In each case, find a quadratic function whose graph passes through the given points. (If there is no solution, enter no solution.)
(0,0),(1,1),(3,-1)
if y = ax^2+bx+c, then we have
c = 0
a+b=1
9a+3b = -1
y = -2/3 x^2 + 5/3 x
To find a quadratic function whose graph passes through the given points, we need to solve a system of equations.
Let's assume the quadratic function is of the form f(x) = ax^2 + bx + c.
Using the given points (0,0), (1,1), and (3,-1), we can substitute these values into the function to obtain three equations:
1) When x = 0, f(0) = 0:
0 = a(0^2) + b(0) + c
0 = c
2) When x = 1, f(1) = 1:
1 = a(1^2) + b(1) + c
1 = a + b + c
3) When x = 3, f(3) = -1:
-1 = a(3^2) + b(3) + c
9a + 3b + c = -1
We can substitute c = 0 (from equation 1) into equations 2 and 3:
2) 1 = a + b + 0
1 = a + b
3) 9a + 3b + 0 = -1
9a + 3b = -1
Now, let's solve this system of equations using the substitution method:
From equation 2, we have a = 1 - b.
Substituting this value into equation 3:
9(1 - b) + 3b = -1
9 - 9b + 3b = -1
-6b = -10
b = 10/6
b = 5/3
Substituting the value of b = 5/3 into equation 2:
a + (5/3) = 1
a = 1 - (5/3)
a = 3/3 - 5/3
a = -2/3
Therefore, the quadratic function that passes through the given points is:
f(x) = (-2/3)x^2 + (5/3)x
Let me know if I can help you with anything else.
To find a quadratic function that passes through the given points, (0,0), (1,1), and (3,-1), you can follow these steps:
Step 1: Set up the general equation of a quadratic function:
A quadratic function can be represented in the form f(x) = ax^2 + bx + c, where a, b, and c are constants to be determined.
Step 2: Plug in the values of the given points.
Using the given points, substitute the x and y values into the quadratic equation to create a system of equations. In this case, we have three equations:
Equation 1: f(0) = a(0^2) + b(0) + c = 0 [Using the point (0,0)]
Equation 2: f(1) = a(1^2) + b(1) + c = 1 [Using the point (1,1)]
Equation 3: f(3) = a(3^2) + b(3) + c = -1 [Using the point (3,-1)]
Step 3: Solve the system of equations.
Using the three equations from Step 2, solve for the variables a, b, and c. To do this, you can use any method of solving systems of equations, such as substitution or elimination.
Let's solve the system of equations:
From Equation 1, we get: c = 0
Substituting this value of c into Equations 2 and 3, we have:
a + b = 1 --(4) [Using Equation 2]
9a + 3b = -1 --(5) [Using Equation 3]
Now we have two equations (4 and 5) with two variables (a and b). We can solve this system of equations by substitution or elimination.
Elimination Method:
Multiplying Equation 4 by 3, we can eliminate b:
3(a + b) = 3(1)
3a + 3b = 3 --(6)
Now subtract Equation 6 from Equation 5:
(9a + 3b) - (3a + 3b) = -1 - 3
6a = -4
a = -4/6
a = -2/3
Substitute the value of a into Equation 4:
(-2/3) + b = 1
b - 2/3 = 1
b = 1 + 2/3
b = 5/3
So, we have found the values of a and b:
a = -2/3
b = 5/3
c = 0
Step 4: Write the quadratic function.
Using the values of a, b, and c, we can write the quadratic function as:
f(x) = (-2/3)x^2 + (5/3)x + 0
f(x) = (-2/3)x^2 + (5/3)x
Therefore, the quadratic function that passes through the points (0,0), (1,1), and (3,-1) is f(x) = (-2/3)x^2 + (5/3)x.