Find the values of a, b, and c such that the equation y=ax^2+bx+c has ordered pair solutions (-3,-22), (-1,2) and (3,2)
a = 2; b = 8; d = 4
a = -2; b = 4; c = 8
a = 2; b = 4; c = 8
a = -2; b = 8; c = 4
lazy way: try the points to see whether they fit.
algebraic way. Plug in the values for x and y to get
9a-3b+c = -22
a-b+c = 2
9a+3b+c = 2
Either way, it's -2,4,8
To find the values of a, b, and c in the equation y = ax^2 + bx + c, we can substitute the x and y values of the three given ordered pairs into the equation and solve for the unknowns.
Let's start by substituting the first ordered pair (-3, -22) into the equation:
-22 = a(-3)^2 + b(-3) + c
-22 = 9a - 3b + c
Next, let's substitute the second ordered pair (-1, 2) into the equation:
2 = a(-1)^2 + b(-1) + c
2 = a - b + c
Lastly, substitute the third ordered pair (3, 2) into the equation:
2 = a(3)^2 + b(3) + c
2 = 9a + 3b + c
Now we have a system of three equations with three unknowns:
-22 = 9a - 3b + c
2 = a - b + c
2 = 9a + 3b + c
We can solve this system of equations using any method, such as substitution or elimination. I'll use the method of elimination to find the values of a, b, and c.
Add the second and third equations together:
2 + 2 = a - b + c + 9a + 3b + c
4 = 10a + 2c
Now, let's solve for one of the variables in terms of the other two equations. Since the second equation is already solved for c, let's solve the first equation for c:
c = -22 - 9a + 3b
Substitute this expression for c into the equation 4 = 10a + 2c:
4 = 10a + 2(-22 - 9a + 3b)
4 = 10a - 44 - 18a + 6b
48 = -8a + 6b
We can simplify this equation further:
48 = -8a + 6b
24 = -4a + 3b
Now, we have two equations with two unknowns:
24 = -4a + 3b
48 = -8a + 6b
We can solve this system of equations using substitution, elimination, or matrix methods. In this case, let's use the elimination method.
Multiply the first equation by 2 and the second equation by 3 to eliminate the b variable:
48 = -8a + 6b
72 = -12a + 9b
Now, subtract the first equation from the second equation:
72 - 48 = (-12a + 9b) - (-8a + 6b)
24 = -12a + 9b + 8a - 6b
24 = -4a + 3b
We can see that the resulting equation is the same as the first equation we obtained in the previous step. This means that the original system of equations is dependent, and there are infinitely many solutions. In other words, there is no unique solution for the values of a, b, and c.
Therefore, none of the given options (a = 2; b = 8; d = 4, a = -2; b = 4; c = 8, a = 2; b = 4; c = 8, a = -2; b = 8; c = 4) are correct.