Worked examples — Graphs of functions — plotting, reading key features

This page is the workout gym for Graphs of functions — plotting, reading key features. The parent note told you the rules. Here we hit every kind of graph a problem can throw at you — every sign, every quadrant, every degenerate input, every limiting value — one worked example at a time.

Before we start: a graph is just the picture made from all the points $(x,f(x))$ — horizontal position is the input $x$, vertical position is the output $f(x)$. When you read a curve left-to-right, you are watching the output change as the input grows. Keep that picture in your head for everything below.

Naming the four regions. The two axes split the plane into four quadrants, always numbered anticlockwise starting top-right: QI = right & up ($x>0,y>0$), QII = left & up ($x<0,y>0$), QIII = left & down ($x<0,y<0$), QIV = right & down ($x>0,y<0$). For each example below we will call out which quadrants the curve visits, so no region is left unaccounted for.

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The scenario matrix

Every graph problem in this chapter lands in one of these cells. Our job is to hit all of them.

Cell	Case class	What makes it tricky	Example
A	Straight line, positive slope	reading intercepts & slope sign	Ex 1
B	Straight line, zero / negative slope	flat and downhill lines	Ex 2
C	Parabola opening up, two real roots	vertex, symmetry, range with a floor	Ex 3
D	Parabola opening down, one repeated root	negative leading sign, a ceiling	Ex 4
E	Rational function, vertical + horizontal asymptote	division by zero, blow-up in both signs	Ex 5
F	Square-root / restricted domain	inputs that are simply not allowed	Ex 6
G	Even vs odd symmetry	folding across $y$-axis vs origin	Ex 7
H	Real-world word problem	translating words into a curve, units	Ex 8
I	Exam twist: solve an equation graphically	intersections = solutions	Ex 9
J	Parabola with no real roots	never touches $x$-axis	Ex 10

Symbols we will use, defined once, in plain words:

• $\mathbb{R}$ means "every real number" — the whole number line.
• $x\to 2^{-}$ means "$x$ creeps toward $2$ from the left (smaller side)"; $x\to 2^{+}$ from the right.
• Quadrants QI–QIV — the four regions defined just above.
• $f^{\prime }(x)$ (used in Ex 9) is read "f-prime of x." The little dash is not multiplication; it is a label meaning "the slope of the curve at the point $x$" — how steeply the graph is tilting right there, positive when uphill, negative when downhill, zero at a flat peak or valley. Think of zooming in on the curve until it looks like a tiny straight line: $f^{\prime }(x)$ is that line's slope. We only mention it in Ex 9 to explain intersections; the full machinery lives in Calculus: derivatives and graphs.

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Cell A — Straight line, positive slope

Forecast: guess before reading — does this line go uphill or downhill, and where does it cross each axis?

1. Domain. No division, no square roots, so every $x$ works: domain $=\mathbb{R}$. Why this step? You must know which inputs are legal before plotting a single point.
2. $y$-intercept — set $x=0$: $f(0)=3(0)-6=-6$. Point $(0,-6)$. Why this step? Plugging $x=0$ tells you where the curve pierces the vertical axis.
3. $x$-intercept — set $f(x)=0$: $3x-6=0\Rightarrow x=2$. Point $(2,0)$. This crossing is the root (the input that makes the output zero). Why this step? The output is zero exactly where the curve crosses the horizontal axis.
4. Slope sign. The number multiplying $x$ is $+3$. Positive → uphill left-to-right, so increasing on $\mathbb{R}$. Why this step? The coefficient of $x$ IS the steepness; its sign is the whole personality of a line.
5. Range. A non-flat line reaches every height eventually → range $=\mathbb{R}$. Why this step? Because the slope is not zero, as $x$ runs from $-\infty$ to $+\infty$ the output $3x-6$ sweeps every real number without skipping any — so no height is left out.
6. Quadrants visited. Left of $x=2$ the output is negative and to the left the input goes negative, so the line passes through QIII; between $x=2$ and where it rises it climbs through QIV (right, below axis) into QI (right, above axis). Why this step? Naming quadrants confirms exactly where the curve lives, not just its intercepts.

Reading the figure: in the sketch the blue line climbs steadily; the red dot marks the $y$-intercept $(0,-6)$ where it cuts the vertical axis, the yellow dot marks the root $(2,0)$ where it cuts the horizontal axis, and the green dot $(1,-3)$ sits exactly between them — proof the rise is uniform.

