Find a polynomial function whose graph passes through (-1,2) (0,1) (1,2) (2,17)
To find a polynomial function that passes through these four points, we can start by considering a general polynomial function of degree three:
f(x) = ax^3 + bx^2 + cx + d
Using the given points, we can set up a system of equations to solve for the coefficients a, b, c, and d.
Let's start with the first point (-1, 2):
2 = a(-1)^3 + b(-1)^2 + c(-1) + d
2 = -a + b - c + d
Next, let's consider the second point (0, 1):
1 = a(0)^3 + b(0)^2 + c(0) + d
1 = d
Now, let's move on to the third point (1, 2):
2 = a(1)^3 + b(1)^2 + c(1) + d
2 = a + b + c + d
Finally, let's consider the fourth point (2, 17):
17 = a(2)^3 + b(2)^2 + c(2) + d
17 = 8a + 4b + 2c + d
Now, we have a system of four equations:
-2a + b - c + d = 2
1 = d
a + b + c + d = 2
8a + 4b + 2c + d = 17
Solving this system of equations, we find:
a = 1
b = -3
c = 0
d = 1
Therefore, the polynomial function that passes through the given points is:
f(x) = x^3 - 3x^2 + 1