Worked examples — Types — constant, linear, quadratic, polynomial, rational, radical, piecewise

This page is the practice arena for Types of Functions. We already met each family. Now we hunt down every scenario each family can throw at you — every sign, every degenerate input, every limiting value — and solve one example for each.

Before a single symbol appears we agree on two words used all over this page:

• Root / zero: an input $x$ where the output is $0$, i.e. the graph touches the horizontal axis.
• Asymptote: an invisible straight line the graph hugs but never quite reaches. Vertical ones sit under a "blow-up"; horizontal ones describe "what the graph settles to far away".

If you need the definitions of the families themselves, keep M02.01 Function Definition and M02.03 Domain and Range in another tab.

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The scenario matrix

Every worked example below is tagged with a cell from this table. Together they touch every cell.

Cell	Family	The tricky scenario
C0	Constant	flat line, slope zero, single-value range
C1	Linear	positive slope vs the sign trap (rising / falling)
C2	Quadratic	$a>0$ min and $a<0$ max, plus the degenerate $a=0$
C3	Quadratic	complex roots (parabola never touches axis)
C4	Polynomial	end behaviour across all four sign/parity combos
C5	Polynomial	repeated root vs "more factors = higher degree" trap
C6	Rational	vertical asymptote + a hole (removable point)
C7	Rational	all three asymptote cases by degree comparison ($y=0$, $y=$const, oblique)
C8	Radical	domain restriction and the $\sqrt{\text{negative}}$ boundary
C9	Piecewise	continuity check at the seam (a limiting-value case)
C10	Word problem / exam twist	real projectile + "find when height $=0$"

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Example 0 — Constant function, the flat line · Cell C0

Forecast: guess — how many $x$-intercepts can a constant function have?

1. Read the rule. $f(x)=5$ means "ignore $x$, always return $5$": no $x$ appears in the formula. Why this step? If the output never mentions the input, it cannot change with $x$.
2. Slope is zero. Between any two points the rise is $0$ (both heights are $5$): $m=\frac{5-5}{x_2-x_1}=0$. Why this step? Slope = rise/run; a flat height gives zero rise, hence zero slope.
3. Range is a single value. Every input maps to $5$, so the range is the set $\{5\}$ (see M02.03 Domain and Range). Why this step? Range = "all outputs actually produced"; here only $5$ is ever produced.
4. $x$-intercept? Solve $5=0$: impossible. A constant $c\ne 0$ never crosses the $x$-axis; the special case $c=0$ is the $x$-axis itself. Why this step? An $x$-intercept needs output $0$, but the output is stuck at $5$.

Verify: $f(-2)=5$, $f(0)=5$, $f(100)=5$ — output unmoved, so slope $0$ and range $\{5\}$ ✓. No solution to $5=0$, so zero $x$-intercepts ✓.

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Example 1 — Linear, the sign trap · Cell C1

Forecast: guess the direction before reading on. Does the $+7$ matter for direction?

1. Identify slope and intercept. Compare $-4x+7$ with $mx+b$: so $m=-4$, $b=7$. Why this step? Direction is decided only by $m$ (the rate of change), never by $b$.
2. Read the sign of $m$. $m=-4<0$, so every $+1$ in $x$ drops $y$ by $4$: the line falls. Why this step? Slope = rise over run; a negative rise means going down as we move right.
3. $y$-intercept: set $x=0$: $f(0)=7$. $x$-intercept (root): set $f(x)=0$: $-4x+7=0\Rightarrow x=\frac{7}{4}$. Why this step? An axis crossing is just "the other coordinate is zero" — solve for the survivor.

Look at figure s01: the red line clearly descends, yet crosses the $y$-axis high up at $7$. That high crossing is the exact bait of the sign trap.

Verify: $f(1)=-4+7=3$, and $3<7=f(0)$, so output dropped as $x$ rose — decreasing confirmed. Root check: $f(7/4)=-4\cdot \frac{7}{4}+7=-7+7=0$ ✓.

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Example 2 — Quadratic: min, max, and the degenerate $a=0$ · Cell C2

Forecast: which of (a),(b) is a bottom of a valley and which is a top of a hill?

