(a) determine the polynomial function whose graph passes through the points, and (b) sketch the

Name: Problem 2, Chapter 1: Elementary Linear Algebra
Uploaded: 2022-02-06T18:28:57-08:00
Duration: 6 min 11 s
Channel: Vaibhav Jain
Description: (a) determine the polynomial function whose graph passes through the points, and (b) sketch the graph of the polynomial function, showing the points. (0,0),(2,-2),(4,0)

(a) determine the polynomial function whose graph passes...

Key Concepts

Graphical Representation of Polynomial Functions

Sketching the graph of a polynomial is an essential aspect of understanding its behavior. This includes identifying intercepts, turning points, and the general end-behavior of the function. Graphical analysis helps in corroborating the algebraic solution and provides visual insights into the nature of the polynomial function.

Polynomial Interpolation

This concept involves finding a polynomial that exactly passes through a given set of data points. It is a common method in numerical analysis where the function is constructed to match specific values at specified input points. It underscores the idea that a unique polynomial of degree n-1 can be determined when given n points that do not all lie on a polynomial of lower degree.

Determining the Degree of the Polynomial

Determining the minimum degree necessary for the polynomial is a key step in interpolation. Typically, if there are n points, a polynomial of degree at most n-1 will be sufficient to pass through all points, provided that the points have distinct x-values and do not lie on a lower degree polynomial. This concept is core to understanding the constraints of polynomial fitting.

Setting Up and Solving a System of Equations

Once the general form of the polynomial is assumed, substituting the coordinates of the known points allows the construction of a system of linear equations. Solving these equations yields the coefficients of the polynomial. This process is a practical application of linear algebra in constructing functions based on discrete data.

Factoring and Root Analysis

When some of the given points correspond to roots (zeros) of the polynomial, the polynomial can often be expressed in a factored form. This makes it easier to study the behavior of the function and can simplify the overall process of determining the correct polynomial. Recognizing when a point is a zero of the function is important for efficient solution techniques.

Transcript

0:00 Hello students.

00:01 Today we are given a question like function, polynomial function passes through the points.

00:05 0 .0.

00:07 We will write it 0 .0 and 2 comma minus 2 and a third point is 4 .0.

00:14 So we are given here the three points.

00:16 So we will consider it like a polynomial function in the form of degree 2.

00:20 It will be like y is equals to a0 plus a1 x plus a2 x squared.

00:28 Okay now what we will do is we will do like we will put the value of x called value of x from the points and compare it with the y values so we will come out with the point 0 .0 we will put it here it will be 0 it will also be 0 so basically it will be 0 so basically it will be 0 is equal to a 0 we will get the value of a 0 which will be 0 now we will put the value of x which is 2 okay and the y value will be minus 2 so we will write like a minus 2 is equals to a 0 plus a 1 times 2 so it will be 2 a1 plus a 2 a 2 times x square what is x it is 2 so basically 2 square which will be 4 so 4 multiplied by a 2 it will be 4 times of a 2 okay now we will put the value of a 0 is a 0 is a 0 is nothing but 0 so we will not write here we will just simply write like minus 2 is equal to twice of a 1 plus 4a2 now we will divide it by 2 we will get like a 1 plus 2 a 2 a 2 is equals to minus 1 we will make it consider it equation number 1 equation number 1 now what we will do like we will just put that value 4 comma 0 in the given particular equation what we will get is like a 0 is nothing but 0 so we don't need to write it and a 1 times of 4 it will be 4 a 1 plus a 2 times of 4 square 4 is nothing but 16 16 times of a 2 okay students 16 times of a 2 is equal to y is nothing but 0 so we will write it as 0 now what will do is like we will divide it by 4 so a 1 plus 4 times of a 2 is equal to 0 it will be considered as considered as equation 2 what we will do like we will subtract it like equation two minus equation one now we will get like a one will be cancelled out by a1 and 4 a2 minus 2 a 2 is equal to 2 a2 is equal to 0 minus minus 1 which will be plus 1 and a 2 will be 1 upon 2 now we will put it in the equation equation what we will get like a 1 a 1 plus twice of twice of a 2 is nothing but 1 so we will just put it like a simply 1 okay twice of a 2 is equal to is equals to minus 1 now we will get the value of a 1 which will be minus 2 we will put those all the 3 values of a 0 a 1 and a 2 in the equation equation equation will be y is equal to a 0 is nothing but 0 so we don't need to write it and a 2 is nothing but 1 by 2 so 1 by 2 x is square plus a 1 a 1 times of x okay so what is a 1 a 1 is nothing but minus 2 so it will be minus 2 times of x this is the required equation of the polymer function and it is the answer of the a part now we need to we need to sketch the graph so what we will do like so basically what we can write is like a y equals to when we take x as common x times of 1 by 2 times of x minus 2 okay now we can say that like there will be the 2 roots of that particular equation one will be the 0 and another will be 1 by 2 x minus 2 is equal to 0 then x is equal to 1 sorry 4 it we can say that the polymer function will cut the x x x equals to 0 and x equal to 4 okay now we will draw it so what we will do like we will just draw the graph of that particular okay how we will draw that we can say that like the graph will intersect the x xx at particular x equal to 0 okay here and the x equal to another one is 4 so basically it will be 4 we can so first of all we will mark it 1 2 3 and 4 and 5 okay it will be minus 1 minus 2 minus 3 and minus 4 okay it will be minus 1 minus 2 minus 3 and it will be nothing but 1 2 3 4 and it will be a why positive y x xx it will be considered as positive x xx and it will be minus y and it is minus x now the graph will intersect the x x x if the x equal to 0 and x equal to 4 so we can say that these are the 2 points where the graph will intersect the xxx and another points are given are already given 0 .0 2 .2...

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