can be 4 0s. There can be less as well, which is what multiplicity helps us determine. If a term has multiplicity more than one, it "takes away" for lack of a better term, one or more of the 0s.
  
  So for instance (x-1)(x+1)(x-2)(x+2) will have four zeros and each binomial term has a multiplicity of 1 Now, if you make one of them have a multiplicity of 2 that takes away one of the zeroes. so (x-1)(x-1)(x+2)(x-2), here there are two (x-1) terms so it has multiplicity 2, this means there is one less zero. So now there are only three zeroes at 1, 2 and -2. ALSO if a term has an even multiplicity it means it touches the x axis rather than crosses it.
  
  Let me know if that didn't help.
  Button navigates to signup page
  (5 votes)
Anthony
Posted 4 years ago. Direct link to Anthony's post “What if there is a proble...”
What if there is a problem like (x-1)^3 (x+2)^2 will the multiplicity be the addition of 3 and 2 or the highest exponent will be the multiplicity?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- kubleeka
  Posted 4 years ago. Direct link to kubleeka's post “A polynomial doesn't have...”
  A polynomial doesn't have a multiplicity, only its roots do. The roots of your polynomial are 1 and -2. 1 has multiplicity 3, and -2 has multiplicity 2.
  Comment on kubleeka's post “A polynomial doesn't have...”
  (4 votes)
Bodhi
Posted 6 years ago. Direct link to Bodhi's post “Why's it called a 'linear...”
Why's it called a 'linear' factor?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Linear equations are degr...”
  Linear equations are degree 1 (the exponent on the variable = 1).
  This same terminology is being used for the factor. It is a linear factor because it is degree = 1.
  Hope this helps.
  Button navigates to signup page
  (5 votes)
kslimba1972
Posted 5 years ago. Direct link to kslimba1972's post “why the power of a polyno...”
why the power of a polynomial can not be negative or in fraction?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Elammen
  Posted 4 years ago. Direct link to Elammen's post “If you found the zeros fo...”
  If you found the zeros for a factor of a polynomial function that contains a factor to a negative exponent, you’d find an asymptote for that factor with the negative power. For example: f(x)=(x+3)^2+(x-5)(x-3)^-1
  
  That (x-3)^-1 is equal to 1/(x-3), so it would be undefined at x=3 because that would make the denominator equal to 0, so a vertical asymptote is created at x=3.
  
  It doesn’t necessarily mean you can’t have negative powers in polynomial fractions, but it does mean that for negative exponents you will find an asymptote where the zero would be in the case of a positive exponent.
  Button navigates to signup page
  (3 votes)
devarakonda balraj
Posted 6 years ago. Direct link to devarakonda balraj's post “how to find weather the g...”
how to find weather the graph is (odd or even)
Button navigates to signup pageComment on devarakonda balraj's post “how to find weather the g...”
(2 votes)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “A function is even when i...”
  A function is even when it's graph is symmetric about the y-axis.
  A function is odd when rotating it 180º about the origin leaves it unchanged.
  Button navigates to signup page
  (5 votes)

Algebra 2

Course: Algebra 2 > Unit 5

Zeros of polynomials & their graphs

What you will learn in this lesson

Fundamental connections for polynomial functions

Check your understanding

Zeros and multiplicity

Check your understanding

The graphical connection

Check your understanding

Challenge problem

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