Sketch the graphs of the following functions: (i) y=-dfrac13(x³-3x+2) (ii) y=x√(4-x) (iii) y=(x²+1)/(x²-4) (iv

Exercise 7.9 · Q2

Q. Sketch the graphs of the following functions:

(i)

y=-\dfrac13(x^3-3x+2)

(ii)

y=x\sqrt{4-x}

(iii)

y=\dfrac{x^2+1}{x^2-4}

(iv)

y=\dfrac{1}{1+e^{-x}}

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Concept understanding — Curve Sketching and Asymptotes

Symmetry. A curve $f(x,y)=0$ is:

symmetric about the $y$ -axis if $f(x,y)=f(-x,y)$ for all $(x,y)$ on it (i.e. $(x,y)$ on the curve $\Rightarrow(-x,y)$ is too);
symmetric about the $x$ -axis if $f(x,y)=f(x,-y)$ (i.e. $(x,y)$ on it $\Rightarrow(x,-y)$ is too);
symmetric about the origin if $f(x,y)=f(-x,-y)$ (i.e. $(x,y)$ on it $\Rightarrow(-x,-y)$ is too — equivalently, the curve is unchanged by a $180^\circ$ rotation about the origin).

Asymptotes. An asymptote is a straight line the curve approaches (the gap shrinking to $0$ ) as the point on the curve runs off to infinity. Three kinds:

Horizontal asymptote $y=L$ : holds if $\displaystyle\lim_{x\to+\infty}f(x)=L$ or $\displaystyle\lim_{x\to-\infty}f(x)=L$ (the two one-sided limits may give different horizontal asymptotes).
Vertical asymptote $x=a$ : holds if $\displaystyle\lim_{x\to a^{-}}f(x)=\pm\infty$ or $\displaystyle\lim_{x\to a^{+}}f(x)=\pm\infty$ — typically where a rational function's denominator vanishes while the numerator does not.
Slant (oblique) asymptote $y=mx+c$ : occurs for a rational function when the numerator's degree is exactly one more than the denominator's. Found by polynomial long division: writing $\dfrac{p(x)}{q(x)}=(\text{quotient})+\dfrac{\text{remainder}}{q(x)}$ , the quotient (a linear expression) is the slant asymptote, since the remainder term $\to0$ as $x\to\pm\infty$ .

Sketching a curve $y=f(x)$ — the seven-point checklist (used, in this order, throughout Examples 7.69–7.72 and Exercise 7.9 Q2):

Domain and range of $f$ .
Intercepts — set $y=0$ for $x$ -intercepts, $x=0$ for the $y$ -intercept (where each exists).
Critical points — solve $f'(x)=0$ and note where $f'$ fails to exist.
Local extrema — classify each critical point (first or second derivative test) and record the extreme value. …

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Each part works through domain, intercepts, monotonicity ( $f'$ ), concavity/inflection ( $f''$ ), and asymptotes — the seven-point checklist of §7.10 — to build a full analytical description of the curve.

Step 1 (i). $y=-\dfrac13(x^3-3x+2)=-\dfrac13(x-1)^2(x+2)$ .

Domain: all reals. Intercepts: $y=0$ at $x=1$ (double root) and $x=-2$ ; $y$ -intercept at $x=0$ : $y=-\dfrac13(0-0+2)=-\dfrac23$ .

$y'=-\dfrac13(3x^2-3)=1-x^2=0\Rightarrow x=\pm1$ are the critical numbers. $y''=-2x$ .

Sign of $y'=1-x^2$ : negative for $x<-1$ (decreasing), positive for $-1<x<1$ (increasing), negative for $x>1$ (decreasing). So $x=-1$ is a local minimum and $x=1$ is a local maximum — confirmed by $y''(-1)=2>0$ (concave up $\Rightarrow$ min) and $y''(1)=-2<0$ (concave down $\Rightarrow$ max).

$y(1)=-\dfrac13(1-3+2)=0$ (local max value); $\ y(-1)=-\dfrac13(-1+3+2)=-\dfrac43$ (local min value).

Concavity: $y''=-2x=0$ at $x=0$ ; concave up on $(-\infty,0)$ , concave down on $(0,\infty)$ — a genuine sign change, so $x=0$ is a point of inflection: $y(0)=-\dfrac23$ , point $\left(0,-\dfrac23\right)$ .

Asymptotes: none (polynomial). End behaviour: since the leading term is $-\tfrac13x^3$ , $y\to-\infty$ as $x\to+\infty$ and $y\to+\infty$ as $x\to-\infty$ .

Step 2 (ii). $y=x\sqrt{4-x}$ .

Domain: $4-x\ge0\Rightarrow x\le4$ . Intercepts: $y=0$ at $x=0,4$ .

$y'=\sqrt{4-x}-\dfrac{x}{2\sqrt{4-x}}=\dfrac{2(4-x)-x}{2\sqrt{4-x}}=\dfrac{8-3x}{2\sqrt{4-x}}$ . Critical number: $x=\tfrac83$ (interior); $x=4$ is an endpoint with a vertical tangent.

