SOLUTION: A. Evaluate the given expression B. Find the domain of the function C. Find the Range h(z)= 1/z+7 Find h(-5) Having some issues with this problem. I'm not sure if I am

Question 890437: A. Evaluate the given expression
B. Find the domain of the function
C. Find the Range
h(z)= 1/z+7
Find h(-5)
Having some issues with this problem. I'm not sure if I am placing the -5 in the correct place. Answers are always wrong for what the book says. Please help.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

$$$

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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