15b. Curve Sketching — Practice

Drill sheet for 15. Curve Sketching. Book-style examples above each worksheet; regenerate for new numbers. Proofs use guided choice.

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1. Concavity and inflection (factored $y''$)

Book. $y=3x^{4}-10x^{3}-12x^{2}+12x-7$ with $y''=12(3x+1)(x-2)$. Candidates $x=-\tfrac13$ and $2$. Sign chart: up on $(-\infty,-1/3)$, down on $(-1/3,2)$, up on $(2,\infty)$. Inflection at both (sign changes): $\bigl(-\tfrac13,-\tfrac{322}{27}\bigr)$ and $(2,-63)$.

Fig. 15-4 — two inflection points; up–down–up

Concavity + inflection from y″

Factor y″, sign-chart, confirm sign changes

(1)Suppose $y''=c(x-(-1))(x-(1))$ with $c>0$ and -1<1$. State concavity intervals and inflection$x$-values.

Roots $-1$, $1$. Leading positive ⇒ $y''>0$ left of $-1$, then alternates.

Concave up on $(-\infty,-1)$ and $(1,\infty)$; down on $(-1,1)$.

Inflection at $x=-1$ and $x=1$ (sign changes).

Inflection at $-1,\,1$; up–down–up pattern.

(2)Suppose $y''=c(x-(-1))(x-(2))$ with $c>0$ and -1<2$. State concavity intervals and inflection$x$-values.

Roots $-1$, $2$. Leading positive ⇒ $y''>0$ left of $-1$, then alternates.

Concave up on $(-\infty,-1)$ and $(2,\infty)$; down on $(-1,2)$.

Inflection at $x=-1$ and $x=2$ (sign changes).

Inflection at $-1,\,2$; up–down–up pattern.

(3)Suppose $y''=c(x-(-1))(x-(1))$ with $c>0$ and -1<1$. State concavity intervals and inflection$x$-values.

Roots $-1$, $1$. Leading positive ⇒ $y''>0$ left of $-1$, then alternates.

Concave up on $(-\infty,-1)$ and $(1,\infty)$; down on $(-1,1)$.

Inflection at $x=-1$ and $x=1$ (sign changes).

Inflection at $-1,\,1$; up–down–up pattern.

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2. $y''=0$ but no inflection

Book. $y=x^{4}-6x+2$: $y''=12x^{2}\ge 0$ — $(0,2)$ is not an inflection. Critical $x=\sqrt[3]{3/2}$ is an absolute minimum ($y''>0$, unique critical number).

Fig. 15-5 — concave up everywhere; absolute min only

False inflection + absolute min

Same-sign y″ ⇒ no inflection; unique crit + y″>0 ⇒ abs min

(1)For $y=x^{4}+kx$ with $k\neq 0$: explain why $x=0$ is not an inflection, and where the absolute min occurs.

$y''=12x^{2}\ge 0$ — never changes sign.

$y'=4x^{3}+k=0\Rightarrow x=\sqrt[3]{-k/4}$; unique critical number with $y''>0$ ⇒ abs min.

No inflection at $0$; unique critical number is an absolute minimum.

(2)For $y=x^{4}+kx$ with $k\neq 0$: explain why $x=0$ is not an inflection, and where the absolute min occurs.

$y''=12x^{2}\ge 0$ — never changes sign.

$y'=4x^{3}+k=0\Rightarrow x=\sqrt[3]{-k/4}$; unique critical number with $y''>0$ ⇒ abs min.

No inflection at $0$; unique critical number is an absolute minimum.

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3. Inflection where $y''$ is undefined

Book. $y=3x+(x+2)^{3/5}$: $y''=-6/(25(x+2)^{7/5})$. Sign change across $x=-2$ ⇒ inflection $(-2,-6)$. Also $y'>0$ off $-2$ ⇒ strictly increasing.

Fig. 15-6 — inflection at a point where y″ blows up

Inflection at undefined y″

Check sign change across the singular point

(1)Examine $y=3x+(x+2)^{3/5}$ for concavity and inflection. (Book: $y''=-6/(25(x+2)^{7/5})$.)

