Question 739462
What you wrote, h(x)=2x+1/x-3, is {{{h(x)=2x+1/x-3}}} , {{{graph(200,200,-4,4,-8,6,2x+1/x-3,2x-3)}}}
But maybe you meant h(x)=(2x+1)/(x-3), which is {{{h(x)=(2x+1)/(x-3)}}} , {{{graph(200,200,-9,15,-15,20,(2x+1)/(x-3),2)}}}
Both functions get as low as {{{-infinity}}}
and as high as {{{infinity}}},
but we have to figure out if there are some in-between values that the functions cannot achieve. 
I will help you discover what each of those functions does.
 
EASIEST FIRST:
{{{h(x)=(2x+1)/(x-3)=(2x-6+7)/(x-3)=(2x-6)/(x-3)+7/(x-3)=2+7/(x-3)}}}
When {{{x}}} has a very large absolute value, the last term gets very small and the value of the function gets very close to {{{2}}}.
You can get as close as you want, by making that last term as small (in absolute value) as necessary.
So the line {{{y=2}}} is a horizontal asymptote.
However, {{{(x)=(2x+1)/(x-3)=2+7/(x-3)}}} can not equal {{{2}}} , because the last term can never be zero.
So you could say that the range of that function is all the real numbers, except 2.
 
THE OTHER FUNCTION:
{{{h(x)=2x+1/x-3}}} does something different.
When {{{x}}} has a very large absolute value, the term {{{1/x}}} gets very small and the value of the function gets very close to the value of {{{y=2x-3}}}.
You can get as close as you want, by making that term {{{1/x}}} as small in absolute value as necessary.
So the line {{{y=2x-3}}} is an oblique asymptote.
For {{{x<0}}}, {{{h(x)<-3}}} and
for {{{x>0}}}, {{{h(x)>-3}}}
so the function never takes the value {{{-3}}}, but there are other forbidden values.
You can figure out that {{{h(x)=2x+1/x-3}}} cannot take any values between {{{-3-2sqrt(2)}}} and {{{-3+2sqrt(2)}}} and that means that
THE RANGE IS all the real numbers {{{y}}} such that
{{{y<=-3-2sqrt(2)}}} or {{{y>=-3+2sqrt(2)}}}
You may want to express that in whatever notation is expected of you,
maybe as a union of intervals:
(-infinity,-3-2sqrt(2)] U [-3+2sqrt(2),infinity)
 
If you know about derivatives, you would find that the derivative of {{{h(x)}}} is
h'(x)={{{2-1/x^2}}} , which has zeros at {{{x=sqrt(2)/2}}} and {{{x=-sqrt(2)/2}}}
That means that the function will have a maximum with {{{h(x)<-3}}} at {{{x=-sqrt(2)/2}}}
and a minimum with {{{h(x)>-3}}} at {{{x=sqrt(2)/2}}}
You could calculate them as
{{{h(-sqrt(2)/2)=2(-sqrt(2)/2)+1/(-sqrt(2)/2)-3=-sqrt(2)-sqrt(2)-3=-2sqrt(2)-3}}} and
{{{h(sqrt(2)/2)=2(sqrt(2)/2)+1/(sqrt(2)/2)-3=sqrt(2)+sqrt(2)-3=2sqrt(2)-3}}}
 
Without mentioning derivatives, you could figure out what values of
{{{y=2x+1/x-3}}} are possible for some real number {{{x}}} by solving that equation for {{{x}}}
{{{y=2x+1/x-3}}} --> {{{yx=2x^2+1-3x}}} --> {{{2x^2-3x-yx+1=0}}} --> {{{2x^2-(3+y)x+1=0}}}
You know that there will be one or two solutions wherever the discriminant is not negative, meaning when
{{{(3+y)^2-8>=0}}} --> {{{9+6y+y^2-8>=0}}} --> {{{y^2+6y+1>=0}}}
The solutions to {{{y^2+6y+1>=0}}} are {{{y=-3 +- 2sqrt(2)}}} and
the solution for {{{y^2+6y+1>=0}}} is the range of {{{h(x)=2x+1/x-3}}} :
{{{y<=-3-2sqrt(2)}}} and {{{y>=-3+2sqrt(2)}}}