Verify: midpoint check — at $x=1$, $f(1)=3-6=-3$, and $(1,-3)$ sits exactly halfway between $(0,-6)$ and $(2,0)$ ✓. Slope between the two intercepts: $\frac{0-(-6)}{2-0}=3$ ✓.

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Cell B — Zero and negative slope

Forecast: one of these is perfectly flat. Which one, and what is its slope?

1. Domain of both. Neither has division or a square root, so both accept every input: domain $=\mathbb{R}$ for $p$ and for $q$. Why this step? Same first move as always — confirm the legal inputs before reading anything off.
2. $p(x)=4$: the flat case. No $x$ appears, so the output is $4$ no matter the input. Slope $=0$. Why this step? A missing $x$ means the height never changes — a horizontal line.
3. $p$'s features. $y$-intercept $(0,4)$; no $x$-intercept (it never touches the $x$-axis); neither increasing nor decreasing; range $=\{4\}$ (a single value). Because $y=4>0$ everywhere, the line lives only in QII (left half) and QI (right half), never dipping below the axis. Why this step? This is the degenerate "line" — worth seeing so you recognise it instantly.
4. $q(x)=-\frac{1}{2}x+1$: negative slope. Coefficient $-\frac{1}{2}<0$ → downhill, so decreasing on $\mathbb{R}$. Why this step? The coefficient of $x$ is the change in output per step right; a negative value means each step right lowers the output, which is exactly what "decreasing" means.
5. $q$'s intercepts & quadrants. $y$-intercept: $q(0)=1\Rightarrow (0,1)$. $x$-intercept: $-\frac{1}{2}x+1=0\Rightarrow x=2\Rightarrow (2,0)$. Falling from upper-left it sweeps QII into QI (above axis, left of $x=2$) and continues into QIV (right of $x=2$, below axis). Why this step? Same recipe as Ex 1 — same recipe every time, only the signs differ.

Verify: $q(4)=-\frac{1}{2}(4)+1=-1$; going right by $4$ the height dropped by $2$, i.e. slope $-2/4=-\frac{1}{2}$ ✓. $p$ is constant so $p(4)=p(-99)=4$ ✓.

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Cell C — Upward parabola, two roots

Forecast: does this bowl open up or down, and how low does its bottom go?

1. Domain. A polynomial has no division and no roots, so every input works: domain $=\mathbb{R}$. Why this step? Even for a "nice" parabola we state the domain first — the habit prevents surprises later.
2. Opening direction. Coefficient of $x^2$ is $+1>0$ → opens upward (a valley, a floor exists). Why this step? The sign in front of $x^2$ decides bowl-up vs bowl-down; everything else follows.
3. Roots. Factor: $x^2-2x-3=(x-3)(x+1)$, zero when $x=3$ or $x=-1$. Roots $(-1,0)$ and $(3,0)$. Why this step? Factoring finds where output $=0$ without plotting — the two $x$-crossings.
4. Axis of symmetry. A parabola is a mirror; the mirror line sits halfway between the roots: $x=\frac{-1+3}{2}=1$. Why this step? Symmetry means the two roots are equidistant from the turning point.
5. Vertex. At $x=1$: $g(1)=1-2-3=-4$. Vertex $(1,-4)$ — the lowest point. Why this step? Plug the axis-$x$ back in to get the floor height.
6. Range & quadrants. Lowest output is $-4$, curve rises forever → range $=[-4,\infty )$. The dip below the axis between the roots sits in QIII (left of $x=0$) and QIV (right of $x=0$); the arms climb above the axis into QII and QI. Why this step? An upward bowl has a minimum but no maximum, and its below-axis belly visits the two lower quadrants.

Reading the figure: the blue curve dips to the red vertex $(1,-4)$, crosses the axis at the two yellow roots $(-1,0)$ and $(3,0)$, and the green dashed line $x=1$ is the mirror — fold the picture along it and the two halves land on top of each other.

Verify: completing the square, $x^2-2x-3=(x-1{)}^2-4$, confirming vertex $(1,-4)$ and floor $-4$ ✓. $g(-1)=1+2-3=0$ and $g(3)=9-6-3=0$ ✓.

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Cell D — Downward parabola, repeated root

Forecast: this one opens the other way. Where is its single peak, and how many times does it touch the $x$-axis?