$y''$ undefined at $x=-2$; sign: $y''>0$ for $x<-2$, $y''<0$ for $x>-2$.

Inflection at $(-2,-6)$. Also $y'>0$ off $-2$ ⇒ strictly increasing, no relative extrema.

Inflection at $x=-2$; always increasing; down for $x>-2$, up for $x<-2$.

(2)Examine $y=3x+(x+2)^{3/5}$ for concavity and inflection. (Book: $y''=-6/(25(x+2)^{7/5})$.)

$y''$ undefined at $x=-2$; sign: $y''>0$ for $x<-2$, $y''<0$ for $x>-2$.

Inflection at $(-2,-6)$. Also $y'>0$ off $-2$ ⇒ strictly increasing, no relative extrema.

Inflection at $x=-2$; always increasing; down for $x>-2$, up for $x<-2$.

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4. Proof: $f''(x_0)=0$ and $f'''(x_0)\neq 0$ ⇒ inflection

Idea. Nonzero $f'''$ ⇒ $f''$ is mono at $x_0$; with $f''(x_0)=0$ the signs of $f''$ flip ⇒ concavity changes.

Third-derivative inflection test

State the argument in steps

(1)Prove: if $f''(x_0)=0$ and $f'''(x_0)\neq 0$, then $x_0$ is an inflection point.

$f'''(x_0)\neq 0$ ⇒ $f''$ is strictly mono at $x_0$ (Thm 13.7 on $f''$).

With $f''(x_0)=0$, $f''$ has opposite signs on the two sides.

Opposite concavity ⇒ inflection at $x_0$.

Sign change of $f''$ forced by nonzero $f'''$ at a zero of $f''$.

(2)Prove: if $f''(x_0)=0$ and $f'''(x_0)\neq 0$, then $x_0$ is an inflection point.

$f'''(x_0)\neq 0$ ⇒ $f''$ is strictly mono at $x_0$ (Thm 13.7 on $f''$).

With $f''(x_0)=0$, $f''$ has opposite signs on the two sides.

Opposite concavity ⇒ inflection at $x_0$.

Sign change of $f''$ forced by nonzero $f'''$ at a zero of $f''$.

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5. Tangents at inflection points

Book. $f(x)=x^{4}-6x^{3}+12x^{2}-8x$: $f''=12(x-1)(x-2)$, $f'''=12(2x-3)\neq 0$ at $1,2$. Points $(1,-1)$, $(2,0)$; tangents $y=2x-3$ and $y=0$.

Tangent lines at inflection

Confirm f‴≠0, then point-slope with f′

(1)For $f(x)=x^{4}-6x^{3}+12x^{2}-8x$, find tangent lines at inflection points (use $f'''\neq 0$ test).

$f''=12(x-1)(x-2)$; $f'''=12(2x-3)\neq 0$ at $1$ and $2$.

Points $(1,-1)$, $(2,0)$.

$f'(1)=2\Rightarrow y=2x-3$; $f'(2)=0\Rightarrow y=0$.

Tangents $y=2x-3$ and $y=0$.

(2)Given inflection at $x=c$ with $f(c)=d$ and $f'(c)=m$, write the tangent line. Example: $c=1$, $d=-1$, $m=2$.

Point-slope: $y-d=m(x-c)$.

Here $y+1=2(x-1)\Rightarrow y=2x-3$.

$y-d=m(x-c)$ (example $y=2x-3$).

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6–8. Cubic sketches

Book 6. $f(x)=2x^{3}-5x^{2}+4x-7$: inflection at $x=\tfrac56$; rel max at $\tfrac23$, rel min at $1$.

Book 8. $g(x)=2x^{3}-9x^{2}+36$: max at $0$, min at $3$, inflection at $\tfrac32$; one real zero $\approx -1.70$.

