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What must be a factor of the polynomial function f(x) graphed on the coordinate plane below? mc018-1.jpg x – 6 x – 3 x + 1 x + 6

source : peeranswer.com

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below? mc018-1.jpg x – 6 x – 3 x + 1 x + 6

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0
mathematics

PDF  Chapter 5 Section 1

PDF Chapter 5 Section 1 – Polynomial Functions and Their Graphs. In this section we begin the study of functions defined by Guidelines for Graphing Polynomial Functions: 1. Zeros: Factor the polynomial to find all its real Include test points to determine whether the graph of the polynomial lies above or below the…What must be a factor of the polynomial function f(x) graphed on the coordinate plane below? Which polynomial function could be represented by the graph below?A polynomial function is simply a function that is made of one or more mononomials. It will have zeros if the domain is extended to the complex field.In the coordinate plane, a quadratic without Since we cannot factor the given polynomial, let's find the two roots of the equation 4×2 + 3x – 6 = 0…

Writing Polynomial Functions from Complex Roots Flashcards | Quizlet – …r2, r3 then the factors of the polynomial are(x-r1), (x-r2), (x-r3)so since it passes through x=-6, x=-5, and x=0the factors… function f(x) graphed on the coordinate plane below? x – 6 x – 3 x + 1 x + 6. apologiabiology apologiabiology. If the polynomial passes through r1,r2, r3 then the factors of…The coordinate plane is a two-dimension surface formed by two number lines. Then determine what the coordinates of the fourth point, D, would be. All right, let's plot these on a graph, as they tell us to do. And notice, both of these have the exact same y. They're both at the same level below the x-axis.The above polynomial is the original polynomial. This demonstrates how the graph of a polynomial is related to its zeros and its factors. For our second example, we should look at the cubic polynomial we saw in a previous section. Viewing this polynomial on a coordinate plane yields this picture below.

Writing Polynomial Functions from Complex Roots Flashcards | Quizlet

What are the factors of a polynomial function with zeros at -2 and 7? – The following graphs of polynomials exemplify each of the behaviors outlined in the above table. Notice that an odd degree polynomial must have at least one real root since the function approaches – ∞ at one end and + ∞ at the other; a continuous function that switches from negative to positive…% Polynomial factors and graphs. 3 1. P(x) =2x −18x PART 6 Given the polynomial function P 2. Which of the following functions could represent the graph BELOW in the xy-plane, where y=P(x)? 23. The graph of the polynomial equation y=α(t) is shown BELOW. Which of the following must be…Polynomial Factors – SAT. Related Topics: More Lessons for Passport to Advanced Math More Lessons for SAT Math More Resources for SAT The function f is defined by f(x) = 2×3 + 3×2 + cx + 8, where c is a constant. In the xy-plane, the graph of f intersects the x-axis at the three points (-4, 0)…

