determine the quadratic function whose graph contains the points(-2,-1), (-1,0), and (-3,0).
Assume the quadratic function is of the form
y(x) = ax^2 + bx + c
The three points give you three equations in the three unknowns.
-1 = 4a -2b +c
0 = a -b +c
0 = 9a -3b +c
Subtract the second from the first and the second from the third to get two independent equations without x.
-1 = 3a -b
0 = 8a -2b
-2 = 6a -2b
2a = 2
a = 1 b = 4 c = b-a = 3
y = (x^2 + 4x +3) = (x+1)(x+3)
I meant to write "Subtract the second from the first and the second from the third to get two independent equations without c"
One could also write one equation for a circle that passes through those three points, but it would contain a y^2 term as well as the x^2, x and constant terms
To determine the quadratic function whose graph contains the given points, we can start by using the general form of a quadratic function:
y = ax^2 + bx + c
Plugging in the coordinates of the given points will give us a system of three equations that we can solve to find the values of a, b, and c.
Let's start by substituting the first point (-2, -1):
-1 = a(-2)^2 + b(-2) + c [Equation 1]
Next, substitute the second point (-1, 0):
0 = a(-1)^2 + b(-1) + c [Equation 2]
Finally, substitute the third point (-3, 0):
0 = a(-3)^2 + b(-3) + c [Equation 3]
Now we have a system of three equations:
-1 = 4a - 2b + c [Equation 1 rewritten]
0 = a - b + c [Equation 2 rewritten]
0 = 9a - 3b + c [Equation 3 rewritten]
To make the elimination method easier, let's subtract Equation 2 from Equation 1 and Equation 3:
-1 - 0 = 4a - 2b + c - (a - b + c)
-1 = 3a - 3b [Equation 4]
0 - 0 = 9a - 3b + c - (a - b + c)
0 = 8a - 4b [Equation 5]
Now we have two equations (Equation 4 and Equation 5) with two variables (a and b). We can solve this system of equations to find the values of a and b.
Let's eliminate the variable b by multiplying Equation 4 by 4 and Equation 5 by 3:
(-1) * 4 = (3a - 3b) * 4
-4 = 12a - 12b [Equation 6]
(0) * 3 = (8a - 4b) * 3
0 = 24a - 12b [Equation 7]
Now subtract Equation 6 from Equation 7:
0 - (-4) = 24a - 12b - (12a - 12b)
4 = 12a
a = 4/12
a = 1/3
Now that we have the value of a, we can substitute it into one of the original equations to find the value of b. Let's use Equation 4:
-1 = 3(1/3) - 3b
-1 = 1 - 3b
-3b = -2
b = 2/3
Finally, we can substitute the values of a and b into one of the original equations to find the value of c. Let's use Equation 2:
0 = (1/3)(-1)^2 + (2/3)(-1) + c
0 = 1/3 - 2/3 + c
0 = -1/3 + c
c = 1/3
Therefore, the quadratic function whose graph contains the points (-2,-1), (-1,0), and (-3,0) is:
y = (1/3)x^2 + (2/3)x + 1/3