Fig. 15-7 — cubic sketch (Problem 6 style)

Fig. 15-9 — cubic g (Problem 8)

Cubic sketch data

Critical numbers, second-derivative test, inflection from f″

(1)For $f(x)=2x^{3}-3x^{2}+c x+d$, locate the inflection $x$-value from $f''$.

$f''=12x-6=0\Rightarrow x=\dfrac{1}{2}$.

Confirm a sign change of $f''$ (linear $f''$ always changes).

Inflection at $x=\dfrac{1}{2}$.

(2)Sketch data for $f(x)=2x^{3}-5x^{2}+4x-7$: inflection, critical numbers, max/min.

$f''=12x-10$: inflection at $x=5/6$.

Critical $x=2/3$ (rel max), $x=1$ (rel min).

Concave down left of $5/6$, up to the right.

Inflection $5/6$; rel max at $2/3$; rel min at $1$.

(3)For $f(x)=2x^{3}-5x^{2}+c x+d$, locate the inflection $x$-value from $f''$.

$f''=12x-10=0\Rightarrow x=\dfrac{5}{6}$.

Confirm a sign change of $f''$ (linear $f''$ always changes).

Inflection at $x=\dfrac{5}{6}$.

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7. Oblique asymptote

Book. $y=x^{2}/(x-2)=x+2+4/(x-2)$. Rel max at $0$, rel min at $4$; VA $x=2$; OA $y=x+2$; no inflection.

Fig. 15-8 — vertical + oblique asymptotes

Polynomial division → oblique asymptote

Rewrite, classify extrema, list asymptotes

(1)For $y=x^{2}/(x-2)$: rewrite by division, find extrema, asymptotes (vertical + oblique).

$y=x+2+4/(x-2)$.

Critical $x=0$ (rel max, $y=0$), $x=4$ (rel min, $y=8$).

Vertical $x=2$; oblique $y=x+2$; no inflection ($y''$ never $0$).

VA $x=2$; OA $y=x+2$; max at $0$, min at $4$.

(2)Divide $x^{2}/(x-3)$ and identify the oblique asymptote.

$x^{2}=(x-3)(x+3)+9$ ⇒ $y=x+3+9/(x-3)$.

Oblique asymptote $y=x+3$; vertical $x=3$.

OA $y=x+3$; VA $x=3$.

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9. Two vertical asymptotes

Book. $y=x^{2}/((x-2)(x-6))$: VA at $2$ and $6$; HA $y=1$; min at $0$, max at $3$; inflection $\approx -1.70$.

Fig. 15-10 — two poles and y=1

Two poles + horizontal asymptote

Asymptotes, critical numbers, link to cubic for inflection

(1)For $y=x^{2}/((x-2)(x-6))$: list vertical/horizontal asymptotes and classify critical numbers $0$ and $3$.

VA at $x=2$, $x=6$; HA $y=1$ (equal degree, leading ratio $1$).

Rel min at $x=0$ ($y=0$); rel max at $x=3$ ($y=-3$).

Inflection near $x\approx -1.70$ from $y''=0$ (linked to $2x^{3}-9x^{2}+36=0$).

VA $2,6$; HA $y=1$; min at $0$, max at $3$.

(2)For $y=x^{2}/((x-2)(x-6))$: list vertical/horizontal asymptotes and classify critical numbers $0$ and $3$.

VA at $x=2$, $x=6$; HA $y=1$ (equal degree, leading ratio $1$).

Rel min at $x=0$ ($y=0$); rel max at $x=3$ ($y=-3$).

Inflection near $x\approx -1.70$ from $y''=0$ (linked to $2x^{3}-9x^{2}+36=0$).

VA $2,6$; HA $y=1$; min at $0$, max at $3$.

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10. Implicit algebraic curve

Book. $y^{2}(x^{2}-4)=x^{4}$: domain $x^{2}>4$ plus $(0,0)$; full axial/origin symmetry; Q1 min at $(2\sqrt{2},4)$; VA $x=\pm 2$; OA $y=\pm x$.