1. Domain. Again a polynomial — no division, no root — so domain $=\mathbb{R}$. Why this step? Start every graph by fixing the legal inputs, even when they are "all of them."
2. Opening. The leading sign is $-1<0$ → opens downward (a hill, a ceiling). Why this step? $(x-2{)}^2$ is never negative and grows as you leave $x=2$; putting a minus in front flips that growth into a drop, so both arms fall away from the top — a downward bowl.
3. Vertex first (it is already in "vertex form"). $-(x-2{)}^2$ is largest when $(x-2{)}^2$ is smallest, i.e. zero, at $x=2$. Vertex $(2,0)$. Why this step? A squared term is never negative, so subtracting it from $0$ peaks where the square vanishes.
4. Roots. Set $-(x-2{)}^2=0\Rightarrow x=2$ only — a repeated root. The curve touches the axis at $(2,0)$ and turns back, it does not cross. Why this step? A repeated root means the vertex sits on the axis — the degenerate "one root" case.
5. Range & quadrants. Highest output is $0$, curve falls forever → range $=(-\infty ,0]$. Since $y\le 0$ everywhere, both arms live below the axis: the left arm in QIII ($x<0$) and near-vertex in QIV, the right arm entirely in QIV ($x>0,y<0$). Why this step? A downward bowl has a maximum (here $0$, at the vertex) but no minimum — the arms plunge to $-\infty$ — so the outputs fill everything from $0$ downward and stay in the lower quadrants.

Verify: $m(1)=-(1-2{)}^2=-1$ and $m(3)=-(3-2{)}^2=-1$ — equal heights either side of $x=2$, confirming symmetry and downward opening ✓.

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Cell E — Rational function, two asymptotes

Forecast: one input is forbidden. Which? And what happens to the curve as $x$ sneaks up to it from each side?

1. Domain. Denominator $x+1\ne 0\Rightarrow x\ne -1$. Domain $=\mathbb{R}\setminus \{-1\}$. Why this step? Division by zero is undefined — that forbidden input becomes the vertical asymptote.
2. Vertical asymptote — both signs.
   • As $x\to -1^{-}$ (from the left), $x+1$ is a tiny negative, so $\frac{1}{\text{tiny neg}}\to -\infty$.
   • As $x\to -1^{+}$ (from the right), $x+1$ is a tiny positive, so $\frac{1}{\text{tiny pos}}\to +\infty$. Vertical asymptote at $x=-1$. Why this step? You must check both sides — the sign of the denominator flips across the forbidden point.
3. Horizontal asymptote — both ends. As $x\to +\infty$ or $x\to -\infty$, denominator is huge, so $\frac{1}{\text{huge}}\to 0$. Horizontal asymptote $y=0$. Why this step? Dividing $1$ by an ever-larger number gives an ever-tinier result — the output is squeezed toward $0$ but never reaches it, which is exactly what a horizontal asymptote describes.
4. Sample the branches & quadrants. $h(0)=\frac{1}{1}=1$ → point $(0,1)$, the right branch, which lives in QI ($x>-1,y>0$). $h(-2)=\frac{1}{-1}=-1$ → point $(-2,-1)$, the left branch, which lives in QIII ($x<-1,y<0$). Why this step? One point on each side of the forbidden line pins down which way each branch bends and which quadrant it occupies — so you don't guess the shape.
5. Range. The output is $\frac{1}{\text{something}}$ — it can be any value except $0$. Range $=\mathbb{R}\setminus \{0\}$. Why this step? For the output to equal $0$ you would need $1=0\times (x+1)=0$, which is impossible — no input can force a fraction with numerator $1$ to be zero, so $0$ is the one height the curve never reaches.

Reading the figure: the red dashed vertical line at $x=-1$ is the wall the two blue branches race up and down along; the yellow dashed horizontal line $y=0$ is the floor/ceiling they flatten toward far left and far right; the green dots $(0,1)$ and $(-2,-1)$ confirm the right branch lives above the axis (QI) and the left branch below (QIII).

Verify: $h(-1.01)=\frac{1}{-0.01}=-100$ (large negative, matching $-1^{-}\to -\infty$) and $h(-0.99)=\frac{1}{0.01}=100$ (large positive, matching $-1^{+}\to +\infty$) ✓.

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Cell F — Restricted domain (square root)

Forecast: where does this curve begin? It cannot exist for all $x$ — why not?

