Which polynomial has a graph that passes through the given points? (
Solution
The equation will be in the form y = ax³ + bx² + cx + d Plugging the various points into the equation ... firstly (-3,1 0): 0 = -27a + 9b - 3c + d ........ \"equation 1\" Then (-1, 0): 10 = -a + b - c + d ........ \"equation 2\" Then (0, -5): 0 = d ........ \"equation 3\" Then (3, 52): 0 = 64a + 16b + 4c + d ........ \"equation 4\" - - - - - - - - - - - - - - - - From equation 3, we can \"lose\" d altogether; therefore, from equation 1: 0 = -27a + 9b - 3c 0 = -9a + 3b - c ........ \"equation 5\" From equation 2: 10 = -a + b - c ........ \"equation 6\" From equation 4: 0 = 64a + 16b + 4c 0 = 16a + 4b + c ........ \"equation 7\" - - - - - - - - - - - - - - - - Equation 5 + equation 7: 0 = -9a + 3b - c + 0 = 16a + 4b + c –––––––––––––– 0 = 7a + 7b 0 = a + b b = -a From equation 5: 0 = -9a - 3a - c 0 = -12a - c c = -12a From equation 6: 10 = -a - a - c 10 = -2a - c c = -2a - 10 Combining these two equations: -12a = -2a - 10 -10a = -10 a = 1 b = -1 c = -12 SOLUTION: y = x³ - x² - 12x
.
Mixin Classes in Odoo 17 How to Extend Models Using Mixin Classes
Which polynomial has a graph that passes through the given points-.docx
1. Which polynomial has a graph that passes through the given points? (
Solution
The equation will be in the form y = ax³ + bx² + cx + d Plugging the various points into the
equation ... firstly (-3,1 0): 0 = -27a + 9b - 3c + d ........ "equation 1" Then (-1, 0): 10 = -a + b - c
+ d ........ "equation 2" Then (0, -5): 0 = d ........ "equation 3" Then (3, 52): 0 = 64a + 16b + 4c
+ d ........ "equation 4" - - - - - - - - - - - - - - - - From equation 3, we can "lose" d altogether;
therefore, from equation 1: 0 = -27a + 9b - 3c 0 = -9a + 3b - c ........ "equation 5" From equation
2: 10 = -a + b - c ........ "equation 6" From equation 4: 0 = 64a + 16b + 4c 0 = 16a + 4b + c ........
"equation 7" - - - - - - - - - - - - - - - - Equation 5 + equation 7: 0 = -9a + 3b - c + 0 = 16a + 4b +
c –––––––––––––– 0 = 7a + 7b 0 = a + b b = -a From equation 5:
0 = -9a - 3a - c 0 = -12a - c c = -12a From equation 6: 10 = -a - a - c 10 = -2a - c c = -2a - 10
Combining these two equations: -12a = -2a - 10 -10a = -10 a = 1 b = -1 c = -12 SOLUTION: y =
x³ - x² - 12x
