find a polynomial function whose graph passes through the points (-1,8),(0,4),(1,2), and (2,14)
To find a polynomial function that passes through the given points, we can start by considering a general polynomial function of degree 3: f(x) = ax^3 + bx^2 + cx + d.
Substituting the x and y values of each point into this function, we get the following system of equations:
-8a + 4b - 2c + d = 8 (Equation 1)
d = 4 (Equation 2)
8a + 4b + 2c + d = 2 (Equation 3)
8a + 16b + 8c + d = 14 (Equation 4)
From Equation 2, we know that d = 4. Substituting this value into Equations 1, 3, and 4, we can simplify the system:
-8a + 4b - 2c = 4 (Equation 5)
8a + 4b + 2c = -2 (Equation 6)
8a + 16b + 8c = 10 (Equation 7)
Now, let's solve this system using any method of choice (for example, substitution or elimination). For convenience, we will use elimination:
Multiply Equation 5 by 2 and add it to Equation 6:
-8a + 4b - 2c = 4
16a + 8b + 4c = -4
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24a + 12b + 2c = 0 (Equation 8)
Multiply Equation 5 by -4 and add it to Equation 7:
32a + 16b + 8c = -16
8a + 16b + 8c = 10
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40a + 0b + 0c = -6 (Equation 9)
From Equation 9, we have 40a = -6, which implies a = -6/40 = -3/20.
Substitute a = -3/20 into Equation 8:
24a + 12b + 2c = 0
24(-3/20) + 12b + 2c = 0
-18/5 + 12b + 2c = 0
12b + 2c = 18/5 (Equation 10)
Now, we have two equations (Equations 10 and 3) with two variables (b and c). We can solve this system:
Multiply Equation 10 by 12 and subtract it from Equation 3:
8a + 4b + 2c = -2
-12b - 2c = -36/5
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8a - 8b = 26/5 (Equation 11)
Simplify Equation 11 by dividing both sides by 8:
a - b = 13/40 (Equation 12)
Now, we can find the value of b by subtracting Equation 12 from Equation 11:
a - b = 13/40
- a + b = -13/40
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0 = 0
The equation 0 = 0 tells us that b can be any value since it does not affect the system. Therefore, b is a free variable.
To find c, we can substitute b = t (where t is any constant) into Equation 10:
12b + 2c = 18/5
12t + 2c = 18/5
2c = 18/5 - 12t
c = 9/5 - 6t
Finally, we can express the polynomial function as:
f(x) = ax^3 + bx^2 + cx + d
= (-3/20)x^3 + tx^2 + (9/5 - 6t)x + 4
In summary, any polynomial function of the form f(x) = (-3/20)x^3 + tx^2 + (9/5 - 6t)x + 4 (where t is any constant) will pass through the points (-1,8), (0,4), (1,2), and (2,14).