$y\!\left(\tfrac83\right)=\tfrac83\sqrt{4-\tfrac83}=\tfrac83\sqrt{\tfrac43}=\tfrac83\cdot\tfrac{2}{\sqrt3}=\tfrac{16}{3\sqrt3}=\tfrac{16\sqrt3}{9}$ — a local maximum (checked via sign of $y'$ : positive for $x<8/3$ , negative for $8/3<x<4$ ).

Concavity: $y''=\dfrac{3x-16}{4(4-x)^{3/2}}$ ; since $x\le4<\tfrac{16}3$ , the numerator $3x-16<0$ throughout the domain, so $y''<0$ everywhere — concave down on the whole domain, no inflection point.

Asymptotes: none (bounded domain, not rational). As $x\to-\infty$ , $y\to-\infty$ .

Step 3 (iii). $y=\dfrac{x^2+1}{x^2-4}$ .

Domain: $x\ne\pm2$ . Symmetry: even function ( $f(-x)=f(x)$ ). Intercepts: no $x$ -intercept ( $x^2+1>0$ always); $y$ -intercept at $x=0$ : $y=-\tfrac14$ .

Vertical asymptotes: $x=2,x=-2$ . Horizontal asymptote: $\displaystyle\lim_{x\to\pm\infty}\frac{x^2+1}{x^2-4}=1\Rightarrow y=1$ .

$y'=\dfrac{2x(x^2-4)-(x^2+1)(2x)}{(x^2-4)^2}=\dfrac{-10x}{(x^2-4)^2}$ . Critical number $x=0$ : sign of $y'$ is positive for $x<0$ , negative for $x>0\Rightarrow$ local maximum at $x=0$ , $y(0)=-\tfrac14$ .

$y''=\dfrac{30x^2+40}{(x^2-4)^3}$ ; numerator always positive, so sign of $y''$ follows sign of $(x^2-4)^3$ : concave up for $|x|>2$ , concave down for $|x|<2$ ; no inflection point (concavity only switches at the excluded points $x=\pm2$ ). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$f(x)=\dfrac1x$
(a) $x=c$
(b) $y=0$
(c) $y=c$
(d) $x=0$
›Reveal solutionSolution
A horizontal asymptote is the limiting value of $f(x)$ as $x\to\pm\infty$ ; here that limit is $0$ .

$f(x)=\dfrac1x$ . As $x\to\infty$ , $f(x)\to0$ ; as $x\to-\infty$ , $f(x)\to0$ as well. …
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CBSE 2018Set ANNUAL1 markMCQ
$y^2(x - 2) = x^2(1 + x)$
(a) asymptotes parallel to both axes
(b) an asymptote parallel to $x$ -axis
(c) no asymptotes
(d) an asymptote parallel to $y$ -axis
›Reveal solutionSolution
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.

Rewrite the given curve $y^2(x-2)=x^2(1+x)$ as $f(x,y) = (x-2)y^2 - x^2(1+x) = 0$ .

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 $(x-2)$ .

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. …
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CBSE 2017Set ANNUAL1 markMCQ
$y^2(x-2) = x^2(1+x)$
(a) an asymptote parallel to $x$ -axis
(b) an asymptote parallel to $y$ -axis
(c) asymptotes parallel to both axes
(d) no asymptote
›Reveal solutionSolution
The curve has a vertical asymptote $x=2$ (parallel to the $y$ -axis); there is no asymptote parallel to the $x$ -axis.

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 $x^3+x^2-xy^2+2y^2=0$ .

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.

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. …
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CBSE 2016Set ANNUAL1 markMCQ
$ay^2=x^2(3a-x)$
(a) $x=-3a,\ x=0$
(b) $x=0,\ x=3a$
(c) $x=0,\ x=a$
(d) $x=0$
›Reveal solutionSolution
The curve touches the axis at the origin and crosses it again at $x=3a$ , giving $x=0$ and $x=3a$ .

The curve is $ay^2=x^2(3a-x)$ , a standard cubic curve traced in the TN Class-12 syllabus.

To find where the curve meets the axis of $x$ (where $y=0$ ), substitute $y=0$ : $a(0)^2=x^2(3a-x)\Rightarrow x^2(3a-x)=0$ .

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). …
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CBSE 2016Set ANNUAL1 markMCQ
$a^2y^2=x^2(a^2-x^2)$
(a) $x\leq a$ and $x\geq -a$
(b) $x<a$ and $x>-a$
(c) $x\leq -a$ and $x\geq a$
(d) $x\leq a$ and $x>-a$
›Reveal solutionSolution
Real $y$ requires the right-hand side to be non-negative, which restricts $x$ to the closed interval $[-a,a]$ .

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).

Since $x^2\geq0$ always, the sign of the product $x^2(a^2-x^2)$ is controlled by $(a^2-x^2)$ whenever $x\neq0$ .

Requiring $a^2-x^2\geq0$ gives $x^2\leq a^2$ , i.e. $-a\leq x\leq a$ .

At $x=0$ the product is automatically $0\geq0$ , which is consistent with (and already included in) the interval $-a\leq x\leq a$ . …
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Q.Sketch the graphs of the following functions:

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