Graphs of functions — plotting, reading key features: Exercises

Prerequisite ideas you may want open in another tab: Domain and range, Asymptotes, Symmetry (even and odd functions), Solving equations graphically, Transformations of functions, Vertical line test, Types of functions, Calculus: derivatives and graphs.

Level 1 — Recognition

Goal: read a feature straight off a picture, no algebra.

Exercise 1.1 — Is it a function?

Look at figure s01. Two curves are drawn, a teal one and a plum one. Using the vertical line test (a vertical line may cross the graph of a function at most once, because one input gives exactly one output), decide which curve is the graph of a function.

Figure — Graphs of functions — plotting, reading key features

Recall Solution 1.1

Drop a vertical dotted line anywhere.

The teal curve (a line-like S that never doubles back) is crossed once by every vertical line → it is a function.
The plum curve (a sideways parabola, $x = y^{2}$ ) is crossed twice for $x > 0$ (a top branch and a bottom branch) → not a function.

Answer: the teal curve is a function; the plum curve is not.

Exercise 1.2 — Read the intercepts

From figure s01, the teal graph crosses the $x$ -axis at one visible point and the $y$ -axis at one point. Recall: an $x$ -intercept is where the curve meets the $x$ -axis (height $y = 0$ ), a $y$ -intercept is where it meets the $y$ -axis (position $x = 0$ ). State both for the teal line, which is $f (x) = x - 1$ .

Recall Solution 1.2

$y$ -intercept: set $x = 0$ , so $f (0) = 0 - 1 = - 1$ → point $(0, - 1)$ .
$x$ -intercept: set $f (x) = 0$ , so $x - 1 = 0 \Rightarrow x = 1$ → point $(1, 0)$ .

Level 2 — Application

Goal: plug into a rule, sample points, read features.

Exercise 2.1 — Plot and describe a line

For $f (x) = - \frac{1}{2} x + 2$ : give the domain, both intercepts, whether it is increasing or decreasing, and the range.

Recall Solution 2.1

Domain: no square roots, no division — every real $x$ works. Domain $= R$ . $y$ -intercept: $f (0) = - \frac{1}{2} (0) + 2 = 2$ → $(0, 2)$ . $x$ -intercept: $- \frac{1}{2} x + 2 = 0 \Rightarrow \frac{1}{2} x = 2 \Rightarrow x = 4$ → $(4, 0)$ . Increasing or decreasing? The slope is $- \frac{1}{2} < 0$ . A negative slope means "downhill left-to-right," so $f$ is decreasing on all of $R$ . Range: a non-flat line reaches every height eventually, so range $= R$ .

Exercise 2.2 — Quadratic features

For $g (x) = x^{2} - 6 x + 5$ : find the roots, the vertex, and state the range.

Recall Solution 2.2

Roots (where $g = 0$ ): factor $x^{2} - 6 x + 5 = (x - 1) (x - 5)$ , so $x = 1$ and $x = 5$ . Vertex $x$ -coordinate: for $a x^{2} + b x + c$ the turning point sits at $x = - \frac{b}{2 a}$ (the axis of symmetry, exactly halfway between the two roots — and indeed $\frac{1 + 5}{2} = 3$ ). Here $x = - \frac{- 6}{2 ( 1 )} = 3$ . Vertex height: $g (3) = 9 - 18 + 5 = - 4$ → vertex $(3, - 4)$ . Since $a = 1 > 0$ the parabola opens upward, so the vertex is the lowest point. Range: $[- 4, \infty)$ .

Exercise 2.3 — Where does a rational function blow up?

For $h (x) = \frac{1}{x + 3}$ : state the domain, the vertical asymptote, the horizontal asymptote, and the $y$ -intercept.

Recall Solution 2.3

Domain: denominator must not be $0$ : $x + 3 \neq = 0 \Rightarrow x \neq = - 3$ . Domain $= R ∖ {- 3}$ . Vertical asymptote: at the forbidden input $x = - 3$ the denominator shrinks toward $0$ , so the fraction grows without bound → vertical asymptote $x = - 3$ . Horizontal asymptote: as $x \to \pm \infty$ the denominator grows huge, so $\frac{1}{huge} \to 0$ → horizontal asymptote $y = 0$ . $y$ -intercept: $h (0) = \frac{1}{0 + 3} = \frac{1}{3}$ → $(0, \frac{1}{3})$ .

Level 3 — Analysis

Goal: reason about behaviour across cases, signs, and limits.

Exercise 3.1 — Sign chart of a rational function

For $h (x) = \frac{1}{x - 2}$ (the parent note's Example 3), determine the sign of $h$ on each side of the asymptote, and describe which quadrants the two branches live in. See figure s02.

Recall Solution 3.1

Split the number line at the only special point, $x = 2$ .

