Find a polynomial function whose graph passes through (5,12) (11,-12) and (0,3)
To find a polynomial function that passes through the given points, we can use the fact that a polynomial of degree n has n+1 unique points on its graph.
Let's consider a polynomial of degree 2 (quadratic) since we have 3 points. The general form of a quadratic polynomial is: f(x) = ax^2 + bx + c.
We can substitute the x and y values of each point into the function:
For (5,12):
12 = a(5)^2 + b(5) + c -> 25a + 5b + c = 12
For (11,-12):
-12 = a(11)^2 + b(11) + c -> 121a + 11b + c = -12
For (0,3):
3 = a(0)^2 + b(0) + c -> c = 3
Now, we have a system of equations:
25a + 5b + 3 = 12 (Eq. 1)
121a + 11b + 3 = -12 (Eq. 2)
c = 3
From Eq. 3, we know c = 3. We can substitute this into Eq. 1 and Eq. 2:
25a + 5b + 3 = 12 -> 25a + 5b = 9 (Eq. 4)
121a + 11b + 3 = -12 -> 121a + 11b = -15 (Eq. 5)
Next, we can solve the system of equations (Eq. 4 and Eq. 5) to find the values of a and b.
Multiplying Eq. 4 by 11 and Eq. 5 by 5, we get:
275a + 55b = 99 (Eq. 6)
605a + 55b = -75 (Eq. 7)
Subtracting Eq. 6 from Eq. 7 to eliminate b:
605a + 55b - 275a - 55b = -75 - 99
330a = -174
a = -174/330
a = -29/55
Substituting the value of a into Eq. 1:
25(-29/55) + 5b = 9
-29/11 + 5b = 9
5b = 9 + 29/11
5b = 99/11 + 29/11
5b = 128/11
b = 128/55
Therefore, the polynomial function that passes through the points (5,12), (11,-12), and (0,3) is:
f(x) = (-29/55)x^2 + (128/55)x + 3