This page is a misconception gym. Each line is a claim or question; the part after ::: is the reveal. Cover the answer, commit to a verdict with a reason, then check. If your reason was "I just remember it that way", you haven't earned it yet — go re-read the relevant section in the parent topic.
Two symbols recur below, so pin them down once here:
The pictures below anchor the four traps students fall into most: slope sign, hole vs. asymptote, crossing an asymptote, and piecewise continuity.
True or false: every constant function is also a polynomial function.
True. f(x)=c is a polynomial of degree 0 (with a0=c) whenever c=0; the special case f(x)=0 is the zero polynomial with no defined degree, but it is still counted as a polynomial. See M02.01 Function Definition.
True or false: a linear function must pass through the origin.
False. Only f(x)=mx (the case b=0) passes through (0,0). In general f(x)=mx+b crosses the y-axis at height b, so any nonzero b shifts the whole line off the origin.
True or false: f(x)=7 is a linear function.
Borderline — it depends on your textbook. It fits f(x)=mx+b with m=0,b=7, so it is affine and its graph is a straight (horizontal) line, but many texts reserve "linear function" for m=0 and call this a constant function instead.
True or false: every quadratic function has two real roots.
False. f(x)=x2+1 has none, since x2+1>0 for all real x. A quadratic has 2, 1 (a repeated root), or 0 real roots depending on the sign of the discriminant b2−4ac.
True or false: a polynomial of degree n always has exactly n real roots.
False. It has at mostn real roots; some may be complex or repeated. It has exactly n roots only if you count complex roots with multiplicity.
True or false: f(x)=x−1x2−1 is the same function as g(x)=x+1.
False. Algebraically f(x)=x+1 for x=1, but f is undefined at x=1 (division by zero) while g(1)=2. They differ at one point, so as functions on their natural domains they are not equal — see M02.03 Domain and Range.
True or false: f(x)=x2−1x2+3x+2 has a vertical asymptote at both x=1 and x=−1.
False. At x=−1 both numerator and denominator vanish (a common factor x+1), giving a hole, not an asymptote; only x=1 is a true vertical asymptote (figure s02).
True or false: a rational function can cross its horizontal asymptote.
True. The horizontal asymptote describes end behaviour (x→±∞); nothing stops the curve from crossing it at finite x (figure s03). What is limited is how often: see the edge-case note below.
True or false: f(x)=x2 equals g(x)=x for all real x.
False. ⋅ returns the non-negative root, so x2=∣x∣. They agree for x≥0 but (−3)2=3=−3.
True or false: every piecewise function is discontinuous.
False. A piecewise definition can still produce a continuous curve if the pieces meet at the boundaries (figure s04, left), e.g. f(x)=x for x<0 and f(x)=x for x≥0 is just the line y=x. See M03.02 Continuity.
Error hunt: "In f(x)=−4x+7 the function increases because 7>0."
The slopem=−4<0 controls increasing/decreasing, not the intercept (figure s01). This function decreases; the 7 only sets where it crosses the y-axis.
Error hunt: "f(x)=−2x2+3 has a minimum at its vertex."
Since a=−2<0 the parabola opens downward, so the vertex is a maximum. Always check the sign of a before naming the extremum.
Error hunt: "f(x)=x2(x−1)(x−2)(x−3) is degree 5 because it has 5 roots."
Degree is the highest power after expanding, which is 5, but there are only 4distinct roots (x=0 is a double root). Root count = degree.
Error hunt: "The vertex of f(x)=ax2+bx+c is at x=2ab."
The formula is x=−2ab; the minus sign is essential. It comes from completing the square, where the squared term vanishes at x=−2ab.
Error hunt: "q(x)p(x) has a horizontal asymptote y=0 whenever the numerator degree is smaller, so x2x3 has one at y=0."
Here the numerator degree (3) is larger than the denominator (2), so there is no horizontal asymptote — there is an oblique one (y=x) instead. The rule was applied backwards.
Error hunt: "f(x)=x−3 has domain all real numbers."
A real square root needs a non-negative radicand, so we need x−3≥0, giving domain x≥3. The radical restricts the domain — see M02.03 Domain and Range.
Error hunt: "f(x)=x1 is continuous because you can draw it in one motion."
You cannot: the curve jumps from −∞ to +∞ across x=0, where the function is undefined. It is continuous on its domain but not on all of R.
