Question

Question #6beba

Answer 1

I'm wondering if you copied the problem incorrectly. I'm getting different answers than you wrote. For the given function, I'm getting a minimum value of `r-4t^2` at `x=2t` (and the graph is symmetrical about `x=2t`).

Explanation:

To complete the square on `f(x)=x^2-4tx+r`, take the coefficient of `x`, which is `-4t`, divide it by 2 to get `-2t`, and then square that to get `4t^2`. Now add and subtract that expression to rewrite the function as `f(x)=(x^2-4tx+4t^2)+r-4t^2`.

The expression in the parentheses is a perfect square. In fact, `f(x)=(x-2t)^2+r-4t^2`. This function is clearly lowest in value when `x=2t`. Furthermore, the minimum value is `f(2t)=4t^2-4t(2t)+r=4t^2-8t^2+r=r-4t^2`. The vertical line `x=2t` is also the location of the axis of symmetry for this upward-opening parabola.