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Quadratic inequalities (visual explanation) | Algebra II | Khan Academy – Добре дошли на презентацията
за квадратни неравенства.
Преди да стигнем до квадратните
неравенства, нека просто започнем с чертането на няколко функции и
ги обясним, и след това бавно ще се придвижим до
неравенствата. Нека кажем, че имах f от х е равно
на х на квадрат, плюс х минус 6. Добре, ако искахме да намерим,
къде тази функция пресича х-оста или нейните корени,
научихме при разлагането на квадрати, че
можем просто да сложим f от х да е равно на 0, нали? Тъй като f от х е равно на 0, когато
пресичате х-оста. Така че, ще кажете, че х на квадрат
плюс х минус 6 е равно на 0. Просто разложихте
този квадрат. х плюс 3 по х
минус 2 е равно на 0. И ще научите, че корените на
тази квадратна функция са х е равно на
минус 3 и х е равно на 2. Как ще изобразим това? Нека начертаем тази
квадратна функция. Това са много кривите ми линии. Корените са х е
равно на минус 3. Така че, това тук, х е при
минус 3, у е 0 – при първото определение един от корените е, където f от
х е равно на 0. Така че, у или f от х
оста тук е 0. Координатата е 0. А тази точка тук
е 2,0. Още веднъж, това е х-оста,
а това е f от х-оста. Знаем също, че у
пресечната е минус 6. Това не е върха,
това е у пресечната. И че графиката ще изглежда
по следния начин – не толкова неравна, колкото съм я начертал,
с което мисля, че получавате основна идея, ако някога сте
виждали добре оформена парабола. Изглежда, че х е минус
3 тук и х е 2 тук. Доста лесно. Намерихме корените,
намерихме как изглежда. Сега, какво ще стане, ако вместо
да искаме да знаем, къде f от х е равно на 0, което е при тези две
точки, какво ако искахме да знаем, къде f от х
е по-голямо от 0? Кои х стойности правят f
от х по-голямо от 0? Или друг начин да го кажем е,
какви стойности правят твърдението вярно? х на квадрат плюс х, минус 6 е
по-голямо от 0, нали, това е просто f от х. Добре, ако погледнем графиката,
кога f от х е по-голямо от 0? Това е оста f от х и
кога сме в положителната част? Ами f от х е по-голямо от
0 тук – нека начертая това с друг цвят – е по-голямо
от 0 тук, нали? Защото тя е над х-оста. А f от х е по-голямо от
0 тук. Така че, просто гледайки го,
кои х стойности правят това вярно? Ами, това е вярно когато х
е по-малко от минус 3, нали, или когато х е по-голямо
от 2. Защото когато х е по-голямо от 2,
f от х е по-голямо от 0, а когато х е по-малко от
минус 3, f от х е по-голямо от 0. Можем да кажем, че решението на
това квадратно неравенство и ние до голяма степен го решихме
визуално, е х е по-малко от минус 3 или х е
по-голямо от 2. И вие можете да го проверите. Може да пробвате с числото
минус 4 и би трябвало да получите f от х е по-голямо от 0. Можете да го изпробвате тук. Или може да пробвате числото 3
и да се уверите, че това работи. И можете просто да се уверите,
че, можете например, да опитате числото 0 и да се
уверите, че 0 не става, нали, защото 0 е
между двата корена. Всъщност се оказва, че
когато х е равно на 0, f от х е минус 6, което е
определено по-малко от 0. Така че, мисля че това ще ви даде
зрителна представа за това, какво означава това квадратно неравенство. Сега, с тази зрителна представа
на заден план, нека направим още няколко задачи и може би
няма да има нужда да минаваме през упражнението като го чертаем, но
може би ще го начертая, просто за да се уверя, че отговаря. Нека ви дам малко
по-сложна задача. Нека кажем, че имах минус х на квадрат
минус 3х плюс 28, нека кажа, че е по-голямо от 0. Добре, искам да се
