Find an equation in standard form of the parabola passing through the points below.
(1,-1) (3,15) (6,54)
To find the equation of a parabola given three points, we can use the general form of the standard equation, y = ax^2 + bx + c, and substitute the three coordinates into the equation. This will create a system of three equations with three unknowns (a, b, c), and we can solve for the unknowns.
Using the point (1, -1):
-1 = a(1)^2 + b(1) + c
-1 = a + b + c ...(1)
Using the point (3, 15):
15 = a(3)^2 + b(3) + c
15 = 9a + 3b + c ...(2)
Using the point (6, 54):
54 = a(6)^2 + b(6) + c
54 = 36a + 6b + c ...(3)
To solve this system of equations, we can use elimination or substitution. Let's use the elimination method.
First, we subtract equation (1) from equation (2):
15 - (-1) = 9a + 3b + c - (a + b + c)
16 = 8a + 2b ...(4)
Next, we subtract equation (2) from equation (3):
54 - 15 = 36a + 6b + c - (9a + 3b + c)
39 = 27a + 3b ...(5)
Now, we have a system of two equations with two unknowns.
Simplifying equation (4), we divide both sides by 2:
8 = 4a + b ...(6)
We can subtract equation (5) from equation (6):
(4a + b) - (27a + 3b) = 8 - 39
-23a - 2b = -31 ...(7)
To eliminate the b variable, let's multiply equation (4) by 2:
16 = 8a + 2b ...(8)
Now, we add equation (7) and equation (8) together:
(-23a - 2b) + (8a + 2b) = -31 + 16
-15a = -15
Dividing both sides by -15, we get:
a = 1
Substituting the value of a back into equation (6):
8 = 4(1) + b
8 = 4 + b
b = 4
Substituting the values of a and b back into equation (1):
-1 = (1) + (4) + c
-1 = 5 + c
c = -6
Therefore, the equation of the parabola passing through the points (1, -1), (3, 15), and (6, 54) is:
y = x^2 + 4x - 6