Looking for something else?

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

## More answers

Looking for something else?

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

- anonymous

x^2+y^3+4y=4 ,what is the equation of the tangent to the curve of the above equation at -1,1

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.

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

- anonymous

x^2+y^3+4y=4 ,what is the equation of the tangent to the curve of the above equation at -1,1

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

implicit derivation:
2x + 3(y^2)(y') + 4(y') = 0
y' = (-2x) / (3y^2 + 4)
replace: y'(-1, 1) = 2/7

- TuringTest

The comment was withdrawn, but it was stated earlier that (-1,1) is not on the graph of x^2+y^3+4y=4. That is true.

- anonymous

Although the point is not a point of tangency, doesn't necessarily mean that there does not exist a line tangent to the curve.
In fact if one takes the derivative using implicit differentiation and substitute it into the point slope equation for the slope one can find the point of tangency. One can then use this point of tangency to find the equation of the line.
I used winplot to find the point....
here is the equation of the line tangent to the graph passing through (-1,1).
(y-0.82342576161)=((-2*0.38469416012450)/(3*(0.82342576161)^2+4))(x-0.38469416012450
)

Looking for something else?

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

- anonymous

One could also use the following equation:
(y-1)=((-2*0.38469416012450)/(3*(0.82342576161)^2+4))(x+1)

- anonymous

abermejo was correct if the point had been a point of tangency.
And it is his equation for the slope that is used in my equation of the line.

- anonymous

thax loads all of you

Looking for something else?

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