anonymous
  • anonymous
find the vertices for the ellipse (x+1)^2 + 4(y+3)^2 = 9
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
standard form of an ellipse with vertices at h and k is given by \[(x-h)^{2}/a ^{2} + (y-k)^{2}/b ^{2}= 1\]
anonymous
  • anonymous
so compare and get the answer !
anonymous
  • anonymous
so the vertices are h and k right??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes in that given general equation its, h, k.. so in your equation what is h and k?
anonymous
  • anonymous
it would be 1 and 3
anonymous
  • anonymous
No.. check properly.. if it was 1 and 3.. the equation to the ellipse would be (x-1)^2 + 4(y-3)^2 = 9
anonymous
  • anonymous
so.. what is it??
anonymous
  • anonymous
it would be -1 and -3 ??
anonymous
  • anonymous
Bingo!
anonymous
  • anonymous
so thats the vertices??and thats it??
anonymous
  • anonymous
Yup that's it.. kind of a puny question eh?? Go ask your teachers to give you better problems to solve :P
phi
  • phi
You should check your definitions (google works) See http://www.mathwords.com/v/vertices_of_an_ellipse.htm for what the vertex is. You found the center of the ellipse
phi
  • phi
Your equation is (x+1)^2 + 4(y+3)^2 = 9 in standard form this is \[ \frac{(x+1)^2}{3^2}+\frac{(y+3)^2}{(\frac{3}{2})^2}=1 \] The semi-major axis is the longest axis. Here it is 3 (denominator of x term) the center is (-1,-3) and when y= -3 you are on the semi-major axis. x= -1+3= 2 or x= -1-3= -4 and the vertices are (-4,-3) and (2,-3)
anonymous
  • anonymous
Here is a graph of it
1 Attachment
anonymous
  • anonymous
thank you both :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.