Graphing (x^2)y + 3y^2 = 5x + 9 or the like?

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Homework Help Overview

The discussion revolves around graphing the equation (x^2)y + 3y^2 = 5x + 9, particularly in the context of verifying differentiation results through graphical methods. The original poster expresses difficulty in graphing equations that include both linear and quadratic terms in y.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the possibility of solving the equation as a quadratic in y, with some questioning how to approach this due to the presence of x. Others suggest using software tools like Mathematica for graphing, indicating a shift from traditional methods.

Discussion Status

The conversation is ongoing, with some participants providing algebraic forms for potential solutions while others explore alternative graphing methods. There is no explicit consensus on the best approach, but various lines of reasoning are being examined.

Contextual Notes

The original poster mentions a preference for manual graphing methods and expresses concern about relying on technology, indicating a desire for algebraic practice despite the challenges posed by the equation's structure.

nousername
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Homework Statement



Hey everyone. I have a mathematics exam coming up, and I am always used to checking my answers using an alternative method. We have differentiation questions and i normally check them by graphing the equation on my calc and then graphing my calculated differential equation and see if the slope of the original function at any x-value (my calculator can calculate slopes) is the same as the y-value for my differential equation.

My problem is, sometimes there are equations such as these: x2y + 3y2 = 5x + 9 which i can't graph because they have the y^2 and another y which isn't squared. My question is, does anybody know a way around this?

Homework Equations



x2y + 3y2 = 5x + 9

The Attempt at a Solution



I have no idea, i looked online but i didn't know what to google, so all the things that came up were just irrelevant.

Any help would be appreciated.

Thank you.
 
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Solve it as a quadratic equation for y.

ehild
 
How do you do that? It has an x in it... Can you show me how? Thanks
 
The solution(s) of the quadratic equation will be functions of x.

3 y^2+(x^2)y-(5x+9)=0

y_1=\frac{-(x^2)+\sqrt{x^4+12(5x+9)}}{6}

y_2=\frac{-(x^2)-\sqrt{x^4+12(5x+9)}}{6}

ehild
 
nousername said:
My question is, does anybody know a way around this?

[Any help would be appreciated.

Thank you.


Yeah, I gotta' way around this but you may not like my answer. It is however I feel the best approach to solving your dilemma: time to step into the 21st century and do away with that calculator and begin using Mathematica then just ContourPlot it:

Code:
ContourPlot[x^2 y + 3 y^2 == 5 x + 9, {x, -15, 15}, {y, -15, 15}]

Now I understand you need the algebra practice solving them manually but eventually you run into one you can't like:

x^5 y^4 + 3 y^3 x == 5 x^3 y^3 + 9


and that one too can be plotted nicely via ContourPlot
 

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