Trace the curve $y = x^3$.

![Figure — The stem is an explicit 'Trace the curve y=x^3' instruction, whose deliverable culminates in the sketched cubi](/figures/application-of-derivatives/graph-point-inflection-fig-6-13.svg) > [!TLDR] > Tracing $y=x^3$: it is odd (origin-symmetric), monotonically increasing, has a point of inflection at the origin with horizontal tangent there, no asymptotes, and rises/falls steeply away from the origin. > > 1. **Domain and range:** $y=x^3$ is defined for all real $x$, and $y$ takes all real values, so domain $=$ range $=\mathbb{R}$. > 2. **Symmetry:** $f(-x)=(-x)^3=-x^3=-f(x)$, so the function is odd; the curve is symmetric about the origin (i.e. symmetric under $180^\circ$ rotation about $O$), and in particular has no symmetry about either axis. > 3. **Intercepts:** the curve passes through the origin only ($x=0\Rightarrow y=0$; $y=0\Rightarrow x=0$), touching both axes at $O$. > 4. **Monotonicity:** $\dfrac{dy}{dx}=3x^2 \ge 0$ for all $x$, with equality only at $x=0$. So the curve is strictly increasing on $(-\infty,0)$ and on $(0,\infty)$, and (since $y'$ vanishes only at the single point $x=0$) increasing throughout $\mathbb R$; there is no local maximum or minimum. > 5. **Tangent at the origin:** at $x=0$, $y'=0$, so the tangent to the curve at the origin is the $x$-axis, $y=0$. > 6. **Concavity and inflection:** $\dfrac{d^2y}{dx^2}=6x$. For $x 0$, $y''>0$ (concave up). Since concavity changes sign at $x=0$ (where $y''=0$), the origin is a point of inflection. …

Trace the curve y = x³.

Question 108 of 148

y = x^3

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★

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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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Figure — The stem is an explicit 'Trace the curve y=x^3' instruction, whose deliverable culminates in the sketched cubi

Tracing $y=x^3$ : it is odd (origin-symmetric), monotonically increasing, has a point of inflection at the origin with horizontal tangent there, no asymptotes, and rises/falls steeply away from the origin.

Domain and range: $y=x^3$ is defined for all real $x$ , and $y$ takes all real values, so domain $=$ range $=\mathbb{R}$ .
Symmetry: $f(-x)=(-x)^3=-x^3=-f(x)$ , so the function is odd; the curve is symmetric about the origin (i.e. symmetric under $180^\circ$ rotation about $O$ ), and in particular has no symmetry about either axis.
Intercepts: the curve passes through the origin only ( $x=0\Rightarrow y=0$ ; $y=0\Rightarrow x=0$ ), touching both axes at $O$ .
Monotonicity: $\dfrac{dy}{dx}=3x^2 \ge 0$ for all $x$ , with equality only at $x=0$ . So the curve is strictly increasing on $(-\infty,0)$ and on $(0,\infty)$ , and (since $y'$ vanishes only at the single point $x=0$ ) increasing throughout $\mathbb R$ ; there is no local maximum or minimum.
Tangent at the origin: at $x=0$ , $y'=0$ , so the tangent to the curve at the origin is the $x$ -axis, $y=0$ .
Concavity and inflection: $\dfrac{d^2y}{dx^2}=6x$ . For $x<0$ , $y''<0$ (concave down / convex upward); for $x>0$ , $y''>0$ (concave up). Since concavity changes sign at $x=0$ (where $y''=0$ ), the origin is a point of inflection. …

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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.Trace the curve y=x3y = x^3y=x3.

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Q.Trace the curve $y = x^3$ .

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