Precalculus Practice Tests #2: Polynomial & Rational Function

Overview

blackpenredpen works through 20 precalculus problems on polynomial and rational functions, covering zeros, graph behavior, asymptotes, holes, inequalities, division, and limits. The solutions emphasize reliable procedures—such as factoring before identifying vertical asymptotes, testing signs across critical intervals, and using the Rational Root Theorem—and include concrete examples from a cubic with roots 1 and (-1 ± i√5)/2 to a rational function with a repeated zero and two vertical asymptotes.

Key takeaways

Factor a rational function’s numerator and denominator before identifying vertical asymptotes: a canceled denominator factor creates a hole, while an uncanceled factor gives a vertical asymptote.
A rational function with equal numerator and denominator degrees has a horizontal asymptote equal to the ratio of leading coefficients, including the sign; 4x³/(−5x³) approaches −4/5.
For polynomial and rational inequalities, mark numerator zeros and denominator exclusions, then test signs on the resulting intervals; never multiply by a variable expression unless its sign is known.
For real-coefficient polynomials, nonreal roots occur in conjugate pairs, so a root such as 1−2i entails 1+2i.
A repeated zero with even multiplicity makes a polynomial graph touch the x-axis and turn around instead of crossing it.
When dividing a polynomial by a linear factor x−a, the remainder is P(a), which can be found directly without completing long division.

Chapters

0:00 Factor First: Vertical Asymptotes and the Rational Root Theorem

For (x²−9)/(x²−2x−3), factoring reveals that x−3 cancels, producing a hole at x=3; the remaining denominator gives the vertical asymptote x=−1.
For 3x³−4x²+2x−6, the Rational Root Theorem forms candidates from factors of 6 over factors of 3.
The proposed rational zero ±3/2 is impossible because 2 is not a factor of the leading coefficient 3.

7:05 Find the Range by Accounting for a Removable Hole

Factoring (x²−5x+6)/(x−2) simplifies the graph to y=x−3, except at x=2.
Substituting x=2 into the simplified line locates the missing point at (2,−1).
The range is all real y-values except −1; the domain is all real x-values except 2.

11:35 Use the Leading Term to Determine Polynomial End Behavior

For a polynomial with leading term −2x⁵, the odd degree and negative leading coefficient determine both ends.
As x→∞, the function approaches −∞; as x→−∞, it approaches ∞.
Lower-degree terms do not affect the graph’s behavior at the far left and far right.

15:01 Build a Monic Cubic from a Real Root and a Complex Pair

A real zero at 2 gives the factor x−2; the complex zero 1−i requires its conjugate 1+i for a polynomial with real coefficients.
The factors are (x−2)(x−1+i)(x−1−i), with leading coefficient 1.
Combining the conjugate factors gives (x−2)((x−1)²+1), which expands to x³−4x²+6x−4.

22:50 Find a Polynomial Remainder with Evaluation or Synthetic Division

For P(x)=2x⁴−3x³+5x−7 divided by x+2, the Remainder Theorem says the remainder is P(−2).
Evaluating gives P(−2)=32+24−10−7=39.
Synthetic division uses coefficients 2, −3, 0, 5, −7; the zero coefficient preserves the missing x² term, and the final entry is 39.

28:50 Locate a Rational Function’s Hole Coordinates

Factor (x²−4)/(2x²−3x−2) as (x−2)(x+2)/((2x+1)(x−2)).
Canceling x−2 identifies a removable discontinuity at x=2, rather than a vertical asymptote.
Substitute x=2 into the remaining expression (x+2)/(2x+1) to find the hole at (2, 4/5).

32:55 Read a Horizontal Asymptote from Leading Coefficients

For (4x³−2x+1)/(3−5x³), numerator and denominator have the same degree, so the horizontal asymptote is the ratio of leading coefficients.
The leading-coefficient ratio is 4/(−5), giving y=−4/5.
For rational functions, a lower numerator degree gives y=0, while a higher numerator degree means there is no horizontal asymptote.

34:05 Use Synthetic Division to Find a Slant Asymptote

For (2x²+5x−1)/(x+2), the numerator degree exceeds the denominator degree by one, indicating a slant asymptote.
Synthetic division with −2 gives quotient 2x+1 and remainder −3.
The function can be written as 2x+1−3/(x+2), so the slant asymptote is y=2x+1.

37:00 Solve Polynomial and Rational Inequalities with Sign Charts

Factoring x³−3x²−4x gives x(x−4)(x+1); testing intervals around −1, 0, and 4 yields the solution [−1,0]∪[4,∞) for the expression ≥0.
For (x+2)/(x+3)≤1, subtracting 1 gives −1/(x+3)≤0, with x=−3 excluded because it makes the denominator zero.
The rational inequality’s solution is (−3,∞); multiplying an inequality by x+3 without knowing its sign could incorrectly preserve the inequality direction.

44:53 Sketch Polynomial Graphs from Zeros, Multiplicity, and Intercepts

For (x+2)(x−1)², the graph crosses the x-axis at −2 but touches and turns at 1 because that zero has even multiplicity.
Grouping x³+3x²−4x−12 gives (x+3)(x−2)(x+2), with zeros −3, −2, and 2.
The cubic’s y-intercept is −12; its simple roots alternate the sign of the graph as it crosses each x-intercept.

55:18 Divide a Quartic by a Quadratic Using Polynomial Long Division

Divide x⁴−2x²+3x−5 by x²+1, inserting 0x³ to preserve the missing term.
Long division gives quotient x²−3 and remainder 3x−2.
The result is P(x)=(x²+1)(x²−3)+(3x−2), separating the quotient from the remainder.

59:54 Use Conjugate Roots to Complete a Cubic’s Root Set

For x³−x²+3x+5, the given root 1−2i implies the conjugate root 1+2i when coefficients are real.
Direct substitution verifies that −1 is the remaining root.
The complete root set is −1, 1−2i, and 1+2i.

1:04:06 Find a Rational Function’s Y-Intercept and a One-Sided Limit

Setting x=0 in (3x²−6x+8)/(2x²+4x−4) gives the y-intercept (0,−2).
For f(x)=1/(x−4), the vertical asymptote is x=4.
Approaching 4 from the left makes x−4 a small negative number, so limₓ→₄⁻ 1/(x−4)=−∞.

1:07:23 Graph a Rational Function by Division and Transformations

The function (2x+3)/(x+2) has vertical asymptote x=−2 and horizontal asymptote y=2.
Long division rewrites it as 2−1/(x+2), showing a shifted reciprocal graph reflected across the x-axis and moved up 2 units.
Testing x=−3 gives y=3, placing the left branch above the horizontal asymptote.

1:12:33 Construct a Rational Function from Asymptotes and a Repeated Zero

Vertical asymptotes at x=−2 and x=3 require denominator factors (x+2)(x−3).
An x-intercept at 1 with multiplicity 2 requires the numerator factor (x−1)².
Matching numerator and denominator degrees and setting the leading-coefficient ratio to 2 gives 2(x−1)²/((x+2)(x−3)), with horizontal asymptote y=2.

1:15:49 Solve a Cubic for Its Real and Complex Roots

For 2x³+x−3=0, testing x=1 gives zero, so x−1 is a factor.
Synthetic division yields 2x²+2x+3 as the remaining factor.
The quadratic formula gives the other roots (−1+i√5)/2 and (−1−i√5)/2, alongside the real root 1.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, blackpenredpen.