Polynomial and Rational Functions — Free Pre-Calculus Review Games.

Even degree, positive leading coefficient: both ends go to positive infinity.

Even multiplicity means the graph touches the x-axis but does not cross.

\(x^2 - 4 = (x-2)(x+2) = 0\) at \(x = 2\) and \(x = -2\).

(x-1)^2 contributes degree 2, (x+3) contributes 1: total degree 3.

Same degree: HA = ratio of leading coefficients = \(3/1 = 3\).

\(x(x^2-4) = x(x-2)(x+2) = 0\). Zeros: \(0, 2, -2\).

A degree-n polynomial has at most n-1 turning points.

Numerator degree one more than denominator gives a slant (oblique) asymptote.

Set each factor to 0: x = 1, -2, 5.

Possible zeros = factors of constant (\(4\)) over factors of leading coefficient (\(1\)).

Factor: \(\frac{(x+1)(x+2)}{x+1} = x+2\). This is actually a hole, not a slant asymptote. The simplified form is \(y = x+2\) everywhere except \(x=-1\).

Since f changes sign (negative to positive), there must be at least one zero between 1 and 3.

Complex zeros come in conjugate pairs for polynomials with real coefficients.

\(\frac{5}{(x-1)(x+1)} = \frac{A}{x-1}+\frac{B}{x+1}\). \(A(x+1)+B(x-1)=5\). \(A=5/2, B=-5/2\).

\(x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x-i)(x+i)\).

Since the degree is odd (3) and the leading coefficient is negative, the graph falls to the right and rises to the left, matching the pattern of \(-x^3\). The distractor 'As \(x \to -\infty\), \(f(x) \to -\infty\); as \(x \to \infty\), \(f(x) \to \infty\)' describes a positive odd-degree function like \(x^3\), not one with a negative leading coefficient. Always check both the parity of the degree and the sign of the leading coefficient together to determine end behavior.

A zero with odd multiplicity 3 causes the graph to cross the x-axis, though it flattens near that point because the factor is cubed. The option 'The graph touches the x-axis and turns around' describes even multiplicity, which does not apply here since 3 is odd. Remember: odd multiplicities cross the axis while even multiplicities bounce off it.

Vertical asymptotes occur where the denominator equals zero and the numerator does not, so setting \((x-5)(x+3)=0\) gives \(x=5\) and \(x=-3\). The choice '\(x = -5\) and \(x = 3\)' incorrectly reverses the signs of the roots from the factors. Always solve each factor of the denominator for zero, keeping the sign as written in the factor.

The degree of a product of factors is the sum of the exponents, so \(3 + 2 = 5\). The distractor '6' would result from multiplying the exponents instead of adding them, which is not how polynomial degree works. When a polynomial is given in factored form, always add the exponents on each factor to find the total degree.

When the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is always \(y = 0\). The choice '\(y = 5\)' mistakenly treats the leading coefficient of the numerator as the asymptote, which only applies when the degrees are equal. Compare the degrees of numerator and denominator first to know which horizontal asymptote rule applies.

The end behavior of any polynomial is determined solely by its leading term, so a degree-4 polynomial with positive leading coefficient behaves like \(x^4\), rising on both ends. The option '\(-x^4\)' would apply if the leading coefficient were negative, causing the graph to fall on both ends instead. For end behavior purposes, all lower-degree terms can be ignored since they become insignificant as \(x\) grows large.

The factor \((x-4)\) appears to the first power, giving multiplicity 1, and odd multiplicities always cause the graph to cross straight through the x-axis. The option 'multiplicity 2' is incorrect because the factor is not squared anywhere in the expression. A multiplicity of exactly 1 means the graph passes through the zero with a roughly linear slope, unlike higher even multiplicities.

The Fundamental Theorem of Algebra guarantees that a degree-\(n\) polynomial has exactly \(n\) zeros when complex zeros and multiplicities are counted. The choice 'Up to 6, but could be fewer' is a common misconception since without counting multiplicity some students undercount, but with multiplicity and complex zeros included the count is always exactly 6. Always remember that this theorem counts zeros with multiplicity, including complex ones, not just distinct real zeros.

