Question

I just need a little help with graphing rational functions.

Answer 1

\[f(x)\frac{ 1 }{ x+1 }\] i know if theres no x on top y=0 but does that also mean that theres no point (x,y) to graph

Answer 2

what's the vertical asymptote here?

Answer 3

well , compare to standerd graph 1/x
do some shifting

Answer 4

if you're not sure, set the denominator x+1 equal to zero and solve for x

x+1 =0

x = ??

Answer 5

or ur not allowed to do that /.?

Answer 6

@jim_thompson5910 i know x+1=0 and x=-1 and y=0 because there no x on top

Answer 7

You are correct. So the vertical asymptote is x = -1 and the horizontal asymptote is y = 0.

Answer 8

i did the asymptotes, im figuring out how to do the curves like ikram002p did

Answer 9

once you have those boundaries set up, you plug in various x values to get corresponding y values

Answer 10

for instance,

if x = 0, then

y = 1/(x+1)
y = 1/(0+1)
y = 1/1
y = 1

When x = 0, y = 1

So the point (0,1) is on this curve

Answer 11

You repeat that for other x values to get more points. Once you have enough points, you can plot them all and draw a curve through them all.

The graph is on this PDF (see last page, but be sure to read through it all)

Answer 12

ohhh okay that makes sense.

Answer 13

Also, here's a table of values (generated with geogebra, but you can use any calculator to get this)

Answer 14

I'm glad it's clicking now

Answer 15

@jim_thompson5910 okay how do you solve it when its like this? \[f(x)=\frac{ 1 }{ x-2 }+1 \]

Answer 16

What are the asymptotes?

Answer 17

2 and 0?

Answer 18

x = 2 is the vertical asymptote

y = 0 is NOT the horizontal asymptote

Answer 19

y = 0 is the horizontal asymptote of 1/(x-2)

however, that +1 at the end shifts everything up 1 unit

Answer 20

so the horizontal asymptote of y = 1/(x-2) + 1 is y = 1

Answer 21

ohh okay.

Answer 22

I get it know can i ask you a domain and range question now for these? the first problem i posted. i know domain is all real numbers except x=-1 but would range be all real numbers except y=0? or just all real numbers?

Answer 23

now* not know

Answer 24

you kick out y = 0 because there is no way to have the function produce y = 0 (no matter which x value you pick)

So the range for the first one is the set of all real numbers but y can't equal 0