Left region $x < 2$ : then $x - 2 < 0$ , so $h = \frac{1}{negative} < 0$ . The left branch sits below the $x$ -axis. For $x < 0$ it is in quadrant III; for $0 < x < 2$ it is in quadrant IV. As $x \to 2^{-}$ , $x - 2 \to 0^{-}$ (a tiny negative), so $h \to - \infty$ : the branch dives down along the asymptote.
Right region $x > 2$ : then $x - 2 > 0$ , so $h > 0$ . The right branch sits above the $x$ -axis in quadrant I. As $x \to 2^{+}$ , $x - 2 \to 0^{+}$ , so $h \to + \infty$ : it climbs the asymptote.
Both branches flatten toward $y = 0$ as $x \to \pm \infty$ .

Answer: negative and below the axis for $x < 2$ (quadrants III and IV), positive and above for $x > 2$ (quadrant I).

Exercise 3.2 — Even, odd, or neither?

Classify each by symmetry. Recall: $f$ is even if $f (- x) = f (x)$ (mirror across the $y$ -axis) and odd if $f (- x) = - f (x)$ (rotate $18 0^{\circ}$ about the origin).

(a) $p (x) = x^{2} - 3$ (b) $q (x) = x^{3} - x$ (c) $r (x) = x^{2} + x$

Recall Solution 3.2

(a) $p (- x) = (- x)^{2} - 3 = x^{2} - 3 = p (x)$ → even (symmetric about $y$ -axis). (b) $q (- x) = (- x)^{3} - (- x) = - x^{3} + x = - (x^{3} - x) = - q (x)$ → odd (symmetric about origin). (c) $r (- x) = x^{2} - x$ . This is neither $r (x) = x^{2} + x$ (the $x$ flipped sign) nor $- r (x) = - x^{2} - x$ (the $x^{2}$ term is wrong). → neither.

Exercise 3.3 — Turning points and end behaviour

The polynomial $k (x) = x^{3} - 3 x$ has degree $3$ . Recall a degree- $n$ polynomial has at most $n - 1$ turning points. State the maximum possible turning points, find the actual ones by testing values, and describe the end behaviour as $x \to \pm \infty$ .

Recall Solution 3.3

Max turning points: $n - 1 = 3 - 1 = 2$ . Locate them by sampling: $k (- 2) = - 8 + 6 = - 2$ , $k (- 1) = - 1 + 3 = 2$ , $k (0) = 0$ , $k (1) = 1 - 3 = - 2$ , $k (2) = 8 - 6 = 2$ . The values go up to a peak near $x = - 1$ ( $k = 2$ ), fall to a valley near $x = 1$ ( $k = - 2$ ). So $2$ turning points — the maximum. (Their exact positions $x = \pm \frac{1}{3}$ come from Calculus: derivatives and graphs.) End behaviour: leading term $x^{3}$ dominates for large $∣ x ∣$ . As $x \to + \infty$ , $x^{3} \to + \infty$ ; as $x \to - \infty$ , $x^{3} \to - \infty$ . The curve rises from bottom-left to top-right.

Level 4 — Synthesis

Goal: build a graph and its full description from a rule.

Exercise 4.1 — Full sketch of a transformed reciprocal

Fully analyse $f (x) = \frac{1}{x - 1} + 2$ and sketch it: domain, asymptotes, intercepts, and which quadrants (relative to the asymptote crossing) each branch occupies. See figure s03.

Recall Solution 4.1

Read this as the base curve $\frac{1}{x}$ shifted right by 1 and up by 2. Domain: $x - 1 \neq = 0 \Rightarrow x \neq = 1$ . Domain $= R ∖ {1}$ . Vertical asymptote: $x = 1$ (where the denominator vanishes). Horizontal asymptote: as $x \to \pm \infty$ , $\frac{1}{x - 1} \to 0$ , so $f \to 0 + 2 = 2$ . Asymptote $y = 2$ . $y$ -intercept: $f (0) = \frac{1}{- 1} + 2 = - 1 + 2 = 1$ → $(0, 1)$ . $x$ -intercept: set $f = 0$ : $\frac{1}{x - 1} = - 2 \Rightarrow x - 1 = - \frac{1}{2} \Rightarrow x = \frac{1}{2}$ → $(\frac{1}{2}, 0)$ . Branches (split at $x = 1$ , compare to the new centre $(1, 2)$ ):

For $x < 1$ : $x - 1 < 0$ so $\frac{1}{x - 1} < 0$ , and $f < 2$ . The branch sits below the horizontal asymptote, left of the vertical one; as $x \to 1^{-}$ , $f \to - \infty$ .
For $x > 1$ : $\frac{1}{x - 1} > 0$ so $f > 2$ , above the horizontal asymptote; as $x \to 1^{+}$ , $f \to + \infty$ . Range: $f$ never equals $2$ , so range $= R ∖ {2}$ .

Exercise 4.2 — Recover a quadratic from its graph

A parabola opens upward, touches the $x$ -axis at $x = 4$ (a repeated root), and passes through $(0, 16)$ . Find its formula and its vertex.

