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

Answer 1

then
x^2 - 6x + 14 = -x^2 - 20x - k
2x^2 + 14x + 14+k = 0
in this quadratic
a = 2
b= 14
c = 14+k

to have only one solution (root), the value of the discriminant has to be zero, thus
b^2 - 4ac = 0
14^2 - 4(2)(14+k) = 0

solve for k

Answer 2

Thanks!