Answer
$c = \frac{2}{3}$
Work Step by Step
Clearly the graph is continuous on the intervals $(-\infty,2)$ and $[2,\infty)$
We need to find the value $c$ such that the graph is continuous at the point $x=2$.:
$cx^2+2x = x^3-cx$
$c(2)^2+2(2) = (2)^3-c(2)$
$4c+4 = 8-2c$
$6c = 4$
$c = \frac{2}{3}$
