Pre cal/cal master needed g(x)={ ax-3, x< -2 { x^2 -2x, x> -2 what value would the function g(x) have point discon. at x=-2 how to find?

Hello, I know this is a little late but been busy and noticed you responded on this and my profile. 

Ok so we want to find what value of a would the function g(x) have point discontinuity at x=-2.

We find this by subsituting -2 where the x is located in the functions x^2 -2x
x^2 -2x
-2^2 - 2(-2) =  8
Now to find a we use ax - 3 and insert -2 for x like the follow 
-2a - 3
Next for the limit to exist both limits have to equal the same so we have
-2a - 3 = 8
Now solve for a
-2a - 3 = 8
-2a - 3 + 3 = 8+3
-2a  = 11
-2a / -2 = 11 / -2
a = -5.5

So when a is at -5.5 the function is continuous and anything > or < will make it discontinuous. So when -5.5 < a < -5.5 a will have discontinuity.

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