Question 890437
you are either dealing with h(z) = (1/z) + 7 or you are dealing with h(z) = 1/(z + 7)


we'll do (1/z) + 7 first.


the domain is all real numbers except z = 0 because that makes the denominator equal to 0 and that would make the answer undefined.


the range would be all real numbers except 7 since h(z) can be negative or positive but will never be 7 because 1 / z will never be equal to 0 since the numerator of that fraction will never be equal to 0.  
because 1/x will never be equal to 0, you will never be able to get 0 + 7 which means you will never be able to get h(z) = 7.


when z = -5, h(z) becomes 1/-5 + 7 which becomes 7 - 1/5 which becomes 6 and 4/5 which is equal to 6.8.


the coordinate point of the equation when z = -5 becomes (-5,6.8).


here's the graph of y = 1/x + 7.


you can see that you have a vertical asymptote at x = 0 and a horizontal asymptote at y = 7.


the only difference is I replaced z with x in order to be able to graph it.


h(z) is replaced with y.
z is relaced with x.


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now we'll look at h(z) = 1/(z + 7)


the domain is all real numbers except z = -7 because that would make the denominator equal to 0.


the range is all real numbers except 0 because the numerator of the fraction 1/(z + 7) will never be equal to 0.


h(-5) becomes 1 / (-5 + 7) becomes 1 / 2 which is equal to .5.


the coordinate point when z = -5 becomes (-5,.5).


here's the graph of y = 1/(x+7).


you can see that you have a vertical asymptote at x = -7 and a horizontal asymptote at y = 0.


once again, y = h(z) and x = z for graphing purposes.


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