1. Domain. You cannot square-root a negative in the reals, so we need $x-3\ge 0\Rightarrow x\ge 3$. Domain $=[3,\infty )$. Why this step? The square root filters out illegal inputs — the curve literally does not exist left of $x=3$.
2. Starting point. At $x=3$: $r(3)=\sqrt{0}=0$. Curve starts at $(3,0)$ (a solid endpoint, included). Why this step? The smallest legal input gives the tip where the curve is born.
3. Sample onward. $r(4)=\sqrt{1}=1$, $r(7)=\sqrt{4}=2$, $r(12)=\sqrt{9}=3$. Why this step? Perfect squares under the root give clean points — the curve rises but ever more gently.
4. Range & quadrants. Output is a square root, which is never negative, and grows without bound: range $=[0,\infty )$. Because both $x\ge 3>0$ and $y\ge 0$, the entire curve lives in QI only. Why this step? As $x-3$ runs over every value from $0$ upward, its square root runs over every value from $0$ upward too (nothing is skipped, since squaring is smooth) — so the outputs fill exactly $[0,\infty )$, all inside the first quadrant.

Reading the figure: the red dot at $(3,0)$ is where the blue curve is born; the shaded red region to its left is the forbidden zone where no point can exist; the yellow sample dots show the curve climbing but flattening — each unit of height needs a bigger jump in $x$.

Verify: $r(4)=1$, $r(7)=2$, $r(12)=3$ ✓. Left of $3$, e.g. $x=2$: $x-3=-1<0$, undefined — no point plotted ✓.

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Cell G — Even vs odd symmetry

Forecast: one graph folds neatly across the $y$-axis; the other looks the same after a $180°$ spin. Guess which is which.

1. Domain of both. Both are polynomials → domain $=\mathbb{R}$ each. Why this step? Symmetry tests only make sense once you know $-x$ is always a legal input too.
2. Even test on $u$. Replace $x$ with $-x$: $u(-x)=(-x{)}^2-1=x^2-1=u(x)$. Since $u(-x)=u(x)$, $u$ is even → mirror across the $y$-axis. It dips into QIII/QIV near the middle (where $y<0$) and rises into QII/QI at the sides. Why this step? "Even" literally means flipping the input's sign changes nothing — a left–right mirror.
3. Odd test on $v$. Replace $x$ with $-x$: $v(-x)=(-x{)}^3-(-x)=-x^3+x=-(x^3-x)=-v(x)$. Since $v(-x)=-v(x)$, $v$ is odd → symmetric about the origin (spin $180°$). It threads through all four quadrants as it wiggles up and down. Why this step? "Odd" means flipping the input flips the output too — a rotational symmetry.
4. Sanity on signs. $v$'s roots: $x^3-x=x(x-1)(x+1)=0\Rightarrow x=-1,0,1$ — symmetric roots, exactly what odd symmetry predicts. Why this step? The algebra test can hide a slip; checking concrete points (the roots) against the predicted symmetry is a cheap, independent confirmation that the classification is right.

Reading the figure: on the left, the blue parabola $u$ is a perfect left–right mirror across the yellow dashed $y$-axis. On the right, the green cubic $v$ is not mirrored left-to-right; instead, if you pin the red origin dot and rotate the whole curve by $180°$, every point lands exactly back on the curve — the point at $(2,6)$ swaps with $(-2,-6)$, the bump on the right becomes the dip on the left — and that "rotate-and-it-matches" behaviour is the visual signature of an odd function.

Verify: $u(2)=3$ and $u(-2)=3$ (equal → even) ✓. $v(2)=6$ and $v(-2)=-6$ (opposite → odd) ✓. More on this in Symmetry (even and odd functions).

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Cell H — Real-world word problem

Forecast: this line must stop — water height can't go negative. When does the story end?

1. Read the slope. $-4$ metres per minute → the tank loses $4$ m of height each minute (decreasing). Why this step? In word problems the slope carries units and physical meaning.
2. Start value. $H(0)=20$ m — the tank begins full at $20$ m. Why this step? Evaluating at $t=0$ gives the state "before any time passes" — the initial condition, which is the $y$-intercept dressed in physical units.
3. Empty time. Set $H(t)=0$: $20-4t=0\Rightarrow t=5$ minutes. Why this step? The $x$-intercept is the moment the physical quantity hits zero.
4. Sensible domain & range (and quadrant). Time can't be negative and the model dies when empty: domain $t\in [0,5]$ min. Height stays between full and empty: range $H\in [0,20]$ m. Since both $t\ge 0$ and $H\ge 0$, the whole graph sits in QI — the only quadrant that makes physical sense here. Why this step? The real world clips the mathematically-infinite line to a finite segment confined to one quadrant.

Verify: $H(0)=20$, $H(5)=20-20=0$ ✓. Halfway in time, $H(2.5)=20-10=10$ m — exactly half-drained ✓. Units: (m/min)$\times$(min) $=$ m ✓.

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Cell I — Exam twist: solve an equation graphically

Forecast: two curves, and their crossings are the answers. How many crossings do you expect?

