View Full Version : Quadratic graphs
Imparcticle
May2-04, 05:25 PM
How do I graph parabolas/quadratic equations?
cookiemonster
May2-04, 06:10 PM
Either use a calculator, or just plot points.
cookiemonster
Ick, your making me remember how to graph stuff with out calc?
Ill start off by assuming you mean a straight forward parabola with out a rotation or anything.
First you need to put the parobala in the form:
Y = a(x – h)^2 + k
Or
X = a(y – h)^2 + h
Where a=1/4p
The position of your vertex (the extreme point) is (h,k)
The focus is (h, k + p) and the directrix line is y = k – p
The other 2nd degree quadratics have similar geometric formulas. Maybe you can show us what you need to graph…
Oops forgot to tell you how to find which way it opens.
For
X = a(y – h)^2 + h
If a < 0 then it opens to the left
If a > 0 then it opens to the right
For
Y = a(x – h)^2 + k
If a < 0 then it opens down
If a > 0 then it opens up
Either use a calculator, or just plot points.
cookiemonster
I think it will be easier to just plug in points and go from there.
MathematicalPhysicist
May3-04, 01:27 PM
I think it will be easier to just plug in points and go from there.
to make a parabola graph you only need three points, if less then it's not kosher.
Aside from the simple y=(x-a)^2+b formulas, there's also conics equations for parabolas:
4py=x^2, where p is the distance to your focus or directrix. (the focus is an arbirtrary point, and the directrix is a line whos equation is y=-d, where d is the distance from y=0 to focus).
As far as graphing, either get a simple table of values, or get three points: two X intercepts and one Y intercept. To get the X intercepts, simply substitute 0 into y so that your equation looks like 0 = (x-a)^2 + b, which shouldn't be too hard to solve. For the Y intercept, do the same, but substitute x as 0, so that you end up with y=(0-a)^2+b , which should be pretty easy to get as well.
after you're done that, plot your points, and draw a curve through them.
Imparcticle
May3-04, 08:31 PM
how about graphing something like y=3x+5b+3?
:eek: You have got more variables in there than im am comfortable to deal with!
Parth Dave
May3-04, 09:14 PM
did u mean y = 3x^2 + 5x + 3?????
if u didnt:
thats just a simple linear equation (im assuming that b is a constant - if its a variable u cant graph it) that has a y-intercept of 5b + 3 and a slope of 3..
to make a parabola graph you only need three points, if less then it's not kosher.
You're right, of course. For some reason I was thinking of a line when I said what I did.
Imparcticle
May4-04, 08:41 PM
did u mean y = 3x^2 + 5x + 3?????
Yes. (my bad) :rolleyes:
for that kind i just use a graphics calculator, i think if you want to look at that and draw it you have to play with the equation a bit till its easier. i 4got the format i used last year for this type of thing.
Gokul43201
May6-04, 08:50 PM
If b is a third variable (call it z, for familiarity), then I believe you are looking at the equation of a plane.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.