Fig. 15-11 — four branches, VA ±2, OA y=±x

Implicit curve sketch

Domain · symmetry · first-quadrant extrema · asymptotes

(1)For $y^{2}(x^{2}-4)=x^{4}$: domain, symmetry, first-quadrant min, asymptotes.

Domain $x^{2}>4$ (plus isolated $(0,0)$); symmetric in both axes and origin.

In Q1: critical $x=2\sqrt{2}$, $y=4$ (rel/abs min on that branch); $y''>0$.

VA $x=\pm 2$; oblique $y=\pm x$.

Symmetry all; min $(2\sqrt{2},4)$ in Q1; VA $\pm 2$; OA $y=\pm x$.

(2)For $y^{2}(x^{2}-4)=x^{4}$: domain, symmetry, first-quadrant min, asymptotes.

Domain $x^{2}>4$ (plus isolated $(0,0)$); symmetric in both axes and origin.

In Q1: critical $x=2\sqrt{2}$, $y=4$ (rel/abs min on that branch); $y''>0$.

VA $x=\pm 2$; oblique $y=\pm x$.

Symmetry all; min $(2\sqrt{2},4)$ in Q1; VA $\pm 2$; OA $y=\pm x$.

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Mixed review

Curve sketching — mixed set

Concavity · inflection tests · cubics · rationals · asymptotes

(1)Given inflection at $x=c$ with $f(c)=d$ and $f'(c)=m$, write the tangent line. Example: $c=1$, $d=-1$, $m=2$.

Point-slope: $y-d=m(x-c)$.

Here $y+1=2(x-1)\Rightarrow y=2x-3$.

$y-d=m(x-c)$ (example $y=2x-3$).

(2)For $y=x^{2}/((x-2)(x-6))$: list vertical/horizontal asymptotes and classify critical numbers $0$ and $3$.

VA at $x=2$, $x=6$; HA $y=1$ (equal degree, leading ratio $1$).

Rel min at $x=0$ ($y=0$); rel max at $x=3$ ($y=-3$).

Inflection near $x\approx -1.70$ from $y''=0$ (linked to $2x^{3}-9x^{2}+36=0$).

VA $2,6$; HA $y=1$; min at $0$, max at $3$.

(3)Divide $x^{2}/(x-3)$ and identify the oblique asymptote.

$x^{2}=(x-3)(x+3)+9$ ⇒ $y=x+3+9/(x-3)$.

Oblique asymptote $y=x+3$; vertical $x=3$.

OA $y=x+3$; VA $x=3$.

(4)For $y=x^{2}/((x-2)(x-6))$: list vertical/horizontal asymptotes and classify critical numbers $0$ and $3$.

VA at $x=2$, $x=6$; HA $y=1$ (equal degree, leading ratio $1$).

Rel min at $x=0$ ($y=0$); rel max at $x=3$ ($y=-3$).

Inflection near $x\approx -1.70$ from $y''=0$ (linked to $2x^{3}-9x^{2}+36=0$).

VA $2,6$; HA $y=1$; min at $0$, max at $3$.

(5)Given inflection at $x=c$ with $f(c)=d$ and $f'(c)=m$, write the tangent line. Example: $c=1$, $d=-1$, $m=2$.

Point-slope: $y-d=m(x-c)$.

Here $y+1=2(x-1)\Rightarrow y=2x-3$.

$y-d=m(x-c)$ (example $y=2x-3$).

(6)Examine $y=3x+(x+2)^{3/5}$ for concavity and inflection. (Book: $y''=-6/(25(x+2)^{7/5})$.)

$y''$ undefined at $x=-2$; sign: $y''>0$ for $x<-2$, $y''<0$ for $x>-2$.

Inflection at $(-2,-6)$. Also $y'>0$ off $-2$ ⇒ strictly increasing, no relative extrema.

Inflection at $x=-2$; always increasing; down for $x>-2$, up for $x<-2$.