Recall Solution 4.2

"Touches" at $x = 4$ means $x = 4$ is a double root, so $g (x) = a (x - 4)^{2}$ for some $a > 0$ . Use the point $(0, 16)$ : $16 = a (0 - 4)^{2} = 16 a \Rightarrow a = 1$ . So $g (x) = (x - 4)^{2} = x^{2} - 8 x + 16$ . Vertex: for $a (x - 4)^{2}$ the turning point is where the squared part is $0$ , i.e. $x = 4$ , height $g (4) = 0$ . Vertex $(4, 0)$ — which is exactly the touch point, as expected for a double root.

Level 5 — Mastery

Goal: combine multiple features, cases, and cross-topic reasoning.

Exercise 5.1 — Solve an equation graphically

Use graphs to count the solutions of $x^{2} - 4 = \frac{4}{x}$ for $x \neq = 0$ , and estimate them by testing integer inputs. (Technique: Solving equations graphically — solutions are the $x$ where the two curves intersect.)

Recall Solution 5.1

Let $L (x) = x^{2} - 4$ (upward parabola, vertex $(0, - 4)$ ) and $R (x) = \frac{4}{x}$ (reciprocal, asymptotes $x = 0$ , $y = 0$ ). Solutions of the equation are the $x$ -coordinates where $L$ and $R$ cross. Equivalently, move everything to one side: $x^{2} - 4 - \frac{4}{x} = 0$ . Multiply by $x$ (valid since $x \neq = 0$ ): $x^{3} - 4 x - 4 = 0$ . Test integers: at $x = - 2$ : $- 8 + 8 - 4 = - 4$ ; at $x = - 1$ : $- 1 + 4 - 4 = - 1$ ; at $x = 1$ : $1 - 4 - 4 = - 7$ ; at $x = 2$ : $8 - 8 - 4 = - 4$ ; at $x = 3$ : $27 - 12 - 4 = 11$ . Sign changes: none among negatives shown, but between $x = 2$ ( $- 4$ ) and $x = 3$ ( $+ 11$ ) the value crosses $0$ → one root between $2$ and $3$ . Check negatives further: $x = - 1.5$ : $- 3.375 + 6 - 4 = - 1.375$ ; $x = - 0.5$ : $- 0.125 + 2 - 4 = - 2.125$ . Both branches near $0$ stay negative there; testing $x = - 1$ gave $- 1$ , still no sign change — but the reciprocal branch on the left drives $R \to - \infty$ , so geometrically the left parabola arm and the left reciprocal arm cross once too. Sampling $x = - 3$ : $- 27 + 12 - 4 = - 19$ (negative), $x = - 2$ : $- 4$ (negative). So the cubic stays negative for $x \leq - 2$ and around $x = - 1$ ; the single real sign change of $x^{3} - 4 x - 4$ is between $2$ and $3$ . Answer: exactly one real solution, located between $x = 2$ and $x = 3$ (numerically $x \approx 2.38$ ).

Exercise 5.2 — Piecewise domain, range, and continuity

Let $f (x) = ⎩ ⎨ ⎧ x + 1, x^{2}, 4, x < 0 0 \leq x \leq 2 x > 2$ Find $f (- 1)$ , $f (0)$ , $f (2)$ , $f (3)$ ; state the range; and identify any jump (a break where the left and right values disagree).

Recall Solution 5.2

Values (pick the piece whose condition your $x$ satisfies):

$f (- 1)$ : $- 1 < 0$ → use $x + 1$ : $- 1 + 1 = 0$ .
$f (0)$ : $0 \leq 0 \leq 2$ → use $x^{2}$ : $0$ .
$f (2)$ : $0 \leq 2 \leq 2$ → use $x^{2}$ : $4$ .
$f (3)$ : $3 > 2$ → use $4$ : $4$ . Continuity check at the seams:
At $x = 0$ : left piece approaches $0 + 1 = 1$ , right piece gives $0^{2} = 0$ . They disagree → a jump of size $1$ at $x = 0$ .
At $x = 2$ : left/middle gives $2^{2} = 4$ , right piece gives $4$ . They match → continuous there. Range: left ray $x + 1$ for $x < 0$ covers $(- \infty, 1)$ ; $x^{2}$ on $[0, 2]$ covers $[0, 4]$ ; the constant piece gives ${4}$ . Union: $(- \infty, 1) \cup [0, 4] = (- \infty, 4]$ . Answer: $f (- 1) = 0, f (0) = 0, f (2) = 4, f (3) = 4$ ; range $= (- \infty, 4]$ ; jump discontinuity at $x = 0$ .

Recall Self-test checklist

Read your graph against this before you call it finished ::: domain, both intercepts, increasing/decreasing intervals, turning points, every asymptote (drawn dashed), symmetry, and end behaviour as $x \to \pm \infty$ . First move for a rational function ::: find where the denominator is zero (that is your vertical asymptote and your domain hole). First move for a piecewise value ::: check which inequality the input satisfies, respecting strict vs non-strict.