(y^2)+2y-2x-7=0

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

(y^2)+2y-2x-7=0

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

y^2+2y= 2x+7 (y^2+2y+1)-1=2x+7 (y+1)^2=2x+8 (y+1)^2=2(x+4) which is a right handed parabola
is that u were looking for?
She has posted so many questions about graphing I don't even know what is the question anymore.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

im just not sure how to find the points for them
okay...lets try to draw it.....the above equation can be written as below (y-(1))^2=2(x-(-4)). which means its centre is (-4,-1). can u locate this point now?
yeah. i just dont know how to find it myself if that makes sense well here i have another one for practice, can you go through this one with me..... (y^2)+8y-2x+22=0
check this one out
1 Attachment
This is what I would do... switch the x and y, so that (x^2)+8x-2y+22=0, and also flip the x and y axis. So that you are essentially graphing a parabola, which is easy, and then flip the page back to normal, and you get the equation you want.
did u see that?
too much flipping....:D
ahh im so confused ha
which part r u confused of?
how to get the point and how to determine the height or width of it
i mean i know how to get the vertex but I dont know how to figure out how much it opens up
okay..for that put x=0 that will give u the point where it touches the y axis and vice versa...got it?
do you mean to put x=0 when its like this y=1/2x^2=4x=11 ?
(y+1)^2=2(x+4) is our parabola. lets put x=0 , we get (y+1)^2=2(0+4) or (y+1)^2=8 or y^2+2y-7=0 which is quadratic , solving it will give us two values of y which will give us two points where the parabola intersects the y-axis.. alright?
thank you

Not the answer you are looking for?

Search for more explanations.

Ask your own question