A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?
(1) –119
(2) –159
(3) –110
(4) -180
(5) -105

• ans is 2.
  
  let the quadratic equation be ax^2 + bx + c. So, f(x) = ax^2 + bx + c
  At x = 0, the value of function is 1.
  x = 0, f(x) = 1
  ax^2 + bx + c = a * 0 + b * 0 + c = c
  c = 1.
  At x = 1, f(x) = 3
  x = 1, f(x) = 3
  a *1 + b *1 + c = 3
  Since c = 1, a + b = 2.
  Also, we know that f(x) is maximum when x = 1. If f(x) is maximum, (dx/dt)(f(x)) = 0
  Differentiating f(x), we have d/dt (ax^2 + bx + c) = 2ax + b
  At x = 1, 2ax + b = 0
  2a + b = 0.
  b = -2a
  Substituting we have a + b = 2, or a + -2a = 2. a = -2. So, b = 4. So the equation is -2x^2 + 4x + 1.
  At x = 10, the value is -2 *100 + 4*10 + 1
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