Need help finding a constant k at which the function is continuous everywhere.

Answer:
k = -6/35
Step-by-step explanation:
To make the function continuous
kx^2 = x+k
These must be equal where the function is defined for two different intervals
This is at the point x=-6 so let x=-6
k(-6)^2 = -6+k
36k = -6+k
Subtract k from each side
36k-k = -6+k-k
35k = -6
Divide by 35
35k/35 = -6/35
k = -6/35
Answer: 36k = k - 6, k = -6/35
Step-by-step explanation:
The two parts are continuous by themselves. If either weren't, no value of k could fix it.
The only point of discontinuity is x=-6. The function will be continuous for any and all values of k which make k(-6)^2 = -6+k. 36k = k-6, 35k=-6, k=-6/35.