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Bonus · Supplementary problems

11. Concavity of Ch.14 #23(a–f)

Book answers (pattern). (a) always up, no infl.; (b) always down, no infl.; (c) infl. $-2/3$; (d) infl. $2$ (up to the right); (e) infl. $2$ (down to the right); (f) infl. $\pm 2\sqrt{3}/3$ (up outside, down between).

Supp: concavity from Ch.14 functions

Read cup/cap and inflection from the answer patterns / y″ sign charts

(1)Inflection at $x=2$ with concave up for $x>2$ and down for $x<2$. State the sign pattern of $y''$.

$y''$ changes $-$ to $+$ at $2$.

Inflection at $2$; down on $(-\infty,2)$, up on $(2,\infty)$.

(2)Two inflection $x$-values $\pm 2\sqrt{3}/3$: up outside, down between. Summarize.

Inflection at $\pm 2\sqrt{3}/3$.

Up for $|x|>2\sqrt{3}/3$; down for $|x|<2\sqrt{3}/3$.

Infl. $\pm 2\sqrt{3}/3$; up outside, down in the middle.

(3)Inflection at $x=2$ with concave up for $x>2$ and down for $x<2$. State the sign pattern of $y''$.

$y''$ changes $-$ to $+$ at $2$.

Inflection at $2$; down on $(-\infty,2)$, up on $(2,\infty)$.

12. Cubic: mean of critical numbers = inflection

Claim. For $f(x)=ax^{3}+bx^{2}+cx+d$, the inflection abscissa is $-b/(3a)$. It equals the average of the two critical numbers when both exist, or the single critical number when the discriminant of $f'$ vanishes.

Supp: cubic critical mean = inflection

Compare −b/(3a) with (r+s)/2 from f′

(1)For $f(x)=1x^{3}+9x^{2}+cx+d$, compute the inflection $x$-value and check it equals the average of the roots of $f'=0$ when two real critical numbers exist.

Inflection: $f''=6x+18=0\Rightarrow x=-3$.

Sum of critical numbers from $f'$: $-2b/(3a)$; half is the same abscissa.

Inflection (and mean of crits) at $x=-3$.

(2)For $f(x)=2x^{3}+3x^{2}+cx+d$, compute the inflection $x$-value and check it equals the average of the roots of $f'=0$ when two real critical numbers exist.

Inflection: $f''=12x+6=0\Rightarrow x=-\dfrac{1}{2}$.

Sum of critical numbers from $f'$: $-2b/(3a)$; half is the same abscissa.

Inflection (and mean of crits) at $x=-\dfrac{1}{2}$.

13. Discuss and sketch gallery

(a) $xy=(x^{2}-9)^{2}$ · (b) $y=x^{4}/(1-x^{2})$ · (c) $y=x^{2}+2/x$ · (d) $y^{3}=6x^{2}-x^{3}$ · (e) $y=1+x^{2}/(x-1)$ · (f) $y=x/(x^{2}+1)$ · (g) $y=x\sqrt{x-1}$ · (h) $y=x\sqrt[3]{2-x}$ · (i) $y=(x+1)/x^{2}$.

Supp: discuss & sketch

Symmetry · asymptotes · extrema · inflection · concavity

(1)Discuss $y=x\sqrt{x-1}$ ($x\ge 1$): mono, concavity, inflection.

Increasing on domain; infl. $(4/3,\tfrac49\sqrt{3})$.

Down for $1\le x<4/3$, up for $x>4/3$.

Increasing; infl. at $4/3$; down then up.

(2)Discuss $y=x\sqrt[3]{2-x}$: max, mono, inflection.

Rel max at $x=3/2$; incr. for $x<3/2$.

Down for $x<2$, up for $x>2$; infl. $(2,0)$.

Max at $3/2$; infl. at $2$; down left of $2$.

(3)Discuss $y=x^{2}+2/x$: asymptotes, extrema, inflection, concavity.

VA $x=0$; rel min $(1,3)$; inflection $(-\sqrt[3]{2},0)$.

Concave up for $x<-\sqrt[3]{2}$ and $x>0$.

