Precalculus math practice

Every skill in this set

Average rate of changePolynomial & Rational Functions

AROC on [a, b] = [f(b) − f(a)] / (b − a)

Precalculus's stand-in for the derivative, and the most-tested single computation in Unit 1. The error is almost never the subtraction — it is forgetting to divide by (b − a), which leaves a total change masquerading as a rate.

Remainder TheoremPolynomial & Rational Functions

the remainder on dividing p(x) by (x − a) equals p(a)

Dividing by (x + 3) means a = −3, not +3. That sign flip is the whole difficulty, and it decides whether you conclude the divisor is a factor.

Height of a hole in a rational functionPolynomial & Rational Functions

cancel the common factor, then evaluate the simplified expression

Naming the x value of a hole is only half an answer. Substituting into the ORIGINAL gives 0/0, which tells you nothing — the height comes from the simplified form only.

Solving an exponential equationExponential & Logarithmic Functions

a·bˣ = c ⟹ x = ln(c/a) / ln(b)

Isolate the exponential BEFORE taking logs. Logging a·bˣ directly invites the error ln(a·bˣ) = a·ln(bˣ), which is wrong — a coefficient becomes an added logarithm, not a multiplied one.

Doubling time from a growth rateExponential & Logarithmic Functions

T = ln 2 / ln(1 + r)

The rule of 70 gives a fast estimate, but the exam wants the exact expression. Note that the divisor is ln(1 + r), not ln(r) — a distinction that changes the answer by an order of magnitude.

Amount remaining after a given timeExponential & Logarithmic Functions

A(t) = A₀·(1/2)^(t/h) for half-life h

The exponent t/h counts how many half-lives have elapsed. Computing that count first is the single best check available: 3.4 half-lives cannot leave 60% of the material.

Converting a base to a continuous rateExponential & Logarithmic Functions

bᵗ = e^(kt) with k = ln b

A 7% annual increase corresponds to a continuous rate of about 6.77%, not 7%. Reporting the base percentage as the continuous rate is the standard error, and the continuous value is always the smaller of the two for growth.

Arc length from a central angleTrigonometric & Polar Functions

s = rθ, with θ in radians

s = rθ requires radians and has no conversion constant precisely because a radian is a ratio of lengths. Using degrees directly gives an answer about 57 times too large.

Linear speed from revolutions per minuteTrigonometric & Polar Functions

ω = 2π·rpm/60 and v = rω

Two conversions in one problem, and both are commonly dropped: revolutions to radians (×2π) and minutes to seconds (÷60). Keeping the radius in meters avoids the factor-of-100 error that follows.

Law of Cosines for a missing sideTrigonometric & Polar Functions

c² = a² + b² − 2ab·cos C

The correct law for SAS, where the Law of Sines has no complete side–angle pair to work with. C must be the angle opposite the side you are finding — pairing it wrongly is the usual failure.

Triangle area from two sides and the included angleTrigonometric & Polar Functions

Area = ½ab·sin C

The sine factor is what most often goes missing, leaving ½ab — which is the area only for a right angle. Since sin C ≤ 1, the true area can never exceed ½ab, and that gives you a free upper-bound check.

Angle between two vectorsFunctions Involving Parameters, Vectors & Matrices

cos θ = (u · v) / (|u||v|)

The sign of the dot product independently predicts the answer — positive means under 90°, negative means over. Checking that against your arccos catches arithmetic errors before you report them.

Solving a 2 × 2 system with an inverse matrixFunctions Involving Parameters, Vectors & Matrices

X = A⁻¹B, with A⁻¹ = (1/det A)·[[d, −b], [−c, a]]

The determinant has to be computed first — a zero determinant means no unique solution and the method does not apply. Report the value of x, which is the first component of A⁻¹B.

Common questions