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## What is a quartic function with only the two real zero given x=-4 and x=1?

Do you mean its only real zeros are -4 and 1, or its only zeros are -4 and 1?

If its only zeros are -4 and 1, the function could be

f(x) = a(x + 4)(x – 1)^3,

f(x) = a((x + 4)^2)(x – 1)^2,

or f(x) = (x – 1)(x + 4)^3,

where a ≠ 0.

If its only real zeros are -4 and 1, it can be any of the above as well as

f(x) = a(x + 4)(x – 1)(x – b – ci)(x – b + ci) where a,c ≠ 0.

What is a quartic function with only the two real zero given… – If its only real zeros are -4 and 1, it can be any of the above as well as. If a 10 pepperoni pizza from a pizza parlor costs $9.14 , what should a 12 pizza from the pizza parlor cost?This gives us y = a(x − 1)2. What is the value of "a"? But as in the previous case, we have an infinite number of parabolas passing through (1, 0). Here are some of them Two questions: Is there a way to find the formula for a Quartic equation? Can I use excel and choose polynomial and order 4?clearly all real. just multiply these two results together then remember the arbitrary and you have it, just remember ≠0 or you get 4 real roots instead of Where c>0. This is the case as the third factor will be quadratic and have no real roots, making x=5 and x=1 the only roots of the quartic function.

How to find the equation of a quadratic function from its graph – A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point. Degree: 3 Zeros: -2,2+2√2i Solution Point: f(−1) = −68 Could you please check my answers? Find an nth degree polynomial function with real coefficients satisfying the given conditions. 1. n=3; 3 and i…x = -4 and x = -1.Consider the quartic real polynomial function $P(x)=x^{4}+bx^{3}+cx^{2}+dx+e$. Because of the facts of $P(x_0) = 0$ and $P'(x_0) \neq 0$, so $x_0$ is one root of the function but not its minimum or maximum point.

What is a quartic function with only the two real zeros given… – x = -4 and x = -1. Question 26 optionsIt is given that a quartic function has only two real zeros. Put -1 and -4 for x. If the value of f(x)=0 at x=-4 and x=-1 then the function have two zeros -4 and -1.Find a quartic function with x =2 and x=9 as it's only real zeros. given function b)Find all the zeros of the function and state its graph…

**Find the polynomial function with integer coefficients that has given zeros** – .

**Linear, Quadratic, and Exponential Models** – In this video, we're going

to play a bit of a game.

I have three different data

sets here, and we need to figure out which of these data

sets come from either a linear, quadratic, or

exponential function. And just as a reminder, linear

function is of the form, y is equal to mx plus b. A quadratic is of the form, y

is equal to ax squared plus bx, plus c. And an exponential is of the

form, y is equal to some constant– let's call

it a– times some number to the x power. Now, the way to tell them

apart– and I'm going to actually show you. It'll probably be intuitive for

you for a linear function, and maybe even for

an exponential. The quadratic, the technique I'm

going to use might not be intuitive, and I'll actually

prove it to you in the next video, or at least show you an

example in the general case of why it makes sense. But the way to tell a linear

function, is that when you increment x by a constant

amount, y will always change by a constant amount. So when you always increase x by

1, y should always increase by a certain amount. So just using that as our first

guideline, let's see which of these represent

a linear function. So in all of these examples, as

we go from one data point to the next, we're increasing

x by 1. And it's very important that

you pay attention to that, because if you weren't, you

would have to adjust things a little bit. But in all of these, we're

always increasing x by 1. So let's look at our

change in y's. When we go from the first data

point to the second data point, our change in y is 60. When we go to the second

to the third, our change in y is 90. So that immediately tells us

we're not dealing with a linear function. When our change in x is 1, our

change in y here is 60. Then it's 90, and as you go here

it's like a 135, so this one is clearly not linear. Let's look at this

one over here. Our first change in y

is 2, then our next change in y is 0. Once again, not linear. As we increment x by a constant

amount, our y does not increase or decrease

at a constant amount. So once again, this

one is not linear. Now let's look over here. Once again, we're increasing

x by 1 on every data point. And when we to go from this

first data point to the second data point, y decreases by 3. Then we go from the second to