Error hunt: "Since a constant function never changes, its derivative is undefined."
Its derivative is defined and equals 0. Feeding f(x)=c into the difference quotient gives hf(x+h)−f(x)=hc−c=h0=0 for every h=0, so the limit as h→0 is 0 — "no change" means zero rate of change, not an undefined one. See M04.01 Derivatives.
Why does an odd-degree polynomial always have at least one real root?
Its two ends run to opposite infinities (−∞ and +∞), so by continuity the graph must cross the x-axis somewhere in between (Intermediate Value Theorem), guaranteeing a real root.
Why does the leading term anxn decide end behaviour and not the other terms?
For large ∣x∣, xn grows faster than every lower power, so the ratio of lower terms to anxn shrinks to zero and the leading term dominates the sign and size — this is why limx→±∞f(x)=limanxn (see M03.01 Limits).
Why is a=0 required in the definition of a quadratic?
If a=0 the x2 term disappears and f(x)=bx+c becomes linear (or constant). The x2 term is exactly what makes the rate of change itself change.
Why do rational functions "blow up" at zeros of the denominator?
Near such a point the numerator is (usually) nonzero while the denominator shrinks toward 0, so the quotient grows without bound — division by an ever-tinier number gives an ever-larger result, producing a vertical asymptote.
Why can a quadratic be written in three forms (standard, vertex, factored) yet be the same function?
Each form is the same polynomial rearranged: expanding vertex or factored form returns the standard coefficients. Each just makes a different feature obvious — intercept, vertex, or roots respectively.
Why does completing the square reveal the vertex?
It rewrites f as a(x−h)2+k; a squared quantity is smallest (zero) when its inside is zero, i.e. at x=h, forcing f to its extreme value k there. So (h,k) is the turning point.
Why does a radical function like x have range y≥0 rather than all reals?
The principal square root symbol returns only the non-negative root by convention, so its outputs never dip below 0 even though x can be large. Its inverse y=x2 (restricted) recovers the other branch — see M02.06 Inverse Functions.
It is simultaneously the constant function c=0, a linear function with m=b=0, and the zero polynomial. Its degree is conventionally left undefined, though some authors assign deg(0)=−∞ so that the degree-addition rule deg(pq)=degp+degq keeps working.
Edge case: is f(x)=x1/2 a polynomial?
No. Polynomials require non-negative integer exponents; the exponent 21 makes this a radical function, with a restricted domain x≥0.
Edge case: is f(x)=32x+1 a rational function or a linear one?
Both descriptions fit, but the denominator is a constant polynomial, so it simplifies to f(x)=32x+31 — an ordinary linear function with no asymptote or domain restriction.
Edge case: can a horizontal or oblique asymptote be crossed infinitely often?
No — only finitely often. A crossing at x=t means f(t)−L=0 (or f(t)−(mx+b)=0); clearing the denominator turns this into a polynomial equation, and a nonzero polynomial has finitely many roots. Vertical asymptotes, by contrast, mark points outside the domain, so the curve can never touch them at all.
Edge case: what happens to f(x)=ax2+bx+c as a→0?
The parabola flattens and widens without bound; in the limit the x2 term vanishes and the graph degenerates into the straight line y=bx+c. This is why a=0 is excluded from "quadratic".
Edge case: can a piecewise function be defined by the same rule on every piece?
Yes, but then the pieces stitch into one ordinary function and the piecewise notation is redundant. Piecewise structure only matters when the rule genuinely changes across boundaries.
Edge case: does f(x)=x2+11 have any vertical asymptotes?
No. The denominator x2+1 is never zero for real x (it is always ≥1), so the function is defined and continuous everywhere — a rational function without vertical asymptotes.
Edge case: what is the domain of f(x)=x−2x?
You need both x≥0 (for the radical) andx=2 (nonzero denominator), so the domain is [0,2)∪(2,∞). Two restrictions must be combined, not chosen between.
Recall Quick self-test
The vertex x-coordinate of ax2+bx+c ::: x=−2ab
Sign of a that opens a parabola downward ::: a<0
A hole (not asymptote) occurs when ::: numerator and denominator share a common factor that cancels
x2 simplifies to ::: ∣x∣, not x
Number of times a horizontal asymptote can be crossed ::: finitely many, since it reduces to a polynomial equation