A polynomial function consists of terms with nonnegative integer exponents on the variable, which \(3x^4 - 2x + 7\) satisfies. The option \(\dfrac{x+1}{x-2}\) is a rational function because it is a ratio of two polynomials, not a single polynomial expression. Recognizing the structural difference between polynomial and rational functions is essential before analyzing asymptotes or end behavior.

Since the numerator's degree (1) is less than the denominator's degree (3), the function approaches zero as \(x \to \pm\infty\), giving horizontal asymptote \(y = 0\). The distractor '\(y = 4\)' incorrectly assumes the degrees are equal, which they are not here. Whenever the denominator's degree is strictly greater, the horizontal asymptote is always the x-axis.

The y-intercept is found by evaluating \(f(0)\), which gives \(2(0)^3 - 5(0)^2 + 0 - 6 = -6\). The choice '\(2\)' mistakenly uses the leading coefficient instead of substituting \(x = 0\) into the full expression. To find any y-intercept, always substitute \(x = 0\) directly into the function.

When a factor cancels between numerator and denominator, the discontinuity becomes a removable hole rather than a vertical asymptote, which is what happens at \(x = -3\) here. The option 'the denominator never equals zero at \(x=-3\)' is false since \((x-1)(x+3)=0\) when \(x=-3\), but the cancellation prevents an asymptote from forming. Always simplify a rational function fully before identifying vertical asymptotes, since common factors indicate holes, not asymptotes.

With an even degree (6) and a positive leading coefficient, both ends of the graph rise toward positive infinity, matching the shape of \(x^6\). The option describing opposite behavior on each end applies only to odd-degree polynomials, not even-degree ones like this. Even-degree polynomials always have matching end behavior on both sides, unlike odd-degree polynomials.

Factoring \(2x^2 - 5x - 3\) gives \((2x+1)(x-3) = 0\), so the zeros are \(x = -\dfrac{1}{2}\) and \(x = 3\). The option '\(x = 3\) and \(x = \dfrac{1}{2}\)' incorrectly flips the sign on the second root, which does not satisfy the original equation when checked. Always verify factored zeros by substituting back into the original quadratic to confirm accuracy.

A polynomial of degree \(n\) can have at most \(n-1\) turning points, so a degree-5 polynomial has at most 4. The option '5' incorrectly equates the number of turning points with the degree itself rather than subtracting one. Remember this rule as \(n - 1\) whenever estimating the maximum number of direction changes in a polynomial graph.

When the degrees of numerator and denominator are equal, the horizontal asymptote equals the ratio of leading coefficients, \(\dfrac{6}{2} = 3\). The option '\(y = 6\)' mistakenly uses only the numerator's leading coefficient without dividing by the denominator's. Whenever degrees match, always divide the leading coefficients to find the horizontal asymptote.

The factor \((x-1)^3\) gives multiplicity 3, an odd number, so the graph crosses the x-axis while flattening near the zero due to the higher power. The option 'Multiplicity 2; the graph touches and turns around' actually describes the behavior at \(x = -2\), not \(x = 1\). When analyzing multiple zeros in one function, match each multiplicity to its correct factor and corresponding root before describing the graph's behavior.

Factoring the denominator gives \((x-2)(x-3)\), and since neither factor cancels with the numerator \((x-1)\), both \(x=2\) and \(x=3\) are vertical asymptotes. The option '\(x = 1\) and \(x = 6\)' misreads the numerator's zero and the constant term of the denominator as if they were asymptotes. Always factor the denominator completely and check for cancellation with the numerator before identifying vertical asymptotes.

The domain excludes any value that makes the denominator zero, and solving \(x^2 - 9 = 0\) gives \(x = 3\) and \(x = -3\). The option 'except \(x = -4\)' incorrectly excludes the numerator's zero, but a zero in the numerator only creates an x-intercept, not a domain restriction. Domain restrictions for rational functions come only from denominator zeros, never from numerator zeros.

