Graphing f(x,y) = 1 - y^2

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In summary, the graph of f(x,y) = 1 - y^2 is a symmetrical parabola that opens downwards along the y-axis. Changing the value of x will shift the parabola horizontally, but the shape will remain the same. Increasing the value of y will shift the parabola downwards and result in a narrower shape. The vertex of the graph can be found at (0,1) by setting y = 0 and solving for x. The graph is also symmetrical along the y-axis.
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joemama69
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Homework Statement



Sketch the graph of the function f(x,y) = 1 - y^2

Homework Equations





The Attempt at a Solution



see pic
 

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  • #2
That's a two dimensional graph- it looks like x= 1- y2. But you are asked to graph z= f(x,y)= 1- y2 in 3 dimensions. The graph is a parabolic cylinder. It's cross section is the parabola z= 1- y2 in the yz-plane but it extends infinitely along the x-axis.
 
  • #3
ahh brain fart


hows the new pic
 

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1. What is the shape of the graph of f(x,y) = 1 - y^2?

The graph of f(x,y) = 1 - y^2 is a parabola that opens downwards along the y-axis. It is a symmetrical curve that extends infinitely in both the positive and negative x-directions.

2. How does changing the value of x affect the graph of f(x,y) = 1 - y^2?

Changing the value of x will shift the parabola horizontally, but the shape will remain the same. A larger x-value will result in a rightward shift, while a smaller x-value will result in a leftward shift.

3. What happens to the graph of f(x,y) = 1 - y^2 when the value of y is increased?

As the value of y increases, the parabola will shift downwards along the y-axis. This results in a narrower shape as the curve approaches the x-axis.

4. How can we determine the vertex of the graph of f(x,y) = 1 - y^2?

The vertex of the graph can be found by setting y = 0 and solving for x. In this case, the vertex is at (0,1).

5. Is the graph of f(x,y) = 1 - y^2 symmetrical?

Yes, the graph is symmetrical along the y-axis. This means that if you were to fold the graph in half along the y-axis, the two halves would match up perfectly.

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