Graph Parabolas: Quadratic Equations Explained

[Imparcticle]

How do I graph parabolas/quadratic equations?

 

[cookiemonster]

Either use a calculator, or just plot points.

cookiemonster

 

[JonF]

Ick, your making me remember how to graph stuff without calc?

Ill start off by assuming you mean a straight forward parabola without 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…

 

[JonF]

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

 

[Chrono]

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]

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.

 

[Evgeny]

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]

how about graphing something like y=3x+5b+3?

 

[vanuatu]

You have got more variables in there than I am am comfortable to deal with!

 

[Parth Dave]

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 can't graph it) that has a y-intercept of 5b + 3 and a slope of 3..

 

[Chrono]

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]

did u mean y = 3x^2 + 5x + 3?

Yes. (my bad)

 

[BryceG]

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]

If b is a third variable (call it z, for familiarity), then I believe you are looking at the equation of a plane.