Question Number 25656 by rita1608 last updated on 12/Dec/17
$${trace}\:{the}\:{curve}\: \\ $$$${y}^{\mathrm{2}} \left({x}+\mathrm{1}\right)={x}^{\mathrm{2}} \left(\mathrm{3}−{x}\right) \\ $$$${clearly}\:{stating}\:{all}\:{the}\:{properties}\:{used} \\ $$$${for}\:{tracing}. \\ $$
Commented by ajfour last updated on 12/Dec/17
Commented by ajfour last updated on 12/Dec/17
![y^2 =((x^2 (3−x))/(x+1)) > 0 ⇒ (x−3)(x+1) < 0 so x∈ (−1, 3] as x→ −1 , y→±∞ double root at x=0 . symmetry about x-axis coz of y^2 .](https://www.tinkutara.com/question/Q25668.png)
$${y}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} \left(\mathrm{3}−{x}\right)}{{x}+\mathrm{1}}\:\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left({x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)\:<\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{so}\:{x}\in\:\left(−\mathrm{1},\:\mathrm{3}\right] \\ $$$${as}\:{x}\rightarrow\:−\mathrm{1}\:\:\:,\:{y}\rightarrow\pm\infty \\ $$$${double}\:{root}\:{at}\:{x}=\mathrm{0}\:. \\ $$$${symmetry}\:{about}\:{x}-{axis}\:{coz}\:{of}\:{y}^{\mathrm{2}} . \\ $$
Commented by rita1608 last updated on 13/Dec/17
$${its}\:{a}\:\mathrm{10}\:{mark}\:{question}\:{plss}\:{state}\:{all} \\ $$$${properties}\: \\ $$