VA $0$; min $(1,3)$; infl. $-\sqrt[3]{2}$; up on $(-\infty,-\sqrt[3]{2})\cup(0,\infty)$.

(4)Discuss $y=(x+1)/x^{2}$: asymptotes, extrema, inflection.

VA $x=0$, HA $y=0$; rel min $(-2,-1/4)$; infl. $(-3,-2/9)$.

Up on $(-3,0)$ and $x>0$; $y\to+\infty$ as $x\to 0$.

VA $0$, HA $0$; min $-2$; infl. $-3$.

14. Even + odd decomposition

Hint. $E(x)=\tfrac12\bigl(F(x)+F(-x)\bigr)$, $O(x)=\tfrac12\bigl(F(x)-F(-x)\bigr)$.

Supp: unique even + odd split

Construct E, O and argue uniqueness

(1)Show any $F$ defined for all $x$ is uniquely $E+O$ with $E$ even and $O$ odd. [Hint: $E(x)=\tfrac12(F(x)+F(-x))$.]

Set $E(x)=\tfrac12(F(x)+F(-x))$, $O(x)=\tfrac12(F(x)-F(-x))$.

Check $E(-x)=E(x)$, $O(-x)=-O(x)$, and $E+O=F$.

Uniqueness: if $F=E_1+O_1=E_2+O_2$, then $E_1-E_2=O_2-O_1$ is both even and odd ⇒ $0$.

$E=\tfrac12(F(x)+F(-x))$, $O=\tfrac12(F(x)-F(-x))$ — unique.

(2)Show any $F$ defined for all $x$ is uniquely $E+O$ with $E$ even and $O$ odd. [Hint: $E(x)=\tfrac12(F(x)+F(-x))$.]

Set $E(x)=\tfrac12(F(x)+F(-x))$, $O(x)=\tfrac12(F(x)-F(-x))$.

Check $E(-x)=E(x)$, $O(-x)=-O(x)$, and $E+O=F$.

Uniqueness: if $F=E_1+O_1=E_2+O_2$, then $E_1-E_2=O_2-O_1$ is both even and odd ⇒ $0$.

$E=\tfrac12(F(x)+F(-x))$, $O=\tfrac12(F(x)-F(-x))$ — unique.

15. Reflect an algebraic curve

Book. $x^{2}-3xy+2y^{2}=1$ → after $x$- or $y$-axis reflection: $x^{2}+3xy+2y^{2}=1$; origin reflection recovers $C$.

Supp: reflect F(x,y)=0

Substitute (−x,y), (x,−y), or (−x,−y)

(1)Curve $C$: $x^{2}-3xy+2y^{2}=1$. Find the equation after reflection in (a) $x$-axis (b) $y$-axis (c) origin.

(a) Replace $y$ by $-y$: $x^{2}+3xy+2y^{2}=1$.

(b) Replace $x$ by $-x$: same as (a).

(c) Replace $(x,y)$ by $(-x,-y)$: recovers $C$ (even total degree).

(a)(b) $x^{2}+3xy+2y^{2}=1$; (c) $C$ itself.

(2)Curve $C$: $x^{2}-3xy+2y^{2}=1$. Find the equation after reflection in (a) $x$-axis (b) $y$-axis (c) origin.

(a) Replace $y$ by $-y$: $x^{2}+3xy+2y^{2}=1$.

(b) Replace $x$ by $-x$: same as (a).

(c) Replace $(x,y)$ by $(-x,-y)$: recovers $C$ (even total degree).

(a)(b) $x^{2}+3xy+2y^{2}=1$; (c) $C$ itself.

16. Symmetry ⇒ even / odd

Book hints. $y$-axis symmetry ⇒ $(-x,f(x))$ on the graph ⇒ $f(-x)=f(x)$. Origin symmetry ⇒ $(-x,-f(x))$ ⇒ odd.

17. Prove Theorem 15.1 (concavity)

Idea. $f''>0$ ⇒ $f'$ increasing near $x_0$; MVT compares $f(x)$ with the tangent $f(x_0)+f'(x_0)(x-x_0)$; the curve lies above the tangent. Part (b) via $-f$.