the third, once again, y decreases by 3. Third to the fourth,

y decreases by 3. 1 to negative 2, once again,

decreases by 3. And we just keep decreasing by

3, so whenever we change x by 1, when we increase x by 1,

y goes down by negative 3. So we actually have

a constant slope. We can say our change in y over

change in x is equal to when x goes up by 1,

y goes down by 3. This is a definition of slope. This is the m in this

equation right here. Our changes are constant, so

this one right here is linear. Now quadratic, there's an

interesting, we could call it a trick right now, because

I haven't really showed you why it works. But in a quadratic, your changes

in y are not going to be constant, as was the

case in the linear. But the change in the changes

of y will be constant. Let me write that down. It probably makes no sense to

you right now, but when you see an example it'll

be kind of fun. Change in change

in y constant. Let me show you what I mean. So over here, let's look at

this first one over here. So we went from 120 to 180, that

was an increase of 60. 180 to 270, that was

an increase of 90. Then we went from 270 to 405,

which is an increase of 135. Now, when I talk about the

change in the change of y, these are the change in y's,

what are the change in the change of y's? Well, we had a change of 60,

then we had a change of 90, so that was a change in the

change of y of 30, or a difference of the difference,

you could view them that way if you like. Then we had a change of 90, and

then we had a change of a 135, so the change in the

change, let's see, that is 45. So here, the change in the

change in y, or the difference of the difference

is not constant. We grew by 60, and then we

grew by 90, which was 30 more than 60. Then we grew by 135, which

is 45 more than 90. So this is not constant

over here. So this tells us this is not

a quadratic function. Now let's look at

this over here. The change in y– let's

write that down. This clearly was not linear. We increased by 2 here,

increased by 0 here, increased by negative 2 here, increased by

negative 4 here, increased by negative 6 there. The change in y's are definitely

not constant. But let's look at the change

in the change of y's. So we increased by 2, then we

increased by 0, so this was a negative 2 in the change in

the change of y, or the difference of the difference. Then when you go from 0 to

negative 2, we changed by 0, then we changed by negative 2,

that is a change in the change of negative 2. Then you go from negative 2 to

negative 4, the change in the change of negative 2. This one is starting

to look quadratic. We changed by negative 4 than we

changed by negative 6, once again, a change in the

change of negative 2. So this data right here fits our

requirement for quadratic, and I'm just kind of giving it

to you as a trick right now. In the next video I'll

actually show you why this works. I'll show you an example, just

so you can get the gut feeling of why it should work. So this one right here

is quadratic. So you might say, hey, Sal has

given us three examples, one of them is going to be linear,

one of them's going to be quadratic, one's going to be

exponential, this one's probably going to

be exponential. And you are probably right. But let me give you the way

that we can figure out if something is exponential. So when x increases by 1 in an

exponential, you should always have a constant multiplying

factor. So when you go from one term to

the next term, they should be a similar ratio from

one to the next. So instead of just subtracting–

in this case, 120 from 180, and 180 from

270– to test for an exponential function, what you

want to do is check whether the ratio of when you increment

x by a constant amount, whether the ratio

of the y's are the same. So when we increased x by 1, y

went from 120 to 180, so it increased by 18 over

12 is 1.5. So we went from this

term to this term. Let me clear all of the

stuff out here. So when we went from here to

here, this was times 1.5. And then when we go from 180 to

270, once again, that looks like we're going times

1.5, right? 1/2 of 180 is 90, add that to

270, so once again, we're increasing by 1.5. 270 to 405, 1/2 of 270 is 135,

that's how much we're increasing by. So once again, we're

increasing by 1.5. So every step of this way, we

are multiplying the previous term by 1.5 to get the next. And that actually tells us that

this right here is a 1.5. But our point here isn't to

actually figure out the formulas, the point here

is to classify them. So now we know definitively that

this right here, at least given the data we have, is

an exponential function. Anyway, hopefully you

found that fun. .

**Given complex zeros find the polynomial – Online Tutor** – .