отърва от този отрицателен знак пред х на квадрат. Просто не го харесвам като е там,
защото го прави да изглежда по-объркващо, за да го разложим. Ще умножа всичко
по минус 1. И двете страни. Получавам х на квадрат плюс 3х, минус
28 и когато умножите или разделите с отрицателно, за всяко
едно неравенство, трябва да обърнете знака. Така че, сега това ще бъде
по-малко от 0. И ако трябваше да разложим това,
получаваме х плюс 7, по х минус 4 е по-малко от 0. Ако това е равно на 0, щяхме да
знаем, че двата корена на тази функция – нека
определим функцията f от х – нека определим функцията като f
от х е равно на – можем да я определим като това или това,
защото те са едно и също нещо. Но за улеснение нека я определим
като х плюс 7, по х минус 4. Това е f от х, нали? Добре, преди да го разложим, знаем
че корените на това, корените са х е равно на
минус 7 и х е равно на 4. Сега, това, което искаме да знаем
е, каква х стойност прави това неравенство да е вярно? Ако това беше някакво
уравнение, щяхме да сме готови. Но ние искаме да знаем, кое
прави това неравенство вярно. Ще ви дам малка подсказка,
това винаги ще бъдат числата между двата корена
или извън двата корена. Така че това, което правя, независимо
дали го правя на някакъв тест или нещо такова, просто проверявам числата, които са
или между корените или извън двата корена. Нека изберем число, което е
между х равно на минус 7 и х равно на 4. Добре, нека опитаме х равно на 0. f от 0 е равно на – можем
да го направим тук – f от 0, е 0 плюс 7, по 0 минус 4
е просто 7 по минус 4, което е минус 28. Така че, f от 0 е минус 28. Сега, дали това – това е
функцията с която работим – това по-малко ли е от 0? Ами да, то е. Така че, всъщност се оказва, че
число, х стойност между двата корена отговаря. Всъщност аз веднага знам,
че отговорът тук са всички х-ове, които са
между двата корена. Така че, можем да кажем, че
решението на това е минус 7 е по-малко от х,
което е по-малко от 4. Защото – сега по другия начин. Може да пробвате число,
което е извън корените, или по-малко от минус 7, или
по-голямо от 4 и да го пробвате. Нека кажем, че опитате
с 5. Опитваме с х равно на 5. Тогава f от 5 ще бъде
12 по 1, нали, което е равно на 12. f от 5 е 12. Това по-малко ли е от 0? Не. Така че, това не става. Още веднъж, това ни дава
увереността, че сме получили правилния интервал. И ако искахме да разгледаме това
визуално, защото имаме този отговор, когато го направите
визуално, това всъщност има, мисля, голям смисъл, но може би съм предубеден.
Нека изтрия това. Ако го разгледате визуално,
то ще изглежда това. О, това е твърде дебело. Ако го начертаете визуално и това
е парабола, това е f от х, корените тук са минус 7,0
и 4,0, казваме че за всички х стойности между тези
две числа, f от х е по-малко от 0. И в това има смисъл, защото
кога f от х е по-малко от 0? Ами това е графиката
на f от х. Това е f от х. И кога f от х е по-малко от 0? Ето тук. Така че, какви х стойности ни дава това? х стойностите, които ни дават това
са ето тук. Надявам се, че не ви обърквам
прекалено много с тези нагледни графики. Вероятно казвате,
добре, от къде знам, че не включвам 0? Можете да я пробвате, но
ако – о, добре, от къде на къде не включвам корените? Ами при корените f
от х е равно на 0. Ако това беше това, ако това
беше по-малко от или равно на 0, тогава отговорът щеше да бъде
минус 7 е по-малко от или равно на х, е по-малко
или равно на 4. Надявам се, че така добивате представа. Вие просто трябва да
пробвате с число между корените и да пробвате с число
извън корените, и това ще ви каже кой интервал прави
неравенството да е вярно. Ще се видим в
следващата презентация. .

