COMEDK2025MathematicsQuadratic EquationActual
If a quadratic function in x has the value 19 when x=1 and has a maximum value 20 when x=2 , then the function is
Options
- Af(x)=x^2-4 x+16
- Bf(x)=-x^2+4 x+16
- Cf(x)=-x^2+5 x+16
- Df(x)=x^2+4 x+16
Correct answer
B. f(x)=-x^2+4 x+16
Step-by-step solution
Let the quadratic function be f(x) = ax^2 + bx + c . Since the function has a maximum value at x=2 , the coefficient a must be negative, and the vertex is at x = - b 2a = 2 , which implies b = -4a . The maximum value is f(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 20 . Substituting b = -4a into the equation for the maximum value: 4a + 2(-4a) + c = 20 -4a + c = 20 c = 20 + 4a . Given f(1) = 19 , we have a(1)^2 + b(1) + c = 19 a + b + c = 19 . Substituting b = -4a and c = 20 + 4a into this equation: a - 4a + 20 + 4a = 19