The notation makes this one sneaky. You need a value when x = 1 to equal 2.
You need a value at x = 2 that makes 4x = 8
2 equations, 2 unknowns. This think has to have an answer.
cx^2 + d = 2 when x = 1
c(1)^2 + d = 2
c + d = 2
cx^2 + d = 8 when x = 2
4c + d = 8
c + d = 2 Subtract
3c = 6
c = 2
Now go back and solve for d
c + d = 2
d = 0
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That should make it continuous at every point.
You should take note that h(x) has a problem at x = 2. The graph is well behaved at less than x =2 and it is well behaved at x > 2. It is just x = 2 that's the double ugly part of the problem. Check your givens above to make sure you understand what I'm saying. If you don't give me a shout.