What effect does changing the function f(x)=2sin(x/2)−1 to the function g(x)=2sin(x)−5 have on the graph of f(x)?

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Step-by-step explanation:

The first function is a sin function with an amplitude of 2, and a period of 4pi.  The -1 moves it down from its original position 1 unit.  There is no phase shift here.  The normal period of a sin function is 2pi, which means that one revoution of the graph (one Up-down-up to put it in simplest terms) starts at 0 and ends at 2pi.  The x/2 means that the value to consider for the period change is 1/2.  The formula to solve for a period change is 2pi/that value (most books call it b).  So we have 2pi/(1/2), which algebraically solves to 4pi.  That means that one up-down-up of the function starts at 0 and ends at 4pi.

That second function doesn't have a b value, so its period is still 2pi.  The -5, however, moves the whole graph down 5 units from the original position (which is 4 units below the first graph).  The second function also has an amplitude of 2. I hope this helps a bit!

3.how is solving radical equations different than solving linear equations?

step-by-step explanation: