Types — constant, linear, quadratic, polynomial, rational, radical, piecewise: Exercises

This page is a self-test ladder. Read one problem, cover the solution, try it, then open the collapsible [!recall]- to check. Difficulty climbs from just recognise the type up to combine everything at once. Every symbol used here was built in the parent note Types of Functions; if a word feels new, re-read that note first.

Prerequisite ideas you may want open in another tab: M02.01 Function Definition, M02.03 Domain and Range, M04.01 Derivatives, M03.01 Limits.

Level 1 — Recognition

The four types have four instantly recognisable shapes. In the figure, each panel is labelled with its own part-letter (a)–(d) — match the formula to the correctly labelled panel:

Figure — Types — constant, linear, quadratic, polynomial, rational, radical, piecewise

Recall Solution 1.1

What to look at: the shape of the formula, not the numbers (and its labelled panel above).

(a) $f (x) = 7$ — no $x$ appears at all, so the output never changes. This is a constant function. As a polynomial its degree is $0$ . Panel (a): a flat horizontal line (lavender).
(b) $g (x) = 4 x - 9$ — the highest power of $x$ is $x^{1}$ . That is a linear function, polynomial of degree 1 (slope $4$ , intercept $- 9$ ). Panel (b): a straight slanted line (coral).
(c) $h (x) = \frac{x + 1}{x - 2}$ — one polynomial divided by another. That is a rational function. Not a polynomial (division by an $x$ -term), so "degree" does not apply the same way. Panel (c): two pieces split by a vertical asymptote at $x = 2$ (mint).
(d) $k (x) = x$ — the variable sits under a root sign, i.e. $x^{1/2}$ . The exponent is not a whole number, so this is a radical function, not a polynomial. Panel (d): a sideways half-parabola starting at the origin (butter).

Recall Solution 1.2

False. A polynomial only allows non-negative whole-number exponents ( $x^{0}, x^{1}, x^{2}, \dots$ ). Here $x^{- 2} = \frac{1}{x ^{2}}$ has a negative exponent, so $p$ is actually a rational function in disguise: $p (x) = \frac{1 + 3 x ^{2}}{x ^{2}}$ .

Level 2 — Application

Before the algebra, picture what "rise over run" means on the two given points:

Recall Solution 2.1

Type: linear → tool: slope-from-two-points, because a straight line has one constant rate of change and two points pin it down completely. $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3.$ In the figure the run (horizontal, $3 - 1 = 2$ ) and the rise (vertical, $10 - 4 = 6$ ) form a right triangle; the slope is the rise divided by the run. Now use one point to find $b$ . Plug $(1, 4)$ into $y = 3 x + b$ : $4 = 3 (1) + b \Rightarrow b = 1.$ Answer: $f (x) = 3 x + 1$ . Check the other point: $f (3) = 3 (3) + 1 = 10$ ✓.

The parabola and its lowest point make the answer visible before you compute:

Recall Solution 2.2

Type: quadratic → tool: vertex formula $h = - \frac{b}{2 a}$ , because a parabola is symmetric and its turning point sits at that $x$ . Here $a = 2, b = - 8, c = 6$ : $h = - \frac{- 8}{2 ( 2 )} = \frac{8}{4} = 2.$ $k = q (2) = 2 (4) - 8 (2) + 6 = 8 - 16 + 6 = - 2.$ Vertex: $(2, - 2)$ . Since $a = 2 > 0$ the parabola opens upward (a valley, as in the figure), so this is a minimum.

Recall Solution 2.3

Type: radical → tool: the inside of an even root must stay $\geq 0$ (you cannot square-root a negative and stay real). See M02.03 Domain and Range. $x - 5 \geq 0 \Rightarrow x \geq 5.$ Domain: $[5, \infty)$ .

Level 3 — Analysis

The dashed guide lines are exactly what you are solving for — find them in the picture:

Recall Solution 3.1

Type: rational → the standard procedure is factor top and bottom first, cancel any common factor, and only then hunt for asymptotes. This ordering matters (see the L3 Trap below). Step 1 — factor both: $s (x) = \frac{2 x ^{2} - 2}{x ^{2} - x - 6} = \frac{2 ( x - 1 ) ( x + 1 )}{( x - 3 ) ( x + 2 )} .$ Step 2 — cancel common factors: the top factors are $(x - 1), (x + 1)$ ; the bottom factors are $(x - 3), (x + 2)$ . No factor is shared, so nothing cancels — meaning there are no holes, and every zero of the denominator is a true asymptote. Step 3 — vertical asymptotes (tool: denominator $= 0$ after cancelling): $(x - 3) (x + 2) = 0 \Rightarrow x = 3 or x = - 2.$ Both survive, so both are genuine vertical asymptotes (dashed lines in the figure). Step 4 — horizontal asymptote (tool: compare leading terms as $x \to \pm \infty$ ) — for huge $∣ x ∣$ only the top powers matter (see M03.01 Limits). Here top and bottom are the same degree ( $2$ ), so we divide leading coefficients: $lim_{x \to \pm \infty} \frac{2 x ^{2} - 2}{x ^{2} - x - 6} = \frac{2 x ^{2}}{x ^{2}} = 2.$ Horizontal asymptote: $y = 2$ .

