Answer to Question #122440 in Calculus for Bibhuti bhusan dehury

Question

Trace the equation of curve y^2=x^2(x+1)

Answer 1

$\text{The curve has two branches. It can be considered as the union of two functions}: \\[1 em] y^2=x^2(x+1)\rightarrow y=\sqrt{x^3+x^2 }=x\sqrt{x+1 } ~~\text{and}~~ y=-x\sqrt{x+1 } \\[1 em] \text{The domain is the set of non-negative real numbers for both the functions}\\[1 em] \therefore \text{The domain } x+1 \geq 0\rightarrow x\geq -1 \\[1 em] \text{As x } \rightarrow \infty\Rightarrow y=x\sqrt{x+1 }\rightarrow \infty ,y=-x\sqrt{x+1 } \rightarrow -\infty \\[1 em] \therefore \text{The range } =\left ( -\infty ,\infty \right )\\[1 em] \text{Intersection with x axis is determined by a substitution }y=0\rightarrow x=1 ,x=-1\\[1 em] \therefore \text{Intersection with x axis is} \left ( 0,0 \right ),\left ( -1,0 \right )\\[1 em] \text{Intersection with y axis is determined by a substitution }x=0\rightarrow y=0 \\[1 em] \therefore \text{Intersection with y axis } is\left ( 0,0 \right )\\[1 em] \text{ Now the table can be made giving some values of the x }\\[1 em] x=-1\rightarrow y=0~~~~~~~ ,~~~~~~x=3\rightarrow y=\pm6 \\[1 em] x=8\rightarrow y=\pm24~~~~~ ,~~~~~~ x=15\rightarrow y=\pm60~~~~~ ,~~~~~~ x=24\rightarrow y=\pm120\\[1 em]$