How do I create a quadratic equation that passes through the points (-1,2),(1,7),(4,5)?
y = a x ^ 2 + b x + c
In this case:
For x = - 1 , y = 2
2 = a ( - 1 ) ^ 2 + b ( - 1 ) + c
a - b + c = 2
For x = 1 , y = 7
7 = a ( 1 ) ^ 2 + b 1 + c
a + b + c = 7
For x = 4 , y = 5
5 = a ( 4 ) ^ 2 + b 4 + c
16 a + 4 b + c = 5
So you must solve system of 3 equations :
a - b + c = 2
a + b + c = 7
16 a + 4 b + c = 5
The exact solutions are :
a = - 19 / 30
b = 5 / 2
and
c = 77 / 15
y = ( - 19 / 30 ) x ^ 2 + ( 5 / 2 )x + 77 / 15
To create a quadratic equation that passes through the given points (-1,2), (1,7), and (4,5), you can follow these steps:
Step 1: Start with the general form of a quadratic equation:
y = ax^2 + bx + c
Step 2: Substitute each of the given points into the equation to create a system of equations. Using (-1,2) gives:
2 = a(-1)^2 + b(-1) + c
which simplifies to:
2 = a - b + c (Equation 1)
Using (1,7) gives:
7 = a(1)^2 + b(1) + c
which simplifies to:
7 = a + b + c (Equation 2)
Using (4,5) gives:
5 = a(4)^2 + b(4) + c
which simplifies to:
5 = 16a + 4b + c (Equation 3)
Step 3: Solve this system of equations to find the values of a, b, and c.
First, subtract Equation 1 from Equation 2 to eliminate c:
7 - 2 = a + b + c - (a - b + c)
5 = 2b (Equation 4)
Next, subtract Equation 1 from Equation 3 to eliminate c:
5 - 2 = 16a + 4b + c - (a - b + c)
3 = 15a + 5b (Equation 5)
Step 4: Solve Equation 4 for b:
5 = 2b
b = 2.5
Step 5: Substitute the value of b into Equation 5 to solve for a:
3 = 15a + 5(2.5)
3 = 15a + 12.5
-9.5 = 15a
a = -9.5/15
a = -0.6333
Step 6: Substitute the values of a and b into Equation 1 to solve for c:
2 = (-0.6333) - 2.5 + c
2 = -3.1333 + c
c = 2 + 3.1333
c = 5.1333
Step 7: Substitute the values of a, b, and c into the general quadratic equation form to get the final equation:
y = (-0.6333)x^2 + (2.5)x + 5.1333
So, the quadratic equation that passes through the given points is y = (-0.6333)x^2 + (2.5)x + 5.1333.