Find the equation of the parabola that passes through the points (-1,10), (4,15), and (2,1)

Answer 1

A parabola is a quadratic equation: y = ax² + bx + c

Since we have 3 unknowns, a, b, c, we need three equations to solve.

Use the three given points to create three equations:

1. P(-1, 10): 10 = a(-1)² + b(-1) + c ---> 10 = a - b + c
2. P(4, 15): 15 = a(4)² + b(4) + c ---> 15 = 16a + 4b + c
3. P(2, 1): 1 = a(2)² = b(2) + c ---> 1 = 4a + 2b + c

Use elimination to solve. Eliminate "c" first:

Equation 4: Subtract equation 2 from equation 1: 5 = 15a + 5b

Equation 5: Subtract equation 3 from equation 1: -9 = 3a + 3b

To eliminate another variable "a", multiply equation 5 by -5: ---> 45 = -15a - 15b

Add equation 4 and this new equation: 50 = -10b

Divide both sides by -10: b = 50/-10 = -5

Substitute b = -5 into equation 4:

5 = 15a + 5(-5)

5 = 15a - 25

Add 25 to both sides: 30 = 15a

Divide by 15 on both sides: a = 30/15 = 2

Substitute "a" and "b" into equation 1:

10 = 2 -(-5) + c

10 = 2 + 5 + c

10 = 7 + c

c = 10 - 7 = 3

CHECK: 16(2) + 4(-5) + 3 = 32 - 20 + 3 = 15

4(2) + 2(-5) + 3 = 8 - 10 + 3 = 1

Answer: y = 2x² -5x + 3

Answer 2

Ax^2 + Bx + C = f(x)

F(-1) = 10 ---> 10 = A-B+C

F(5)=15 ---> 15 = 16A + 4B + C

f(2) = 1 ---> 1 = 4A + 2B + C

Uses the first equation to eliminate A from the second and third equations:

-145 = 20B - 15C <--- -16 times equation 1 plus equation 2

-39 = 6b - 3c <--- -4 times equation 1 plus equation 3

Next eliminates the C by multiplying the bottom equation by -5 and adding to top equation:

50 = -10b

b = -5

-39 = 6(-5) - 3c

-39 = -30 - 3c

-9 = -3c

c=3

Finally

A-B+C = 10

A - -5 + 3 = 10

A + 5 + 3 = 10

A + 8 = 10

A = 2

f(x) = 2x^2 -5x + 3