With odd degree 5 and negative leading coefficient, the graph mirrors \(-x^5\), rising on the left and falling on the right. The choice describing the graph falling on both ends applies only to even-degree functions with negative leading coefficients, not odd-degree ones. For odd-degree polynomials, the two ends of the graph always point in opposite directions.

The Rational Zero Theorem states possible rational zeros are \(\pm\dfrac{p}{q}\) where \(p\) divides the constant term 4 and \(q\) divides the leading coefficient 3, so valid numerators are 1, 2, or 4 and valid denominators are 1 or 3; \(\dfrac{2}{3}\) is actually valid since 2 divides 4 and 3 divides 3, so let's check '\(\dfrac{1}{3}\)' instead, since fractions like \(\dfrac{2}{3}\) are valid, making \(\dfrac{2}{3}\) correct and not the answer — the true non-candidate must be checked against divisors of 4 (1,2,4) over divisors of 3 (1,3).

Since the factor \((x-2)\) cancels from both numerator and denominator, the discontinuity at \(x=2\) is removable, appearing as a hole rather than an asymptote. The option 'A vertical asymptote' would be correct only if the factor did not cancel, which is not the case here. A common factor between numerator and denominator always signals a hole, while an uncancelled denominator factor signals a vertical asymptote.

X-intercepts of a rational function occur where the numerator equals zero (and the denominator does not), so solving \(x^2 - 4 = 0\) gives \(x = 2\) and \(x = -2\). The option '\(x = -5\)' is actually the vertical asymptote from the denominator, not an x-intercept. Always locate x-intercepts from the numerator's zeros and vertical asymptotes from the denominator's zeros, keeping the two separate.

Higher odd multiplicities like 3 still cause the graph to cross the x-axis, but the curve flattens near the zero due to the cubic-like local behavior of the factor. The option 'The graph touches but does not cross' describes even multiplicity, not odd multiplicity of any value. Recognize that all odd multiplicities cross the axis, but higher odd multiplicities produce a flatter crossing than multiplicity 1.

Each zero corresponds to a factor set equal to zero, so the zeros \(0\), \(2\), and \(-1\) give factors \(x\), \((x-2)\), and \((x+1)\) respectively, multiplying to \(f(x) = x(x-2)(x+1)\). The option \(f(x) = x(x+2)(x-1)\) has the signs of the last two factors reversed, producing zeros at \(-2\) and \(1\) instead of the required \(2\) and \(-1\). To build a polynomial from given zeros, write each zero \(r\) as the factor \((x - r)\) and multiply them together.

As \(|x|\) grows large, lower-degree terms become insignificant compared to the leading term, so \(f(x)\) and \(g(x)=5x^7\) share identical end behavior. The option 'They are different because of the lower-degree terms' misunderstands that these terms only affect the middle of the graph, not the far left and right behavior. Any polynomial's end behavior is governed entirely by its leading term, regardless of the other terms present.

By the Fundamental Theorem of Algebra, a degree-4 polynomial has exactly 4 zeros counting multiplicity, so it can have at most 4 real zeros if none are complex. The option '5' exceeds the total number of zeros permitted by the degree, which is not possible for any polynomial. A polynomial can never have more real zeros than its degree, since complex zeros (if any) always come in addition up to that total.

For distinct linear factors in the denominator, each factor gets its own constant numerator over that factor, giving \(\dfrac{A}{x-2} + \dfrac{B}{x+4}\). The option with \((x+4)^2\) incorrectly assumes a repeated factor, but \((x+4)\) appears only once in the original denominator. Always match the decomposition structure to the actual factorization and multiplicity of each denominator factor.

