Q.Find the slant (oblique) asymptote for the function $f(x)=\dfrac{x^2-6x+7}{x+5}$.

A horizontal asymptote is the limiting value of $f(x)$ as $x\to\pm\infty$; here that limit is $0$.

1. $f(x)=\dfrac1x$. As $x\to\infty$, $f(x)\to0$; as $x\to-\infty$, $f(x)\to0$ as well. …

Treating the curve's equation as a polynomial in $y$ shows the coefficient of $y^2$ vanishes at $x=2$, giving a genuine vertical asymptote, while treating it as a polynomial in $x$ shows the leading coefficient never vanishes, so there is no horizontal asymptote.

1. Rewrite the given curve $y^2(x-2)=x^2(1+x)$ as $f(x,y) = (x-2)y^2 - x^2(1+x) = 0$.
2. For an asymptote parallel to the $y$-axis, examine $f$ as a polynomial in $y$: the highest power of $y$ is $y^2$, with coefficient $(x-2)$.
3. Setting the coefficient of the highest power of $y$ to zero, $x-2=0 \Rightarrow x=2$. Checking that the curve genuinely becomes unbounded there: at $x=2$, the right side $x^2(1+x)=4(3)=12\neq0$, so as $x\to2$, $y^2=\dfrac{x^2(1+x)}{x-2}\to\pm\infty$ — confirming $x=2$ is a true vertical asymptote. …

The curve has a vertical asymptote $x=2$ (parallel to the $y$-axis); there is no asymptote parallel to the $x$-axis.

1. Write the curve as $y^2(x-2)=x^2(1+x)$, i.e. $y^2=\dfrac{x^3+x^2}{x-2}$, or in full polynomial form $x^3+x^2-xy^2+2y^2=0$.
2. Asymptote parallel to the $y$-axis: set the coefficient of the highest power of $y$ (here $y^2$, coefficient $x-2$) to zero: $x-2=0\Rightarrow x=2$. As $x\to2$, $y\to\pm\infty$, confirming $x=2$ is a genuine vertical asymptote.
3. Asymptote parallel to the $x$-axis: this requires the coefficient of the highest power of $x$ (here $x^3$, coefficient $1$) to vanish for some value — it is a nonzero constant and never vanishes, so there is no asymptote parallel to the $x$-axis. …

The curve touches the axis at the origin and crosses it again at $x=3a$, giving $x=0$ and $x=3a$.

1. The curve is $ay^2=x^2(3a-x)$, a standard cubic curve traced in the TN Class-12 syllabus.
2. To find where the curve meets the axis of $x$ (where $y=0$), substitute $y=0$: $a(0)^2=x^2(3a-x)\Rightarrow x^2(3a-x)=0$.
3. This factorises to $x^2=0$ or $3a-x=0$, giving $x=0$ (a repeated/double root, so the curve touches the axis and has a node/cusp at the origin) and $x=3a$ (a simple crossing). …

Real $y$ requires the right-hand side to be non-negative, which restricts $x$ to the closed interval $[-a,a]$.

1. The curve is $a^2y^2=x^2(a^2-x^2)$. For real values of $y$, we need $y^2\geq0$, so the left side $a^2y^2\geq0$ automatically — but for the equation to have a real solution for $y$ at a given $x$, the right side $x^2(a^2-x^2)$ must also be $\geq0$ (it must equal a non-negative quantity).
2. Since $x^2\geq0$ always, the sign of the product $x^2(a^2-x^2)$ is controlled by $(a^2-x^2)$ whenever $x\neq0$.
3. Requiring $a^2-x^2\geq0$ gives $x^2\leq a^2$, i.e. $-a\leq x\leq a$.
4. At $x=0$ the product is automatically $0\geq0$, which is consistent with (and already included in) the interval $-a\leq x\leq a$. …