More Graphs of Polynomial Functions – Problem 1 – >> Sketch a graph of f of x equals negative x to
the fourth plus 4x squared by finding its zeros, picking test points in each interval,
and determining its end behavior.
First, we'll find the zeros by
setting the function equal to 0. Now, we can factor out the GCF first and the
GCF is x squared, but I'm actually going to pull out a negative x squared to
make things look a little nicer. That gives us leftovers x squared minus 4. And, by factoring out that negative, this
is now clearly a difference of squares, so we can factor that even further
into x plus 2 times x minus 2. And, then we can set each factor equal
to 0, so negative x squared equals 0, x plus 2 equals 0, or x minus 2 equals 0. And, this gives us solutions x equals 0, x
equals negative 2, or x equals positive 2. So, those zeros, which also give us intercepts,
are going to be 0, negative 2, and positive 2. Now, we also want to pick test points
in between these intervals to figure out if the graph's going to be above
and then below or below and then above or above and above or below and below. So, if we pick positive 1 and negative 1, that
will give us the information that we need. [ Writing ] So, if we evaluate our function at positive
1, it's going to be the opposite of 1 to the fourth plus 4 times 1 squared, it's going to be negative 1 plus
4, which gives us positive 3. And, then when we plug in negative 1 — [ Writing ] — we end up getting the exact same
thing, because when you take a negative to the fourth power or a negative
squared, it becomes positive, so we're also going to have 3 for our y value. So, we're going to have negative 1 comma 3 and
positive 1 comma 3, which means that we're going to be above the x axis on
both of these intervals. Now, we need to look at the end behavior to
see what happens as x approaches infinity. Will y approach positive infinity or negative
infinity and what will happen on the left? Now, for this, we have to just look at the
leading term, which is negative x to the fourth. Because it's an even degree, that means we're
either going to be going up in both directions or down in both directions, and because
the leading coefficient is negative, that means that we're going to be
going down in each of the directions. So, as we graph this, we know for each of these
intervals, we're going to be above the x axis. And, then on both of the outsides,
we're going to be going down. Y is going to approach negative infinity,
so it's going to be something like this. This is our graph, f of x equals the
opposite of x to the fourth plus 4x squared. And, again, there's only so
much accuracy that we have here. We have the intercepts being accurate and
whether we're above or below the x axis, however, we don't know that these
are actually the maximum points. It could go even higher first and then go down,
or it could be going higher here and then down. We don't really know that. Again, we need calculus for that, so right now,
we can only get some accuracy for our graphs. .

