Question 675431: Suppose we have a function: f(x,y)=(x+y)^2
Graph: f(3,y) & f(x,2)
I know its going to be a parabolic cylinder but I am struggling with showing the work?
Answer by Edwin McCravy(20067)   (Show Source):
You can put this solution on YOUR website!
Whew! You challenged me to draw it on here using HTML!!! That was
a lot of work, simulating 3D coordinates using 2D HTML coordinates!
Here's the first one graphed. It's easy to draw it on paper, but
with HTML it's pretty tedious.
f(x,y)=(x+y)² is a parabolic cylinder but f(3,y)=(3+y)² is only a cross
section of that cylinder, which is the red parabola which has its vertex
at (3,-3,0). It is in the plane x=3. The green rectangle is in the plane of
the parabola f(3,y) and it is 3 units in front of and parallel to the
yz-plane, That was a lot of work!
I'm too tired right now after that to do the f(x,2) but maybe I'll do it
tomorrow. f(x,2) = (x+2)² is also a parabola with vertex (-2,2,0) parallel
to and two units right of the xz-plane. Maybe you can figure out how to draw
that one yourself. Come back here tomorrow and see if I've done it. It's
going to be challenging to draw too, because its vertex (-2,2,0) is BEHIND
the yz-plane!!!. To add it to the graph above we'll have to extend the
x-axis back BEHIND the yz-plane. The one above was easier to draw because its
vertex is in front of the yz-plane. I'll see what I can do tomorrow -- maybe!
:) Luck!
Edwin
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