Question 128507
# 1


d)


{{{y=(7)/(x+9)}}} Start with the given function



{{{x+9=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.




{{{x=0-9}}}Subtract 9 from both sides



{{{x=-9}}} Combine like terms on the right side






Since {{{x=-9}}} makes the denominator equal to zero, this means we must exclude {{{x=-9}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq-9\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-9}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -9\right)\cup\left(-9,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -9 from the domain


---------------------------------------------------------





Now to find the range, simply graph the function to get



note: ignore the vertical lines, they are not part of the graph and are asymptotes

{{{graph(500,500,-15,5,-10,10,(7)/(x+9))}}}



Looking at the graph, we can see that y can be any number <b>except</b> 0. So {{{y<>0}}}. 





So our range is:  *[Tex \LARGE \textrm{\left{y|y\in\mathbb{R} y\neq0\right}}]


which in plain English reads: y is the set of all real numbers except {{{y<>0}}}


So our range looks like this in interval notation

*[Tex \Large \left(-\infty, 0\right)\cup\left(0,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> 0 from the range



<hr>



e)






{{{y=(3)/(x^2-4)}}} Start with the given function



{{{x^2-4=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.





{{{(x-2)(x+2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)





Now set each factor equal to zero:


{{{x-2=0}}} or {{{x+2=0}}}


{{{x=2}}} or {{{x=-2}}}  Now solve for x in each case



So our solutions are {{{x=2}}} or {{{x=-2}}}




Since {{{x=-2}}} and {{{x=2}}} make the denominator equal to zero, this means we must exclude {{{x=-2}}} and {{{x=2}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq-2 and x\neq2\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-2}}} or {{{x<>2}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -2\right)\cup\left(-2,2 \right)\cup\left(2,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -2 and 2 from the domain




---------------------------------------------------------





Now to find the range, simply graph the function to get





note: ignore the vertical lines, they are not part of the graph and are asymptotes


{{{graph(500,500,-15,5,-10,10,(3)/(x^2-4))}}}



Looking at the graph, we can see that y can be any number <b>except</b> 0. So {{{y<>0}}}. 





So our range is:  *[Tex \LARGE \textrm{\left{y|y\in\mathbb{R} y\neq0\right}}]


which in plain English reads: y is the set of all real numbers except {{{y<>0}}}


So our range looks like this in interval notation

*[Tex \Large \left(-\infty, 0\right)\cup\left(0,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> 0 from the range