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You can't divide by zero right? It's undefined and in terms of a graph it becomes a vertical asymptote so you could say x is restricted when your denominator is 0, right?...
well if x-6=0, x=6... So what happens when we actually plug that in to our function? y=(6-4)/(6-6)=2/0. so the value of y(6) is undefined. Meaning x is restricted by that value. It can be any other real number, just not 6.
well if we plug in -6... y(-6)=(-6-4)/(-6-6)=-2/-12=1/6, which is a real value. Not restricted where t=-6. Only restriction is t=-6
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