We call the vertex $(x_v,y_v)$, where $x_v$ is its $x$-coordinate and $y_v$ is its $y$-coordinate — the height of the tip. For any quadratic $f$, $x_v=-\frac{b}{2a}$ and $y_v=f(x_v)$.

1. Vertex $x$-coordinate uses $x_v=-\frac{b}{2a}$. For (a): $x_v=-\frac{-4}{2\cdot 1}=2$; then $y_v=g(x_v)=g(2)=4-8+3=-1$. Vertex $(2,-1)$. Why this step? Completing the square shows the parabola is symmetric about $x=-\frac{b}{2a}$; the vertex sits on that mirror line.
2. Sign of $a$ decides min vs max. For (a) $a=1>0$: opens up ⇒ $(2,-1)$ is a minimum. Why this step? Positive $a$ means the $x^2$ term eventually dominates upward, so the tip is the lowest point.
3. For (b): $b=0$ so $x_v=0$; then $y_v=q(0)=3$. Vertex $(0,3)$, and $a=-2<0$ ⇒ maximum. Why this step? Same rule, opposite sign — always check $a$'s sign, never assume "vertex = minimum".
4. Degenerate case $a=0$. Then $a{x}^2+bx+c=bx+c$: it is no longer a parabola, just a line with no vertex. Why this step? The definition requires $a\ne 0$; killing it collapses the family to linear.

Figure s02 stacks the up-parabola (min) and down-parabola (max) so the sign-of-$a$ effect is unmissable.

Verify: (a) $g(1)=1-4+3=0$ and $g(3)=9-12+3=0$ — symmetric roots about $x=2$ ✓. (b) $q(0)=3$ is above $q(1)=-2+3=1$ and above $q(-1)=-2+3=1$, so $(0,3)$ is indeed highest ✓.

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Example 3 — Quadratic with complex roots · Cell C3

Forecast: how many times will this parabola cross the horizontal axis — 2, 1, or 0?

1. Compute the discriminant $D=b^2-4ac$. Here $D=2^2-4(1)(5)=4-20=-16$. Why this step? $D$ is the quantity under the square root in the quadratic formula; its sign counts real crossings.
2. Interpret $D<0$. Negative ⇒ the square root is imaginary ⇒ no real roots. The parabola never touches the axis. Why this step? A real root needs a real $\sqrt{D}$; a negative $D$ forbids it.
3. Where does it sit? $a=1>0$ (opens up) and vertex $y_v=p(-1)=1-2+5=4>0$, so the whole curve floats above the axis. Why this step? An upward parabola whose lowest point is positive can never reach zero.

Verify: the complex roots are $x=\frac{-2\pm \sqrt{-16}}{2}=-1\pm 2i$. Plug $x=-1+2i$: $(-1+2i{)}^2+2(-1+2i)+5=(1-4i-4)+(-2+4i)+5=0$ ✓.

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Example 4 — Polynomial end behaviour, all four combos · Cell C4

Forecast: which two have "both ends up together"?

1. Only the leading term matters far out. For huge $|x|$, lower powers are dust; keep $a_n{x}^n$. Why this step? ${\lim }_{x\to \pm \infty }f(x)=\lim a_n{x}^n$ because $x^n$ outgrows everything below it.
2. Parity of $n$ decides "same or opposite ends". Even $n$: $x^n\ge 0$ both sides ⇒ ends agree. Odd $n$: sign flips with $x$ ⇒ ends disagree.
3. Sign of $a_n$ flips the whole picture. Multiply the above by the leading sign.

case	left ($x\to -\infty$)	right ($x\to +\infty$)
(i) $x^3$	$-\infty$	$+\infty$
(ii) $-x^3$	$+\infty$	$-\infty$
(iii) $x^4$	$+\infty$	$+\infty$
(iv) $-x^4$	$-\infty$	$-\infty$

Figure s03 draws all four skeletons so the "parity chooses shape, sign flips it" rule is one glance.

This uses M03.01 Limits language ("what does the output approach"). Verify: at $x=10$: $x^3=1000>0$, $-x^3=-1000<0$, $x^4=10000>0$, $-x^4=-10000<0$ — matches the right column ✓.

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Example 5 — Repeated root, dodging the counting trap · Cell C5

Forecast: is the degree $3$ or $4$? Are there $4$ roots or fewer?

