find a quadratic function in standard form for each set of points.
(-1,2),(0,1),(1,-4)
y=ax^2+bx+c
a) 2=a-b+c
b) 1= c
c) -4=a+b+c
from equation b< c=1
add a)+c)
-2=2a+2*1
a=-2
then in equation c)
-4=-2+b+1
b= -3
y=ax^2+bx + c
y=-2x^2-3x+1
thank you very much
To find a quadratic function in standard form that passes through the given set of points (-1,2), (0,1), and (1,-4), we can use the general quadratic equation:
y = ax^2 + bx + c
Substituting the coordinates of the three points into this equation, we can set up a system of equations to solve for the coefficients a, b, and c.
For the point (-1,2):
2 = a(-1)^2 + b(-1) + c
2 = a - b + c ----(1)
For the point (0,1):
1 = a(0)^2 + b(0) + c
1 = c ----(2)
For the point (1,-4):
-4 = a(1)^2 + b(1) + c
-4 = a + b + c ----(3)
Now, we have a system of three equations:
1) 2 = a - b + c
2) 1 = c
3) -4 = a + b + c
From equation (2), we already know that c = 1. Substituting this into equations (1) and (3):
1) 2 = a - b + 1
2 = a - b ----(4)
3) -4 = a + b + 1
-5 = a + b ----(5)
Now, we can solve equations (4) and (5) simultaneously to find the values of a and b:
Adding equations (4) and (5):
2 + (-5) = (a - b) + (a + b)
-3 = 2a
a = -3/2
Substituting the value of a back into equation (4):
2 = (-3/2) - b
4 = -3 - 2b
7 = -2b
b = -7/2
Therefore, the values of a and b are -3/2 and -7/2, respectively. Substituting these values back into equation (1), we can find the value of c:
2 = (-3/2) - (-7/2) + c
2 = -3/2 + 7/2 + c
2 = 2 + c
c = 0
Finally, substituting the values of a, b, and c back into the quadratic equation, we obtain the quadratic function in standard form:
y = (-3/2)x^2 - (7/2)x + 0
Simplifying the equation, we have:
y = (-3/2)x^2 - (7/2)x
To find a quadratic function in standard form for a set of points, we can use the fact that a quadratic function can be written in the form:
f(x) = ax^2 + bx + c
where a, b, and c are constants.
Let's use the given points (-1,2), (0,1), and (1,-4) to find these constants.
Step 1: Plug in the x and y coordinates of each point into the quadratic equation to get three equations.
For the point (-1,2):
2 = a(-1)^2 + b(-1) + c
2 = a - b + c ---> Equation 1
For the point (0,1):
1 = a(0)^2 + b(0) + c
1 = c ---> Equation 2
For the point (1,-4):
-4 = a(1)^2 + b(1) + c
-4 = a + b + c ---> Equation 3
Step 2: Solve the system of equations (Equations 1, 2, and 3) simultaneously to find the values of a, b, and c.
From Equation 2, we know that c = 1.
Substitute c = 1 into Equations 1 and 3:
2 = a - b + 1 ---> Equation 4
-4 = a + b + 1 ---> Equation 5
Now, we have a system of two equations with two variables (Equations 4 and 5).
Step 3: Solve the system of equations (Equations 4 and 5) to find the values of a and b.
Add Equation 4 and Equation 5:
2 + (-4) = (a - b) + (a + b) + (1 + 1)
-2 = 2a + 2
Divide both sides of the equation by 2:
-1 = a + 1
Subtract 1 from both sides of the equation:
a = -2
Now, substitute the value of a = -2 into Equation 4:
2 = (-2) - b + 1
Rearrange the equation:
1 = -b
Multiply both sides of the equation by -1:
-1 = b
Step 4: Substitute the values of a = -2, b = -1, and c = 1 into the quadratic function equation f(x) = ax^2 + bx + c to find the quadratic function in standard form.
f(x) = (-2)x^2 + (-1)x + 1
Rearrange the equation in ascending order of powers:
f(x) = -2x^2 - x + 1
Thus, the quadratic function in standard form for the given set of points is f(x) = -2x^2 - x + 1.