What is a functions operations algebra 2

Author: Matthew David

What is a functions operations algebra 2

Introduction of what is a function algebra 2:

Function algebra 2 is a  branch of mathematics to find unknown variable from the expression with the help of known values. The algebraic expression deals with variables, the variable are represents by alphabetic letters. In functions algebra 2 numbers are constant, algebraic expression may include  real number, complex number, and polynomials. Functions in  algebra 2 may include in the function of p(y), q(y),… to find the x value of the algebra functions.

For example:

Function algebra 2 may be described in this form p(y) = 2y2+3y + 4. In this function we need to find the function variable y as 2.

Functions operation algebra 2 finding unknown variable from the given expression with the help of known values. The algebraic expression contains variables are represented alphabetic letters.  The functions operations algebra 2 looks f(x), q(x),… to find the x value of the algebra functions.There are several operation of function as show in below

Functions operations algebra 2:

    (f + g)(y) = f (y)+ g (y)
    (f – g)(x) = f (y)–g (y)
    (f .g)(x) = f (y).g (y)
    `(f/g)` (x)=`(f(y))/(g(y)).`
    
Addition and subtraction problems in the functions operations algebra 2:

Problem (i): Adding two functions operations f(x) = x+2 and g(x) = x-3 find (f+g)(x).

Solution: Adding two function using the functions operations algebra 2.

Given f(x) = x+2 and g(x) = x-3 find (f+g)(x).

Using the functions operations of  (f + g)(x) = f (x)+g (x).

(f+g)(x) = (x+2)+(x-3).

(f+g)(x)= x+x+2-3 in this step adding both functions.

(f+g)(x) = 2x-1.

Problem (ii): subtracting two functions operations f(x) = (x+5) and g(x) = (x-8) find (f-g)(x).

Solution: Subtracting two functions using the functions operations algebra 2.

Given f(x) = x+5 and g(x) = x-8 find (f-g) (x).

Using the functions operations of (f - g)(x) = f (x)-g (x).

(f-g)(x) = (x+5)-(x-8).

(f-g)(x)= x-x+5-8 in this step subtracting both functions.

(f-g)(x) = -3

Multiplying and division problems in the functions operations algebra 2:

Problem (i): multiplying two functions operations f(x) = x+5 and g(x) = x-8 find (f.g)(x).

Solution: multiplying two functions using the functions operations algebra 2.

Given f(x) = x+5 and g(x) = x-8 find (f.g) (x).

Using the functions operations of (f . g)(x) = f (x). g (x).

(f.g)(x) = (x+5) . (x-8).

(f.g)(x)= x2-8x+5x-40 = x2  - 3x -40 in this step multiplying both functions.

Problem (i): dividing two functions operations f(x) = 10x and g(x) = 2x find `(f/g)` (x).

Solution: multiplying two functions using the functions operations algebra 2.

Given f(x) = 10x and g(x) = 2x find `(f/g)` (x).

Using the functions operations of `(f/g)` (x) = `(f(x))/(g(x)).`

`(f/g)` (x) = `(10x)/(2x)` in this step x will be cancelled.

`(f/g)` (x)= 5 in this step dividing both functions

Example Problems for what is a function algebra 2

Simplify using the square function algebra 2.

Problem1; Using square functions. What is a function algebra 2 in. p(x) = x2 +6x +14 find the p(6).

Solution :

Find the function p(6). Here substitute the value 5 to the variable.

p(x) = x2 +6x +14 find the f(6).

The value of x is 6 is given

p(6) = 62 +6*6 +14

p(6) = 36 +36 +14 In this step 62 is 36 it is calculate and 6*6 is 36 be added

p(6) = 86.

The functions algebra 2 p(6) = x2 +6x +14 find the p(6) is 86.

Problem2; Using square  function algebra 2. z(x) = x2 +6x +25 what is a function algebra 2 in z(7).

Solution :

Find the function y(7). Here substitute the value 7 to the variable.

z(7) = x2 +6x +25 find the z(7).

The value of x is 3 is given

z(7) = 72 +6*7 +25

z(7) = 49 +42+25.

z(7) = 116.

z(7) = x2 +6x +25 find the y(7) is 116.

Functions algebra 2 problems using the cubic functions

Problems1: using the cubic function algebra 2: f(x) = x3 +2x2 + 2x + 4 What is a function algebra 2 in f(2).

Solution:

Find the function f(2). Here substitute the value 2 to the variable.

f(x) = x3 +2x2 + 2x + 4 find the f(2).

Here the value of x is given as 2

f(2) = 23 + 2*22 + 2*2 +4

f(2) = 8 +8+4 +4 In this step 23 is calculated  as 8 and 2 square is 4

f(2) = 24.

f(x) = x3 +2x2 + 2x + 4 find the f(2) = 24.

Problems 2: what is a function algebra 2 in cubic equations. f(x) = 2x3 +4x2 + 3x + 24 find the functions algebra 2 of f(4).

Solution:

Find the function f(4). Here substitute the value 4 to the variable.

f(x) = 2x3 +4x2 + 3x + 24 find the f(4).

Here the value of x is given as 3

f(4) = 2*43 + 4*42 + 3*4 +24

f(4) = 128 +64+12 +24

f(4) =  228

f(x) = 2x3 +4x2 + 3x + 24 find the f(4) = 228.   

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