1. Split into two graphs. Left side is the parabola $y=x^2$; right side is the line $y=x+2$. Solutions are where they intersect — same $x$, same $y$. Why this step? An equation $\text{LHS}=\text{RHS}$ is solved exactly where the two graphs meet. See Solving equations graphically.
2. Find the intersections algebraically to label them. $x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=2$ or $x=-1$. Why this step? Moving everything to one side turns "where curves meet" into "roots of one function."
3. Read the meeting heights. At $x=2$: $y=x^2=4$. At $x=-1$: $y=x^2=1$. Crossings $(2,4)$ and $(-1,1)$ — both above the axis, in QI and QII respectively. Why this step? At an intersection both curves share the same height, so the $y$-value is the common output — the point $(x,y)$ is the full picture of that solution, and it must satisfy both sides.
4. Twist check — how many crossings? A line can cut a parabola at most twice; here it cuts exactly twice, so both solutions are real. No hidden third root. (This is where the slope idea $f^{\prime }(x)$ from the preamble helps intuition: the parabola's slope keeps changing while the line's stays fixed, so they can align to cross at most twice.) Why this step? Setting line $=$ parabola gives a quadratic (degree $2$), and a quadratic has at most two real roots — so geometrically a line and a parabola can meet at most twice, telling you to stop looking after two crossings.

Reading the figure: the blue parabola and yellow line cross at the two red dots $(-1,1)$ and $(2,4)$; the $x$-coordinates of those dots ($-1$ and $2$) are the solutions of the equation, and you can read them straight off the horizontal axis.

Verify: at $x=2$: LHS $=4$, RHS $=2+2=4$ ✓. At $x=-1$: LHS $=1$, RHS $=-1+2=1$ ✓.

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Cell J — Parabola with no real roots

Forecast: this bowl opens upward like Ex 3 — but where are its roots? Try to find them before reading.

1. Domain. A polynomial once more → domain $=\mathbb{R}$. Why this step? Same opening habit; the "no real roots" surprise comes later, not from the domain.
2. Opening. Coefficient of $x^2$ is $+1>0$ → opens upward, a floor exists. Why this step? The leading sign again fixes bowl-up vs bowl-down before anything else.
3. Hunt for roots. Set $x^2+1=0\Rightarrow x^2=-1$. No real number squares to a negative, so there are no real roots — the curve never crosses the $x$-axis. Why this step? Trying to solve $f(x)=0$ and hitting an impossible equation is precisely how you detect the "no $x$-intercept" case.
4. Vertex. Smallest value of $x^2$ is $0$ at $x=0$, so $k(0)=1$. Vertex $(0,1)$ — the whole bowl sits above the axis. Why this step? The lowest point of an upward parabola is its floor; here the floor is at height $1>0$, which is why it misses the axis.
5. Range & quadrants. Lowest output $1$, rising forever → range $=[1,\infty )$. Since $y\ge 1>0$ always, the curve stays strictly above the axis, living in QII ($x<0$) and QI ($x>0$) only. Why this step? An upward bowl fills everything from its floor upward; the floor $1$ is above $0$, matching "no roots" and keeping it out of the lower quadrants.

Verify: $k(0)=1$ (vertex height) ✓; $x^2+1$ has discriminant $0^2-4(1)(1)=-4<0$, confirming no real roots ✓; $k(2)=5$ and $k(-2)=5$ (symmetric, all outputs $\ge 1$) ✓.

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Recall Scenario matrix — self-quiz

Sign of the $x^2$ coefficient tells you a parabola opens... ::: upward if positive, downward if negative. A forbidden input in a rational function becomes a... ::: vertical asymptote. $f(-x)=f(x)$ means the function is... ::: even (mirror across the $y$-axis). $f(-x)=-f(x)$ means the function is... ::: odd (symmetric about the origin). To solve $\text{LHS}=\text{RHS}$ graphically you look for... ::: the intersection points of $y=\text{LHS}$ and $y=\text{RHS}$. The domain of $\sqrt{x-3}$ is... ::: $x\ge 3$, i.e. $[3,\infty )$. A parabola has no real roots when its floor (vertex) sits... ::: entirely above the $x$-axis (for an upward bowl). The four quadrants, numbered anticlockwise from top-right, are... ::: QI (x>0,y>0), QII (x<0,y>0), QIII (x<0,y<0), QIV (x>0,y<0).

Related vault topics: Domain and range · Types of functions · Vertical line test · Transformations of functions · Asymptotes · Calculus: derivatives and graphs.