Sketching a hyperbola. (1 Viewer)

[Heliotrope]

So sketching a goddamn hyperbola is turning out to be the biggest stumbling block I've had all year. My textbooks tell me how to graph one with a table of values, but neither one tells me how to <i>sketch</i> one with the main points labeled. I've had several people try to explain it to me, but their explanations were all different and none of them made any sense. My Asian tutor who I can't understand taught me this really roundabout way of doing it that involved drawing the basic shape and then SHIFTING the plane, but I was all WTF? So I've come here for help. Can somebody please, please give me a step-by-step account of how the fuck you're supposed to sketch one? For example, this question:

d) Sketch the function y = 1/x+2 + 2, labelling all main features.

Thanks.

 

[-pari-]

lol dw, i still have problems with that.

 

[watatank]

So sketching a goddamn hyperbola is turning out to be the biggest stumbling block I've had all year. My textbooks tell me how to graph one with a table of values, but neither one tells me how to <i>sketch</i> one with the main points labeled. I've had several people try to explain it to me, but their explanations were all different and none of them made any sense. My Asian tutor who I can't understand taught me this really roundabout way of doing it that involved drawing the basic shape and then SHIFTING the plane, but I was all WTF? So I've come here for help. Can somebody please, please give me a step-by-step account of how the fuck you're supposed to sketch one? For example, this question:

d) Sketch the function y = 1/x+2 + 2, labelling all main features.

Thanks.

first put that over a common denominator

f(x) = [2x + 5] / [ x + 2]

find vertical assymptote, where denominator = 0

[ x + 2] = 0, .'. x= -2

horizontal assymptote is the limit where x ---> infinity

[2x + 5] / [ x + 2] so imagine x is a huge number those constants in that fraction isnt going to matter, so its going to tend to (2x/x) = 2

horizontal assymptote: y = 2

test points and sketch your curve.

 

[S1M0]

So sketching a goddamn hyperbola is turning out to be the biggest stumbling block I've had all year. My textbooks tell me how to graph one with a table of values, but neither one tells me how to <i>sketch</i> one with the main points labeled. I've had several people try to explain it to me, but their explanations were all different and none of them made any sense. My Asian tutor who I can't understand taught me this really roundabout way of doing it that involved drawing the basic shape and then SHIFTING the plane, but I was all WTF? So I've come here for help. Can somebody please, please give me a step-by-step account of how the fuck you're supposed to sketch one? For example, this question:

d) Sketch the function y = 1/x+2 + 2, labelling all main features.

Thanks.

its not as hard as your might think.

Okay, assuming that you know how to sketch a normal parabola (1/x), what you have there is simply y=1/x, except that the asymptote will be at -2:

Because...

x+2 = 0
x = -2.
The line can't pass through here because 1/0 is undefined.

To find the horizontal asymptote, you have to use limit x-->Infinity for 1/x+2 + 2

Then sketch and you're set.

 

[ellen.louise]

humb...

okay, for a pretty quick sketch you need to find the assymptotes, and maybe also where (if at all) the graph crosses the axes.

as x + 2 is the denominator, x + 2 cannot equal 0. so x = -2 is the vertical assymptote.

1/ (x + 2) cannot equal zero either, right? cause the denominator can't be zero. so you rearrange the equation so you get

y - 2 = 1/( x + 2)

from this you get that y - 2 cant be zero, so your horizontal assymptote is
y = 2.

Then go on with substituting in x = 0 and then y = 0 to get that the points of intersection with the axes are at

(-5/2, 0) and (0, 5/2)

mark in the assyptotes as dotted lines, mark in your points of intersection, and then sketch the basic hyperbola graph around that.

 

[watatank]

My Asian tutor who I can't understand taught me this really roundabout way of doing it that involved drawing the basic shape and then SHIFTING the plane, but I was all WTF? So I've come here for help. Can somebody please, please give me a step-by-step account of how the fuck you're supposed to sketch one? For example, this question:

d) Sketch the function y = 1/x+2 + 2, labelling all main features.

Thanks.

the first way takes longer, but

the roundabout way might be a bit hard to grasp, but i'll try to explain anyway.

you should know what y = 1/x looks like. hyperbola, assympotes are the x and y axes.

if you have something in the form y = 1/(x-a) your new vertical assymptote is 'a'.

so in the example given y = 1/x+2 is the same as y = 1/[ x-(-2) ] so that's why your assymptote is x = -2

y = 1/x+2 +2 (the bolded part) is just adding a constant and simply means a vertical shift. so it moves two spots up.

i don't suggest you do it this way until you can grasp the basics well enough though

 

[narrowpin]

dude when u add the two in ur first step isnt it (2x+5)/(x+2)?

 

[S1M0]

dude when u add the two in ur first step isnt it (2x+5)/(x+2)?

 

[watatank]

hahaha yeah. *edits post*

it doesn't have a HUGE bearing on what the final graph looks like though.

 

[Heliotrope]

okay, for a pretty quick sketch you need to find the assymptotes, and maybe also where (if at all) the graph crosses the axes.

as x + 2 is the denominator, x + 2 cannot equal 0. so x = -2 is the vertical assymptote.

1/ (x + 2) cannot equal zero either, right? cause the denominator can't be zero. so you rearrange the equation so you get

y - 2 = 1/( x + 2)
0
from this you get that y - 2 cant be zero, so your horizontal assymptote is
y = 2.

Then go on with substituting in x = 0 and then y = 0 to get that the points of intersection with the axes are at

(-5/2, 0) and (0, 5/2)

mark in the assyptotes as dotted lines, mark in your points of intersection, and then sketch the basic hyperbola graph around that.

Best explanation ever. I totally get how to do it now! Thank you.

you should know what y = 1/x looks like. hyperbola, assympotes are the x and y axes.

if you have something in the form y = 1/(x-a) your new vertical assymptote is 'a'.

so in the example given y = 1/x+2 is the same as y = 1/[ x-(-2) ] so that's why your assymptote is x = -2

y = 1/x+2 +2 (the bolded part) is just adding a constant and simply means a vertical shift. so it moves two spots up.

i don't suggest you do it this way until you can grasp the basics well enough though

I get this too! And it makes sense! Thank you.

 

[ellen.louise]

cool. hope that helps. luck!