1. Degree = sum of exponents, not number of factors. $x^2$ contributes $2$, each linear factor $1$: total $2+1+1=4$. Why this step? Degree is the highest power after expanding; multiplicities add up.
2. List roots with multiplicity. $x=0$ (from $x^2$, double), $x=1$, $x=3$. Why this step? Each factor $(x-r)$ gives root $r$; a squared factor still gives one location, just touched twice.
3. Count distinct roots. Locations $\{0,1,3\}$ ⇒ 3 distinct roots, but degree $4$. Why this step? At a double root the curve bounces off the axis instead of crossing — one point, double contact.

Verify: expand $x^2(x-1)(x-3)=x^2(x^2-4x+3)=x^4-4x^3+3x^2$ — leading power $4$ ✓. And $f(0)=0,f(1)=0,f(3)=0$, while $f(2)=16-32+12=-4\ne 0$ ✓.

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Example 6 — Rational: vertical asymptote and a hole · Cell C6

Forecast: both bad points $x=0$ and $x=1$ kill the denominator — will both be asymptotes?

1. Factor top and bottom. $\frac{x^2-1}{x^2-x}=\frac{(x-1)(x+1)}{x(x-1)}$. Why this step? Factoring exposes which zero of the denominator is cancelled by the top.
2. Cancel the shared factor $(x-1)$. Simplified form $\frac{x+1}{x}$, valid for $x\ne 1$. Why this step? A factor that appears top and bottom is a removable problem — a hole, not a wall.
3. Classify each forbidden input. At $x=1$: cancelled ⇒ hole; its height is $\frac{1+1}{1}=2$, so the hole sits at $(1,2)$. At $x=0$: survives in the denominator ⇒ vertical asymptote $x=0$. Why this step? A denominator zero that cannot cancel forces division by zero — a genuine blow-up.

This ties to M03.02 Continuity: a hole is a single point of discontinuity that could be "patched", an asymptote cannot.

Verify: near $x=0$, $r(0.001)=\frac{1.001}{0.001}\approx 1001$ (blows up) ✓. And ${\lim }_{x\to 1}\frac{x+1}{x}=2$, matching the hole height ✓.

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Example 7 — Rational: all three asymptote cases by degree · Cell C7

Forecast: which one flattens to the $x$-axis itself, which to a raised horizontal line, and which drifts along a slant?

1. Compare degrees: top vs bottom. (a) top $1$, bottom $2$ (top lower). (b) both degree $1$ (equal). (c) top $2$, bottom $1$ (top one higher). Why this step? The degree gap decides the type of end behaviour.
2. Top degree lower ⇒ horizontal asymptote $y=0$. (a) far out $\frac{3x+1}{x^2+1}\approx \frac{3x}{x^2}=\frac{3}{x}\to 0$, so the graph hugs the $x$-axis, $y=0$. Why this step? When the bottom grows faster than the top, the ratio is squeezed to zero.
3. Equal degrees ⇒ horizontal asymptote = ratio of leading coefficients. (b) $\frac{2}{1}=2$, so $y=2$. Why this step? Far out, $\frac{2x+1}{x-3}\approx \frac{2x}{x}=2$; the $+1$ and $-3$ become negligible.
4. Top degree exactly one higher ⇒ oblique (slanted) asymptote via division. (c) Divide: $\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}$. The remainder $\to 0$, leaving the line $y=x+1$. Why this step? Polynomial long division peels off the straight-line trend; the leftover fraction vanishes far away.

Verify: (a) $w(1000)=\frac{3001}{1000001}\approx 0.003\to 0$ ✓. (b) $s(1000)=\frac{2001}{997}\approx 2.007\to 2$ ✓. (c) $t(1000)-(1000+1)=\frac{2}{999}\approx 0.002\to 0$, confirming line $y=x+1$ ✓.

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Example 8 — Radical: domain and the $\sqrt{\text{negative}}$ boundary · Cell C8

Forecast: below which $x$ does the graph simply not exist?