Grouping gives \(x^2(x+3) - 1(x+3) = (x+3)(x^2-1) = (x+3)(x-1)(x+1)\), so the zeros are \(x = -3, 1, -1\). The option '\(x = 3, -1, 1\)' incorrectly changes the sign of the first zero, which does not satisfy the factored form \((x+3)=0\). Factoring by grouping is especially useful when a cubic has four terms that share common binomial factors.

Performing polynomial long division of \(x^2+2x-3\) by \(x-1\) gives a quotient of \(x+3\) with remainder \(0\), so the slant asymptote is \(y = x+3\). The option '\(x + 1\)' results from an arithmetic slip in the division process and does not match the correct quotient. To find a slant asymptote, always perform polynomial division and use only the quotient, discarding the remainder term.

When the numerator's degree exceeds the denominator's degree by exactly one, the function has a slant (oblique) asymptote instead of a horizontal one, which applies here since degree 2 is one more than degree 1. The option 'A horizontal asymptote, because the degrees are equal' is false since the degrees are 2 and 1, not equal. Comparing the exact difference in degree between numerator and denominator tells you whether to expect a horizontal, slant, or no asymptote at all.

The factors \((x-3)^2\) and \((x+5)^4\) both have even multiplicities (2 and 4), so the graph touches the x-axis and turns around at both \(x=3\) and \(x=-5\). The option 'Only \(x = -1\)' is incorrect because the factor \((x+1)\) has multiplicity 1, an odd number, causing the graph to cross rather than touch. When multiple zeros exist in one function, check each factor's exponent individually to determine touching versus crossing behavior.

Multiplying through and substituting \(x=1\) gives \(7(1)-4 = A(1-2)\), so \(3 = -A\), meaning \(A = -3\); substituting \(x=2\) gives \(7(2)-4 = B(2-1)\), so \(B = 10\). The option '\(A = 3, B = 4\)' comes from sign errors in the substitution process and does not satisfy the original equation when checked. The cover-up method of substituting the root of each factor is a fast, reliable way to solve for partial fraction constants with distinct linear factors.

The Complex Conjugate Root Theorem states that complex zeros of polynomials with real coefficients always occur in conjugate pairs, so if \(3-2i\) is a zero, its conjugate \(3+2i\) must also be a zero. The option '\(x = -3-2i\)' is not the conjugate of \(3-2i\) but rather its negative, which is a different complex number entirely. Always find a complex conjugate by flipping only the sign of the imaginary part, keeping the real part unchanged.

Adding the multiplicities \(2 + 1 = 3\) gives the minimum degree, and since the graph rises on the right and falls on the left (odd-degree behavior matching \(x^3\)), the leading coefficient must be positive. The option 'Degree 3, negative leading coefficient' would instead produce a graph rising on the left and falling on the right, the opposite of what's described. Always combine the total multiplicity for minimum degree with the observed end behavior to determine the leading coefficient's sign.

Dividing \(2x^2-3x+1\) by \(x-2\) using long division gives a quotient of \(2x+1\) with a remainder, so the slant asymptote is \(y = 2x+1\). The option '\(y = 2x-3\)' arises from misreading the middle coefficient of the numerator rather than completing the actual division. When finding a slant asymptote, perform full polynomial long division and keep only the quotient portion as the asymptote equation.

Since \(f(2) = -2\) is negative and \(f(3) = 13\) is positive, and \(f\) is continuous, the IVT guarantees a zero exists somewhere between \(x=2\) and \(x=3\) where the function crosses from negative to positive. The option 'Between \(x=1\) and \(x=2\)' is not directly supported by the given function values, since only \(f(2)\) and \(f(3)\) were provided and compared for a sign change. The IVT requires identifying a sign change between two function values to guarantee a zero exists in that interval.

Writing \(3x+5 = A(x+1) + B\) and substituting \(x=-1\) gives \(3(-1)+5 = B\), so \(B = 2\); matching the coefficient of \(x\) gives \(A = 3\). The option '\(A = 3, B = 5\)' incorrectly uses the constant term of the original numerator as \(B\) without actually substituting into the equation. For a repeated linear factor, substitute the root to find the highest-power constant first, then match coefficients for the remaining constants.

