Algebra Function Table
Algebra function table finding the value of expression in the unknown variable. The algebraic expression deals with variables, the variable contains only the alphabetic letters. Algebraic expression may include real number, and polynomials. Functions in algebra in the form of p(y), q(y),… To find the x value of the function table with algebra. The expression is in the form of ax2+bx+c. p(y) = 2y2+3y + 4. In this function we need to find the function variable y as 2.
| X | p(x) = ax2 + bx + c |
| a | p(a) = aa2 + ba + c |
| b | p(b) = ab2 + bb + c |
| c | p(c) = ac2 + bc + c |
| d | p(d) = ad2 + bd + c |
Below are the examples on algebra function table -
Example 1:
Using square equation in algebra function table. f(x) = x2 +2x +4 find the f(7),f(8),f(9), and f(10).
Solution: Equation is in the form of square equation using the value of x find the function
| X | f(x)=x2 + 2x + 4 | f(x) |
| 7 | f(7)=72 + 2*7 + 4=49+14+4=67 | f(7)=67 |
| 8 | f(8)=82 + 2*8 + 4=64+16+4=84 | f(8)=84 |
| 9 | f(9)=92 + 2*9 + 4=81+18+4=103 | f(9)=103 |
| 10 | f(10)=102 + 2*10 + 4=100+20+4=124 | f(10)=124 |
Hence the square equation in algebra function table show above. The value of f(7),f(8),f(9),and f(10) are 67,84,103,and 124.
Problem 2:
Using square equation in algebra function table of f(x) = x2 +2x +14 find the f(1),f(2),f(3),and f(4).
Solution : Equation is in the form of square equation using the value of x find the function
| X | f(x)=x2+2x+14 | f(x) |
| 1 | f(x)=12+2*1+14=1+2+14=17 | f(1)=17 |
| 2 | f(x)=22+2*2+14=4+4+14=22 | f(2)=22 |
| 3 | f(x)=32+2*3+14=9+6+14=29 | f(3)=29 |
| 4 | f(x)=42+2*4+14=16+8+14=38 | f(4)=38 |
Hence the square equation in algebra function table as show above. The value of are f(1),f(2),f(3),and f(4) are 17, 22, 29,and 38.
Problems 3:
Using cubic equation of algebra function table of find f(x) = x3 + 2x2 + 2x + 4 the f(3),f(4),f(5), and f(6).
Solution : Equation is in the form of cubic equation using the value of x find the function
| X | f(x)=x3+2x2+2x+4 | f(x) |
| 3 | f(x)=33+2*32+2*3+4=27+18+6+4 | f(3)=55 |
| 4 | f(x)=43+2*42+2*4+4=64+32+8+4 | f(4)=108 |
| 5 | f(x)=53+2*52+2*5+4=125+50+10+4 | f(5)=189 |
| 6 | f(x)=63+2*62+2*6+4=216+72+12+4 | f(6)=304 |
Hence the cubic equation in algebra function table of the value of are f(3),f(4),f(5),and f(6) are 55, 108, 189,and 304.
Problem 4:
Using cubic equation in algebra function table. find f(x) = x3 +2x2 + 2x + 14 the f(2),f(4),f(6), and f(8).
Solution : Equation is in the form of cubic equation using the value of x find the function
| X | f(x)=x3+2x2+2x+14 | f(x) |
| 2 | f(x)=23+2*22+2*2+14 | f(2)=34 |
| 4 | f(x)=43+2*42+2*4+14 | f(4)=118 |
| 6 | f(x)=63+2*62+2*6+14 | f(6)=314 |
| 8 | f(x)=83+2*82+2*8+14 | f(8)=670 |
Hence the cubic eqution in algebra function table as show above. The value of are f(2),f(4),f(6),and f(8) are 34, 118, 314,and 670.
