Find a polynomial function whose graph passes through (7,12) (9,-10) and (0,3)
To find a polynomial function passing through the given points, we can use the fact that a polynomial of degree n must have n+1 distinct points.
Since we have three points, we can use a quadratic function (a polynomial of degree 2) to pass through them. The general form of a quadratic function is:
f(x) = ax^2 + bx + c
Using the three points given, we can create a system of equations to solve for the coefficients a, b, and c:
1) When x = 7, f(x) = 12:
12 = a(7^2) + b(7) + c
84a + 7b + c = 12
2) When x = 9, f(x) = -10:
-10 = a(9^2) + b(9) + c
81a + 9b + c = -10
3) When x = 0, f(x) = 3:
3 = a(0^2) + b(0) + c
c = 3
Using equations 1 and 2, we have the following system of equations:
84a + 7b + 3 = 12
81a + 9b + 3 = -10
Subtracting equation 3 from both of these equations, we can eliminate c:
84a + 7b = 9
81a + 9b = -13
Multiplying equation 1 by 3 and equation 2 by 7, we can eliminate b:
252a + 21b = 27
567a + 63b = -91
Subtracting equation 5 from equation 4:
(567a + 63b) - (252a + 21b) = -91 - 27
315a + 42b = -118
Now, we have the following system of equations to solve:
315a + 42b = -118
81a + 9b = -13
Solving this system of equations, we get:
a = -1
b = 5
Now that we have the values of a, b, and c, we can write the polynomial function:
f(x) = -x^2 + 5x + 3