18–19. Graphing calculator labs

18. $f(x)=x^{3}-3x^{2}+4x-2$: prove increasing ($f'>0$), find inflection, plot $f$, $f^{-1}$, and $y=x$.

19. $y=x^{2}/(x^{3}-3x^{2}+5)$: hand sketch asymptotes, then refine poles with a GC.

Book note on #18

Some printings list an inflection at $(-1,3)$; for this cubic, $f''=6x-6$ vanishes at $x=1$ (with $f(1)=0$). Use the analytic $f''=0$ location.

Supp: GC sketch labs

Analytic mono/inflection + calculator check / inverse symmetry

(1)Analytically: show $f(x)=x^{3}-3x^{2}+4x-2$ is increasing and locate its inflection; then on a GC compare $f$, $f^{-1}$, and $y=x$.

$f'=3x^{2}-6x+4$ has discriminant $36-48<0$, so $f'>0$ everywhere (strictly increasing / invertible).

$f''=6x-6=0$ at $x=1$ with a sign change ⇒ inflection at $(1,0)$.

GC: graphs of $f$ and $f^{-1}$ are mirrors across $y=x$.

Always increasing; inflection at $x=1$; inverse reflects across $y=x$.

(2)Sketch $y=x^{2}/(x^{3}-3x^{2}+5)$ by hand (asymptotes / end behavior), then use a GC to refine vertical asymptotes.

Denominator zeros ⇒ candidate VAs; degree gap ⇒ HA $y=0$.

GC confirms which poles are real and the local shape between them.

HA $y=0$; check real roots of $x^{3}-3x^{2}+5=0$ for VAs; GC for detail.

Bonus mixed

Supplementary mixed set

Problems 11–19 style

(1)Concavity for a cubic-like $y''$ with a single root at $x=-2/3$.

Inflection at $x=-2/3$.

Concave down for $x<-2/3$, up for $x>-2/3$ (if leading $y''$ coeff positive).

Inflection at $-2/3$; down left, up right.

(2)Inflection at $x=2$ with concave up for $x>2$ and down for $x<2$. State the sign pattern of $y''$.

$y''$ changes $-$ to $+$ at $2$.

Inflection at $2$; down on $(-\infty,2)$, up on $(2,\infty)$.

(3)Curve $C$: $x^{2}-3xy+2y^{2}=1$. Find the equation after reflection in (a) $x$-axis (b) $y$-axis (c) origin.

(a) Replace $y$ by $-y$: $x^{2}+3xy+2y^{2}=1$.

(b) Replace $x$ by $-x$: same as (a).

(c) Replace $(x,y)$ by $(-x,-y)$: recovers $C$ (even total degree).

(a)(b) $x^{2}+3xy+2y^{2}=1$; (c) $C$ itself.

(4)Ch.14 #23(a)-style: a graph that is always a cup. Concavity / inflection?

$y''>0$ everywhere (e.g. upward parabola / $x^{4}$-type).

No inflection; concave upward everywhere.

(5)Curve $C$: $x^{2}-3xy+2y^{2}=1$. Find the equation after reflection in (a) $x$-axis (b) $y$-axis (c) origin.

(a) Replace $y$ by $-y$: $x^{2}+3xy+2y^{2}=1$.

(b) Replace $x$ by $-x$: same as (a).

(c) Replace $(x,y)$ by $(-x,-y)$: recovers $C$ (even total degree).

(a)(b) $x^{2}+3xy+2y^{2}=1$; (c) $C$ itself.

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Self-check

Quiz

Curve sketching practice review

1 / 6

If y″ changes from + to − across x₀, the graph…

Pick an answer to see if you are right.

Further Reading

• 16. Review of Trigonometry — radians and unit-circle sine/cosine
• 15. Curve Sketching · Concavity · Symmetry — theory and explorers
• 14b. Max/Min — Practice — extrema / optimization drills
• 14. Maximum and Minimum Values — critical numbers and tests