Question 1116302: Find a polynomial function whose graph passes through (-1,-9),(0,-2),(1,1),(2,12)
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39617)   (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
There are 4 points, so assume a general polynomial of
degree which is one less, or a polynomial of degree 3.
Substitute each point (x,y) in Ax³ + Bx² + Cx + D = y:
A(-1)³ + B(-1)² + C(-1) + D = -9
A(0)³ + B(0)² + C(0) + D = -2
A(1)³ + B(1)² + C(1) + D = 1
A(2)³ + B(2)² + C(2) + D = 12
Simplify and solve that system of 4 equations.
The second equation gives us D = -2, then the
system simplifies to:
-A + B - C = -7
A + B + C = 3
8A + 4B + 2C = 14
Adding the first two equations gives
2B = -4
B = -2
Substituting B = -2 in the 2nd and 3rd equations:
A + (-2) + C = 3
8A + 4(-2) + 2C = 14
which simplifies to
A + C = 5
8A + 2C = 22
Multiplying the 1st by -2
-2A - 2C = -10
8A + 2C = 22
--------------
6A = 12
A = 2
Substitute in
A + C = 5
2 + C = 5
C = 3
So A=2, B=-2, C=3, D=-2 and
Ax³ + Bx² + Cx + D = y becomes
2x³ - 2x² + 3x - 2 = y
or use P(x) for y and write it on the left:
P(x) = 2x³ - 2x² + 3x - 2 = y
Here's the graph. Notice that it passes through
all four given points:
Edwin
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