Question

how can i sketch the curve z=k for the specified values of k? if z = x^2 + y ;
k = -2, -1, 0, 1, 2

Answer 1

yes but actually our topic in that is about contour plotting..

Answer 2

how can i sketch the curve z=k for the specified values of k? if z = x^2 + y ; k = -2, -1, 0, 1, 2

-2 = x^2 + y

-1 = x^2 + y

0 = x^2 + y

1 = x^2 + y

2 = x^2 + y

sketch = http://www.wolframalpha.com/input/?i=plot+-2+%3D+x^2+%2B+y+and+-1+%3D+x^2+%2B+y+and+0+%3D+x^2+%2B+y+and+1+%3D+x^2+%2B+y+and+2+%3D+x^2+%2B+y+from+-3+to+3

Answer 3

how can i plot that equations?

Answer 4

how can i plot those equations?
-2, -1, 0, 1, 2

-2 = x^2 + y

-1 = x^2 + y

0 = x^2 + y

1 = x^2 + y

2 = x^2 + y

Answer 5

-2 = x^2 + y

y = -2 - x^2

It is a parabola

Answer 6

ah..ok tnx but how can i know that it is a parabola?

Answer 7

@jedai17 Because hopefully you took an Algebra II class before taking Calculus III

and you remember studying conic sections and how to graph parabolas

if not, here is a review http://tutorial.math.lamar.edu/Classes/Alg/Parabolas.aspx

Answer 8

im sorry i forgot about it..

Answer 9

that is why i provided a link for you to review

Answer 10

ok thankyou i will review it..

Answer 11

btw it it the same if z = k in z = x^2 +9y^2 ; k = 0, 1, 2, 3, 4 ??

Answer 12

Yes. Except k = x^2 + 9y^2 will be an ellipse rather than a parabola.

Answer 13

ah ok...thank you so much :)

Answer 14

how to find the vertex of this equation?
y = -2 - x^2 ?

Answer 15

Here is the sketch of your ellipsoid variant

http://www.wolframalpha.com/input/?i=plot+0+%3D+x^2+%2B+9y^2+and+1+%3D+x^2+%2B+9y^2+and+2+%3D+x^2+%2B+9y^2+and+3+%3D+x^2+%2B+9y^2+and+4+%3D+x^2+%2B+9y^2+from+-3+to+3

Answer 16

\[y=-x^2-2\]
vertex x value = -b/2a
\[ax^2+bx+c\]

a = -2, b = 0, c = -2

-b/2a = - ( 0 ) / 2( - 2 ) = 0

So vertex occurs at x = 0, or ordered pair (0, -2)

Answer 17

in this graph
http://www.wolframalpha.com/input/?i=plot+0+%3D+x^2+%2B+9y^2+and+1+%3D+x^2+%2B+9y^2+and+2+%3D+x^2+%2B+9y^2+and+3+%3D+x^2+%2B+9y^2+and+4+%3D+x^2+%2B+9y^2+from+-3+to+3
why it has an x interval from -3 to 3?

Answer 18

I made it from -3 to 3. It looks ugly if it is from 0.000001 to - 3250123989.

It is an arbitrary domain I chose to give you a full picture of the graph.

Answer 19

ah ok..tnx

Answer 20

in this
y=−x2−2

vertex x value = -b/2a
ax2+bx+c


a = -2, b = 0, c = -2

-b/2a = - ( 0 ) / 2( - 2 ) = 0

So vertex occurs at x = 0, or ordered pair (0, -2)
why the value of a is -2? i think it is -1..

Answer 21

It is a = -1. My mistake.

Answer 22

But since it is -b / 2a, and b = 0, it is fortunately irrelevant in this case.

Answer 23

its ok thanks again

Answer 24

what do u mean by irrelevant?

Answer 25

If -b / 2a, and b = 0, does it matter what a is ?

No.

0 / anything = 0.

So what is the difference between

0 / 2(-2)

and

0 / 2(-1)

They both equal 0. Even though I made a mistake on my a value, we were fortunately ok because a did not effect the outcome.

irrelevant - Not connected with or relevant to something.

Answer 26

ah ok thank you so much for this infos.. :)