Recall Solution 3.2

Type: polynomial (degree $5 \geq 3$ , so it earns the "polynomial" name) → tool: the leading term $- 3 x^{5}$ dominates for large $∣ x ∣$ . Degree $5$ is odd, leading coefficient $- 3$ is negative.

As $x \to + \infty$ : $x^{5} \to + \infty$ , times $- 3$ gives $t (x) \to - \infty$ .
As $x \to - \infty$ : $x^{5} \to - \infty$ , times $- 3$ gives $t (x) \to + \infty$ . So the graph comes down from the top-left and exits out the bottom-right — the classic odd-degree, negative-leader shape.

Recall Solution 3.3

Type: rational, and the top degree ( $2$ ) is exactly one more than the bottom degree ( $1$ ) — the flag for a slant asymptote. Tool: polynomial long division, because we want the straight line the graph hugs at infinity. Divide $x^{2} + 1$ by $x - 1$ : $\frac{x ^{2} + 1}{x - 1} = x + 1 + \frac{2}{x - 1} .$ As $x \to \pm \infty$ the remainder term $\frac{2}{x - 1} \to 0$ , so the graph approaches the slanted line $y = x + 1.$ Oblique asymptote: $y = x + 1$ .

Level 4 — Synthesis

Recall Solution 4.1

Type: piecewise → tool: compare the value the two pieces approach at the joint $x = 1$ (continuity idea, see M03.02 Continuity).

Left piece as $x \to 1^{-}$ : $x + 3 \to 1 + 3 = 4$ .
Right piece at $x = 1$ : $x^{2} = 1^{2} = 1$ . The left branch lands at height $4$ but the right branch starts at height $1$ . They do not meet, so there is a jump of size $4 - 1 = 3$ . Reading the figure (works even in greyscale): the upper, straight branch (labelled "x + 3") ends at an open circle at the point $(1, 4)$ — hollow, meaning that height is not reached. The lower, curved branch (labelled "x^2") begins at a filled circle at $(1, 1)$ — solid, meaning that height is reached. A solid double-headed arrow marks the gap of $3$ between them. Because the filled point and the hollow point sit at different heights, $f$ is not continuous at $x = 1$ .

Recall Solution 4.2

Type: quadratic → tool: complete the square, because vertex form $a (x - h)^{2} + k$ reads off the minimum directly. $x^{2} + 6 x + 11 = (x^{2} + 6 x + 9) - 9 + 11 = (x + 3)^{2} + 2.$ So $h = - 3, k = 2$ : vertex $(- 3, 2)$ . Since the squared term is never negative, the smallest value is when $(x + 3)^{2} = 0$ . Minimum value: $2$ (attained at $x = - 3$ ).

Level 5 — Mastery

Recall Solution 5.1

Type: quadratic (downward parabola, $a = - 5 < 0$ ). (a) Time of peak — tool: vertex. $t = - \frac{b}{2 a} = - \frac{20}{2 ( - 5 )} = - \frac{20}{- 10} = 2 s .$ (b) Maximum height — plug $t = 2$ back in. $h (2) = - 5 (4) + 20 (2) + 1 = - 20 + 40 + 1 = 21 m .$ (c) Ground time — tool: quadratic formula, because we need the roots of $- 5 t^{2} + 20 t + 1 = 0$ . First multiply both sides by $- 1$ : $5 t^{2} - 20 t - 1 = 0$ . Why is this allowed, and why bother? Multiplying an equation " $= 0$ " by any non-zero number gives the same solution set — if the left side was zero, $- 1$ times zero is still zero, so no roots are gained or lost. We choose $- 1$ only for comfort: it makes the leading coefficient positive so the formula is tidier. With $a = 5, b = - 20, c = - 1$ : $t = \frac{20 \pm ( - 20 ) ^{2} - 4 ( 5 ) ( - 1 )}{2 ( 5 )} = \frac{20 \pm 400 + 20}{10} = \frac{20 \pm 420}{10} .$ $420 \approx 20.4939$ , giving $t \approx \frac{20 + 20.4939}{10} \approx 4.05$ s or $t \approx \frac{20 - 20.4939}{10} \approx - 0.05$ s. A negative time is before we threw the ball, so discard it. The ball lands at $t \approx 4.05$ s.

Recall Solution 5.2

This is a rational function whose numerator is radical — so two conditions must hold at once. Condition 1 (radical, M02.03 Domain and Range): the inside of the square root $\geq 0$ : $x + 4 \geq 0 \Rightarrow x \geq - 4.$ Condition 2 (rational): the denominator $\neq = 0$ : $x - 1 \neq = 0 \Rightarrow x \neq = 1.$ Intersect both: start with $x \geq - 4$ , then punch out the single point $x = 1$ . Domain: $[- 4, 1) \cup (1, \infty)$ .

Recall Quick recall check

Which tool answers "where does a rational function blow up"? ::: Factor top and bottom, cancel common factors, THEN set the denominator equal to zero. What sign of $a$ makes a parabola's vertex a maximum? ::: $a < 0$ (opens downward). Why is $x$ not a polynomial? ::: Its exponent $1/2$ is not a non-negative whole number. When does a rational function have a slant (oblique) asymptote instead of a horizontal one? ::: When the numerator's degree is exactly one more than the denominator's.

Related deep dives: M02.04 Function Transformations, M02.06 Inverse Functions, M04.01 Derivatives, M05.01 Integration.