Ex: Determine Concavity and Points of Inflection – f(x)=x^2*e^(4x) – In this example, we're given the function
f of x equals x squared, times e raised to the power of four x, and asked to determine the open intervals where the function is
concave up or concave down, and also determine any
points of inflection.
Here, because we're
talking about concavity and points of inflection, we're working with the second derivative of the given function. Where any possible points
of inflection will occur, where the second derivative
is equal to zero or undefined, and over an interval with the
second derivative is positive, the function is concave up, and if the second derivative
is negative over an interval, the function is concave down. We want to find the second derivative, but before we find the second derivative, we must first find the first derivative. In looking at the function f of x, notice how we'll have to
apply the product rule in order to find the first derivative. Applying the product rule, f prime of x is equal to the first function
which we'll let be x squared, times the derivative
of the second function, which would be the derivative
of e to the four x, and then we have plus
the second function of e, raised to the power of four x, times the derivative
of the first function, which would be the
derivative of x squared, which means f prime of x is
equal to x squared, times… Now, to find the derivative
of e to the four x, we'll have to apply the chain rule. The derivative of e to the four x would be e to the four x times the derivative of four x, which is four. So, the derivative would
be four e to the four x. Now we have plus e to the four x, times the derivative of
x squared, that's two x. Now we'll simplify. F prime of x is equal to, here we'd have four x squared, e to the four x, plus
two xe to the four x. This is the first
derivative, but remember, our goal is to find the second derivative and determine where it's
undefined or equal to zero. In order to find the second derivative, notice I'll have to apply the product rule and find the derivative of the first term, and then the product rule again to find the derivative of the second term. Let's do this on the next slide. F double prime of x is
going to be equal to the derivative of the first term. We'd have the first
function, four x squared, times the derivative
of the second function, which is the derivative
of e to the four x, plus the second function
of e to the four x, times the derivative
of the first function, which will be four x squared. This gives us a derivative
of the first term, and now we need to find the
derivative of the second term, which would be two x times the derivative of e to the
four x, the second function, plus the second function
of e to the four x times the derivative of the first function, which is two x. F double prime of x is
equal to four x squared, times the derivative of e to the four x, which is four e to the four x, and we have plus e to the four x, times the derivative of four x squared, that'd be eight x, so times eight x, plus two x, times the
derivative of e to the four x, which is four e to the four x, and then plus e to the four x, times the derivative
of two x, which is two. Now we'll find these products, so we have f double prime of x equals, here we're going to have
16x squared e to the four x, plus, here we have eight xe to the four x, plus, here we have another
eight xe to the four x, and plus two e to the four x. Notice how we do have two like terms here, so we'll go ahead and combine those. Let's see, we have f
double prime of x equals 16x squared e to the four x, plus 16xe to the four x, plus two e to the four x. Now we want to determine where the second derivative is going to be undefined or equal to zero. Notice how it'll never be undefined, so we'll set this equal
to zero and solve for x. Again, the solutions would give us the locations of possible
points of inflection. Let's go ahead and do
this on the next slide. Now we'll set this
equal to zero and solve, and we'll start by factoring. Notice how the greatest common factor of the three terms would
be two e to the four x. If we factor out two e to the four x, we'd be left with eight x squared, and then plus eight x, and then plus one. Of course, we can distribute
if we want to check. This product would be equal to zero when two e to the four x equals zero, or when eight x squared, plus
eight x, plus one equals zero. Notice how two e to the four
x is never going to be zero, and therefore we'll only get solutions by solving this quadratic equation here, which unfortunately does not factor. Therefore, we'll have to
apply the quadratic formula, where a equals eight, b equals
eight, and c equals one. The quadratic formula is given,
here, below, for reference. So solving for x, we would
have x equals negative b, which would be negative eight, plus or minus the square
root of b squared, that would be eight squared, minus four times a, which is eight, times c, which is one, all over two times a, or two times eight. Simplifying, we'd have negative eight, plus or minus the square root of, this is going to be 64 minus 32 which is 32, divided by 16. Now we can simplify the square root of 32. Remember, the square root of 32 is equal to the square
root of 16 times two, which simplifies to four square root two, since the square root
of 16 is equal to four. Now we have x equals negative eight, plus or minus four square root two, divided by 16. To simplify this, let's break it up into two separate fractions. We'd have negative 8/16, plus or minus four
square root two over 16, which would be negative 1/2 plus or minus, and here we'd have square
root two divided by four. The two locations of
possible points of inflection would be x equals negative 1/2 minus square root two divided by four, and then the other x value would be x equals negative 1/2 plus square root two divided by four. Now we're going to divide the domain of the original function
which is all real numbers into three intervals using these x values. When we do this, though, it'll be helpful to have
decimal approximations, so negative 1/2 minus square
root two divided by four is approximately negative 0.8536, and negative 1/2 plus square
root two divided by four is approximately negative 0.1464. To save some time, I've
already set this up in a table. Notice how I included
both the exact values of x as well as the decimal approximations. Looking at the approximations, the first interval is from negative infinity to
approximately negative 0.8536. The next interval is from negative 0.8536 to negative 0.1464. The third interval is from negative 0.1464 to positive infinity. Now we'll pick test
values in each interval. For this first interval,
let's use negative one. For this interval here, let's
use, let's say, negative 0.5. For this interval let's use one. Now we're going to test this out on the second derivative in
each of these test values, and the sign of the second
derivative will tell us whether the function is concave up or concave down over the given interval. We'll do this using the graphing calculator to save some time. I've already entered the function
and the second derivative into y one and y two, which we see here. Now we'll use the table feature to check the sign of the second
derivative at these test values, which, remember, is in y two. Before we go to the table though, let's check our set by
pressing second, window. We want to make sure the
independent variable is on ask, and it already is, so
we'll press second, graph. I've already typed in the test values, and the way I did this was I just entered negative one, enter, negative 0.5, enter, and one, enter. Remember, the second
derivative is in y two, so at x equals negative one, the second derivative is positive, at negative 0.5 it's negative, and at positive one it's positive. So the signs are positive,
negative, positive, which means over this first interval the function is concave up, because the second derivative is positive. Here the second derivative is negative, so the function is concave down over this entire interval. And on the right, the second
derivative is positive, so, again, the function is concave up over the entire interval. Notice how this tells us we do have a point of inflection here at approximately negative 0.8536, because the function changes
concavity at that location. Also, it changes concavity at
approximately negative 0.1464. So to finish the problem, we'll list the points of inflection. We'll go and just use the decimal approximations for
the points of inflection. For the first point of inflection, the x-coordinate is
approximately negative 0.8536. We'll have to evaluate
the original function to determine the y-coordinate of this first point of inflection. The second point of
inflection has an x-coordinate of approximately negative 0.1464, and, again, we'll have
to evaluate the function at this x value to
determine the y-coordinate. Let's go ahead and do this
using the graphing calculator. Using the table now we want
the x value of negative 0.8536, enter, and remember the
original function is in y one, so the approximate y-coordinate, if we move the cursor over to y one, we can see more decimal
places down here below. To four decimal places
this would round to 0.0240, because the six indicates to round up, so approximately 0.0240. So 0.0240, that's the approximation of the y-coordinate of the
first point of inflection. And now for the second
point of inflection, we'll press down and then
right, back to the next x value. We'll enter negative 0.1464, enter, and, again, we'll go
ahead and move the cursor over to y one so we can see
more decimal places here. The y-coordinate would
be approximately 0.0119. Of course, if we entered
the exact x values rather than the decimal approximations we would have more accuracy, but for this problem, the decimal approximations work just fine. I hope you found this helpful. .