Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

The curve does not exist for -3<x<3 because there can be no solutions for y when x^2<9.

Because of the presence of x^2 and y^2 the curve is essentially reflected in all quadrants. If we examine QI only we have the same patterns reflected in the other quadrants. x=-3 and x=3 are asymptotic.

As x gets larger, the constant 9 becomes more insignificant and y^2 becomes x^2, in other words, the curve in QI has an asymptote y=x, which is reflected in all the other quadrants. So we know the curve is trapped between the vertical asymptote and the line y=x.

(2ydy/dx)(x^2-9)+2y^2x=4x^3 is the derivative.

When dy/dx=0, there is a turning point. y^2=2x^2.

That is, x^4/(x^2-9)=2x^2. x^4=2x^4-18x^2; x^2=2x^2-18; x^2=18, x=3sqrt(2) and y^2=36, so y=6. In QI then, there is a turning point at (3sqrt(2),6). This is reflected in each quadrant.

This is a minimum (in QI) because the curve is trapped between the above asymptotic.

There is enough information now to trace the curve(s).

The asymptotic resemble a cross intersected by vertical lines.

The asymptotic in QI and QIV has been drawn in to illustrate how close the curve comes to it.