Since the polynomial has real coefficients, complex zeros must occur in conjugate pairs, so \(3i\) requires its conjugate \(-3i\) to also be a zero, completing the degree-4 set \(\{1,-2,3i,-3i\}\). The option listing only three zeros ('\(x = 1, -2, 3i\)') is incomplete because a degree-4 polynomial must have exactly four zeros counting multiplicity. Whenever a non-real zero appears in a real-coefficient polynomial, immediately include its conjugate to satisfy the total zero count required by the degree.

The common factor \((x-2)\) cancels, creating a removable hole at \(x=2\), while the remaining denominator factor \((x-5)\) produces a genuine vertical asymptote since it does not cancel. The option 'Vertical asymptotes at both \(x=2\) and \(x=5\)' fails to account for the cancellation that converts the discontinuity at \(x=2\) into a hole rather than an asymptote. Always simplify a rational expression fully first: canceled factors become holes, while uncanceled denominator factors become vertical asymptotes.

Counting sign changes in \(f(x)\): \(+\) to \(-\), \(-\) to \(+\), \(+\) to \(-\), \(-\) to \(+\) gives 4 sign changes, so the number of positive real zeros is 4, 2, or 0 (decreasing by even numbers). The option '3 or 1' would be appropriate for an odd number of sign changes, but this polynomial has an even number of sign changes (4), ruling out odd possibilities. Descartes' Rule always gives possible counts that decrease from the number of sign changes by multiples of 2.

Writing \(2x^2+3 = A(x^2+1) + (Bx+C)x\) and substituting \(x=0\) gives \(3 = A\); matching \(x^2\) coefficients gives \(A+B=2\), so \(B=-1\), and matching \(x\) coefficients gives \(C=0\). The option '\(A = 2, B = 3, C = 0\)' swaps the values of \(A\) and one coefficient incorrectly without verifying against the \(x^2\) coefficient equation. When an irreducible quadratic factor appears in the denominator, its numerator must include both a linear and constant term, \(Bx+C\), not just a single constant.

Factoring as a quadratic in \(x^2\) gives \((x^2-1)(x^2-4)=0\), so \(x^2=1\) or \(x^2=4\), yielding the four real zeros \(x = 1, -1, 2, -2\). The option '\(x = 1, -1, 2i, -2i\)' incorrectly treats \(x^2=4\) as having complex solutions, but \(x^2=4\) has real solutions \(\pm 2\), not imaginary ones. When solving a quartic that is quadratic in form, substitute \(u=x^2\), solve for \(u\), then carefully back-substitute checking whether each value of \(u\) is positive (real roots) or negative (imaginary roots).

To have a horizontal asymptote at \(y=3\) with vertical asymptotes from a degree-2 denominator, the numerator must also be degree 2 with leading coefficient equal to 3 times the denominator's leading coefficient, and including a factor of \(x\) ensures a zero at \(x=0\), giving \(f(x)=\dfrac{3x^2}{(x-1)(x+2)}\). The option \(f(x) = \dfrac{3x}{(x-1)(x+2)}\) has numerator degree 1 versus denominator degree 2, which produces horizontal asymptote \(y=0\), not \(y=3\). To construct a rational function matching a horizontal asymptote with nonzero value, the numerator and denominator must have equal degree, with the leading coefficient ratio equal to the asymptote's value.

Synthetic division of \(2x^3+x^2-13x+6\) by \((x-2)\) yields the quotient \(2x^2+5x-3\), which factors as \((2x-1)(x+3)\), giving the additional zeros \(x=\tfrac{1}{2}\) and \(x=-3\) along with the known \(x=2\). The option 'Remaining factor \(2x^2-5x+3\)' contains sign errors from an incorrect synthetic division process and does not match the correct division result. After using synthetic division to reduce a cubic by a known zero, always factor the resulting quadratic completely to find any remaining real zeros.