Originally Posted by

**shadow6** "If $\displaystyle f(x) = x^2 - 6x + 14$ and $\displaystyle g(x) = -x^2 - 20x - k$, determine the value of $\displaystyle k$ so that there is exactly one point of intersection between the two parabolas."

This is what I've done so far:

$\displaystyle f(x) + g(x) = 0$

Why plus?! The two parabolas meet when $\displaystyle f(x)=g(x)\Longleftrightarrow f(x)-g(x)=0$ ...!

Tonio

$\displaystyle -26x + 14 - k = 0$

$\displaystyle a = 0, b = -26, c = (14 - k)$

$\displaystyle b^2 - 4ac = 0$

$\displaystyle (-26)^2 - 4(0)(14 - k) = 0$

$\displaystyle 676 - 0 = 0$

Based on the fact that this method cannot solve for $\displaystyle k$, I've come to believe that this attempt at a solution was wrong. Though I can't think of any other way of solving this. Help would be appreciated.