Suppose f(x) is a quadratic polynomial with leading coefficient 1 such that f(x) + f(-x) = 2x^2 for all x and f(2016) = 0. What is the value of f(10) ?
thanks
if f(x) = x^2+bx+c, we have
(x^2+bx+c) + (x^2-bx+c) = 2x^2
2x^2+c = 2x^2
so, c=0 and
f(x) = x^2+bx
f(2016) = 0, so
2016^2 + 2016b = 0
b = -2016
f(x) = x^2 - 2016x = x(2016-x)
f(10) = 10(2016-10) = 10(2006) = 20060
To find the value of f(10), we need to use the given information and properties of quadratic polynomials.
From the given equation, we have f(x) + f(-x) = 2x^2 for all x. This means that the sum of f(x) and f(-x) is equal to 2x^2 for any x. Since f(x) is a quadratic polynomial, it can be written as f(x) = ax^2 + bx + c, where a, b, and c are constants.
Substituting -x for x in f(x), we have f(-x) = a(-x)^2 + b(-x) + c = ax^2 - bx + c.
Now, we can use this information to find the specific coefficients of f(x).
f(x) + f(-x) = (ax^2 + bx + c) + (ax^2 - bx + c) = 2ax^2 + 2c.
Comparing this with the given equation 2x^2, we can conclude that 2a = 2 and 2c = 0.
From 2a = 2, we get a = 1. Hence, the leading coefficient is 1.
From 2c = 0, we get c = 0. Hence, the constant term is 0.
Therefore, f(x) = x^2 + bx + 0 = x^2 + bx.
Since we know that f(2016) = 0, we can substitute 2016 for x and solve for b.
f(2016) = (2016)^2 + b(2016) = 0.
Simplifying this equation, we get 2016^2 + 2016b = 0. Solving for b, we have b = -2016.
Finally, we can substitute 10 for x in f(x) to find f(10).
f(10) = 10^2 + (-2016)(10) = 100 - 20160 = -20060.
Therefore, the value of f(10) is -20060.