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Class 10 (Old)
Course: Class 10 (Old) > Unit 2
Lesson 1: Geometrical meaning of the zeroes of a polynomialPolynomial factors and graphs — Harder example
Watch Sal work through a harder Polynomial factors and graphs problem.
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find a harder example. This is way too easy when all you have to do is count the x intercepts and how many times the graph touches the xaxis(80 votes)
the exercises are then going to be much more difficult(11 votes)
what is the difference between distinct zero and double zero?Why would the graph bounce back?What is the official definition of double zero?Thanks.(15 votes)
A double zero results from a function having a repeated root, for example: roots derived from factors of the form (x-a)^2. We already know that roots occur where the graph touches/cuts the x axis, so if a factor is ofsome squared formthen the corresponding y values of the function would bepositive.At the point of the root, the graphdoesn't cross the x axis(because a quadratic function governs that portion of the graph) but instead bounces back from the x axis. Get it?(17 votes)
- Can’t i just apply the rule that if x power’s even then both negative and positive curves goes in the same direction and if it is negatives then doesn’t(16 votes)
- Just look at the exponents if the exponents are even the graph lines go to the same direction
Since the graph has both arrows down the exponent SHOULD be EVEN(15 votes) - What's the difference between distinct and double zeros?(11 votes)
- When the graph bounces like on the last choice, both those zeros are actually double zeros. When you solve for x values when y=0, you will have two identical answers for every time the graph bounces back.(8 votes)
- one can also use the odd and even aspect right? Like X raised to 1 which is odd hence shud face diff directions and X raised to 2 which is even facing same direction downwards.(10 votes)
Can you find a harder question like from a practice test or something? The hard questions aren't supposed to be this easy.(9 votes)
I don't get why A does not work. Something about end behavior.(3 votes)
You can explain why A doesn't work 2 ways. You could talk about end behavior: You should know that the end behaviour of polynomials is determined by if their degree is an odd or even power: if odd, then the ends go in opposite directions and if even, the ends goin the same direction. Here, our ends are going in the same direction so we need to pick something with an even degree. Adding up all the x's you see, for A) we get a degree of 3 and for B) we get 4. This allows us to rule out A).
Another way you can approach this is by thinking about multiplicity of roots, or how often the root is repeated (or when the polynomial is shown in root form like here, what the exponent on each root is). Roots with multiplicity 1 have the graph passing through them, roots with even multiplicity have the graph touch the root and "bounce" backwards, and roots with odd multiplicity greater than 1 have the graph do a little "squiggle", where it becomes horizontal at the root but then eventually crosses it. Here, at x=0 the graph goes down to 0 but then bounces back, so we need an even multiplicity for that root, which A) does not have and B) does have.(12 votes)
- can't understand the difference between choice a) & b)(5 votes)
A simple way of looking at this would be to remember that any number squared is positive.
So, in option B, the first x value (before negation; x^2), whether initially positive or negative, will always be positive. That number is then negated (-x^2), and so will always be a negative number. (And would show a downward trend, which is what the graph displays).
However, in choice A, a negative x value would immediately become a positive number, as any number negated twice is positive. (And would show an upward trend, which is not what the graph displays.)
Hope that helps, and feel free to ask if anything is unclear!(7 votes)
- He’s not going to answer to your questions.... don’t ask him(7 votes)
Video transcript
- [Instructor] We're asked
which of the following could be the equation of the graph above. So pause this video and see
if you can work through this before we do it together. All right, so first of all, let's just think about where the graph intersects the X axis. We can see it does that
at X equals negative four. It does that at X equals zero. And it does that at X
equals positive three. So we would expect to see an expression that's equal to zero
at these three places. So first of all, X equals zero. You could see if X equals zero. All of these will become zero, 'cause zero times anything else is zero. So they all meet that one. See, X equals negative four. In order for it to become zero
at X equals negative four, we'd want to see an X plus four when we factor things out. We see an X plus four here, we see an X plus four here. These second two, they have X minus four. So we don't like those. So we can immediately rule those out. And then for X equals three to make the entire
expression equal to zero, you'd need to have an X minus three. And we can see we have an X minus three and an X minus three. So now we have to decide
what makes more sense to have a negative X out
here or a negative X squared. Well, let's think about what would happen in either case as X increases. As X increases, as X
becomes larger and larger positive values, you take
the negative of that, and it makes sense in this A choice that the value would get
more and more negative. That's also the case in choice B. As X increases, you square it. You'll get even a more positive value. And then you take the negative of that, it should go down. So both choice A and B is
consistent with this behavior of as X becomes more positive, Y is becoming more and more negative. Now let's look on the other hand. What about when X becomes
more and more negative? Well, when X becomes more and
more negative in choice A, when you take the negative of a negative you would expect this part to
become more and more positive. So you would expect for choice A, that actually the Y values would increase as X becomes more and more negative, which clearly isn't happening. So we can rule that one out, as well. But let's just verify that B works. So when X becomes more and more negative, you take the square of that. So you're going to get a positive value, but then you take the negative of that. So it's going to become,
so Y will keep decreasing and becoming more and more negative, which is exactly what
we see in this graph. So we're liking choice B.