Question

Can someone help me graph this rational function? I'm having trouble understanding it. f(x) = two divided by quantity x squared minus two x minus three

Answer 1

\[f(x) = 2 \div x ^{2} - 3\]

Answer 2

correct?

Answer 3

\[f(x)=\frac{2}{x^2-2x+3}\]

Answer 4

woops sorry

Answer 5

Yes satellite73 that's what it looks like

Answer 6

your sure it is \(+3\) and not \(-3\) at the end right?

Answer 7

Yes

Answer 8

that's too bad, cause that would have been easier

Answer 9

Since when do they like to make it easy for us? XD

Answer 10

first thing to look for is the zeros of the denominator, because that will give you the vertical asymptotes
but this denominator has not zeros

Answer 11

true

Answer 12

so the domain is all real numbers, no vertical asymptotes for this one
that makes things harder

Answer 13

but not too bad
it should be clear that the horizontal asymptote is \(y=0\) right, or no?

Answer 14

Does it have a range?
Why is the horizontal asymptote y = 0?

Answer 15

last question first
the numerator is a fixed number, it is 2
the denominator will get larger and larger the bigger x gets
so you will have a fraction with a numerator of 2, and a denominator that gets large without bound
think of a fraction like \(\frac{2}{1,000,000}\)

Answer 16

another way to check is that the denominator is a polynomial of degree 2 and the numerator is a polynomial of degree zero ( a constant) and since the degree of the top is smaller than the degree of the bottom, the vertical asymptote is \(y=0\)
that is always true

Answer 17

as for the range, if it is clear than
\[x^2-2x+3\] has no zeros, it should also be clear that it is always positive
and since \(2\) is positive that means
\[\frac{2}{x^2-2x+3}\] is always positive
so one part we know about the range is that \(y>0\)

Answer 18

one more step, but let me know if i lost you yet

Answer 19

No you are making sense to me. So the horizontal asymptote is y = 0, there are no vertical asymptotes here, the domain is all real numbers and the range is y >0. Is that all correct?

Answer 20

all but the last part
the domain is not \(y>0\) that is the lower bound for the domain, we have to find the upper one
that is why i said "one more step"
the domain will be
\[0<y\leq something\]

Answer 21

the numerator is fixed the denominator gets bigger as x goes to infinity or to minus infinity
if we find the minimum value of the denominator that will give the maximum value of the function

Answer 22

that is identical to finding the second coordinate of the vertex of
\[y=x^2-2x+3\]
first coordinate is \(-\frac{b}{2a}=-\frac{-2}{2}=1\) and the second coordinate is
\[1-2+3=2\]

Answer 23

so the very largest this can be is when the denominator is 2 and you would get
\[f(1)=\frac{2}{2}=1\]

Answer 24

so the domain is a measly
\[0<y\leq 1\]