Find the quadratic function f(x)=ax^2+bx+c for which f(1)= -10, f(-2)= -22, f(2)= -38.

Question

What is the function?

f(x)=

Answer 1

To start, f(x) is the output of the function, also known as the "y-value".

Plug in the x-values, setting it equal to the f(x) value given:

f(1) = a(1)^2 +b(1) + c = -10

=> a + b + c = -10

f(-2) = a(-2)^2 + b(-2) + c = -22

=> 4a - 2b + c = -22

f(2) = a(2)^2 + b(2) + c = -38

=> 4a + 2b + c = -38

Now we have 3 linear systems of equations to solve for a, b, c

(1) a + b + c = -10

(2) 4a - 2b + c = -22

(3) 4a + 2b + c = -38

Multiply the first equation by 2

=> 2a + 2b + 2c = -20

Add this to the second equation

=> equation (*): 6a + 3c = -42

Add equations 2 and 3 together:

=> equation (**): 8a + 2c = -60

Now multiply equation (*) by 2 and equation (**) by -3 to cancel out the c's:

=> a = -8

Now plug a back in to solve for c:

6(-8) + 3c = -42

=> c = 2

Now solve for b (can plug this back into any of the equations):

a + b + c = -10

=> -8 + b + 2 = -10

=> b = -4

This gives the final quadratic equation of:

f(x) = -8x^2 - 4x + 2

(we can check our answer by plugging any of the x and f(x) values back in)

I will choose plugging in f(1) = -10

=> f(1) = -8(1)^2 -4(1) + 2

=> f(1) = -8 - 4 + 2

=> f(1) = -10