1. A real square root needs a non-negative inside. Require $2x-6\ge 0$. Why this step? $\sqrt{\text{negative}}$ has no real value; the domain is exactly where the inside behaves.
2. Solve the inequality. $2x\ge 6\Rightarrow x\ge 3$. Domain $=[3,\infty )$. Why this step? Isolating $x$ gives the exact cutoff where reality begins.
3. Check the boundary $x=3$ (degenerate edge). $u(3)=\sqrt{0}=0$: allowed, the graph starts here; for $x<3$ nothing is plotted. Why this step? The boundary is included because $\ge$ allows equality — the edge point is real.

See M02.03 Domain and Range for the general "inside $\ge 0$" rule of radicals.

Verify: $u(3)=\sqrt{0}=0$ ✓; $u(11)=\sqrt{16}=4$ ✓; $u(2)$ would be $\sqrt{-2}$ — correctly excluded ✓.

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Example 9 — Piecewise: continuity at the seam · Cell C9

Forecast: do the two pieces meet at $x=2$, or is there a jump?

1. Value approaching from the left. As $x\to 2^{-}$, use $x+1$: limit $=2+1=3$. Why this step? Continuity needs the left-hand approach to head somewhere definite.
2. Value approaching from the right (and the actual value). As $x\to 2^{+}$ and at $x=2$, use $3x-3$: $3(2)-3=3$. Why this step? Continuity also needs the right-hand approach and the actual output at $x=2$ to be pinned down, so we evaluate the piece that owns the point.
3. Compare all three. Left limit $=3$, right limit $=3$, value $=3$: all equal ⇒ continuous, no jump. Why this step? Continuity at a point = "limit exists and equals the value" (see M03.02 Continuity).

Conclusion: since the left limit, right limit, and actual value all equal $3$, the two pieces meet perfectly and $v$ is continuous at $x=2$.

Figure s04 shows the two segments kissing exactly at $(2,3)$ — the red dot is the seam.

Verify: left piece at $2$: $2+1=3$; right piece at $2$: $3\cdot 2-3=3$ — identical, so glued smoothly with no jump ✓.

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Example 10 — Word problem / exam twist: projectile · Cell C10

Forecast: will the ground-hit time be before or after the peak?

1. Max height is the vertex; use $t_v=-\frac{b}{2a}$. $a=-4.9,b=20$: $t_v=-\frac{20}{2(-4.9)}=\frac{20}{9.8}\approx 2.041\text{ s}$. Why this step? $a<0$ ⇒ opens down ⇒ vertex is the highest point.
2. Plug back for the peak height. $H(2.041)=-4.9(2.041{)}^2+20(2.041)+2\approx 22.41$ m. Why this step? The $y$-coordinate of the vertex is the actual maximum value.
3. Ground hit: solve $-4.9t^2+20t+2=0$. Quadratic formula: $t=\frac{-20\pm \sqrt{400+39.2}}{-9.8}=\frac{-20\pm \sqrt{439.2}}{-9.8}$. Take the positive root: $t\approx 4.180$ s. Why this step? Time can't be negative; only one root is physically meaningful — the exam twist.

Verify: peak $H(2.0408)\approx 22.41$ m and it comes before the ground hit at $\approx 4.18$ s (peak is earlier, as physics demands) ✓. Check ground: $H(4.180)=-4.9(4.180{)}^2+20(4.180)+2\approx 0$ ✓.

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Recall Self-test

Slope and range of the constant $f(x)=5$? ::: Slope $0$; range $\{5\}$; no $x$-intercept since $5\ne 0$. Direction of $f(x)=-4x+7$? ::: Decreasing — slope $-4<0$; the $+7$ is only the starting height. Why does $x^2+2x+5$ never cross the axis? ::: Discriminant $=-16<0$, so no real roots (parabola floats above the axis). In $\frac{(x-1)(x+1)}{x(x-1)}$ what is $x=1$? ::: A hole at $(1,2)$ — the $(x-1)$ cancels, so it's removable, not an asymptote. Horizontal-$y=0$, horizontal-const, or oblique for $\frac{3x+1}{x^2+1}$, $\frac{2x+1}{x-3}$, $\frac{x^2+1}{x-1}$? ::: $y=0$ (top lower), $y=2$ (equal), oblique $y=x+1$ (top one higher). Domain of $\sqrt{2x-6}$? ::: $[3,\infty )$ — inside must be $\ge 0$.