ACT Math study guide · Chapter 14
Other Function Families
For a rational function, factor the numerator and denominator first. A zero that remains in the denominator after common factors are canceled creates a vertical asymptote. A factor that cancels creates a hole at that x-value instead. A horizontal asymptote compares the degrees of the numerator and denominator: if the degrees are equal, the asymptote is the ratio of the leading coefficients; if the numerator's degree is smaller, the asymptote is y = 0; if the numerator's degree is larger, there is no horizontal asymptote at all, and the function grows without bound instead of leveling off.
A logarithm answers one question: what exponent produces this result? logₐ(b) asks what power a must be raised to in order to get b. Evaluating a logarithm directly, without a calculator, means recognizing b as a power of a.
Exponential growth and decay models take the form A(t) = A₀(1 + r)ᵗ, where A₀ is the starting amount and r is the growth or decay rate as a decimal, positive for growth and negative for decay. A geometric sequence follows the same structure in discrete steps: each term is the previous term multiplied by a constant ratio r.
Where this appears as a model: an investment growing at a fixed percentage rate. $1,000 invested at 5% annual interest grows to 1000(1.05)^10≈$1,628.89 after 10 years.
▸ Finding asymptotes
Worked Example Find the horizontal asymptote of f(x) = (4x + 1) ⁄ (2x − 3).
Compare the degrees of the numerator and denominator.
Both have degree 1, so the asymptote is the ratio of leading coefficients.
y = 4/2 = 2
▸ Exponential growth and geometric sequences
Worked Example A geometric sequence has a₁ = 4 and r = 3. Find a₅.
Each term multiplies the previous term by r, applied (n−1) times from the first term.
a₅ = 4 · 3⁴ = 4 · 81 = 324
Worked Example Solve (x + 1)/(x − 3) = 5/(x − 3) + 2.
Recognize. Both sides share the same denominator, x − 3. Clear it by multiplying every term by that denominator, then check the result for extraneous solutions.
Set up.
(x + 1) = 5 + 2(x − 3)
Solve.
x + 1 = 5 + 2x − 6
x + 1 = 2x − 1
2 = x
Check. x = 2 does not make the original denominator zero (2 − 3 = −1 ≠ 0), so the solution is valid.
Best route. Hand algebra clears the fractions directly. Desmos can graph both sides to confirm the intersection, but it does not set up the multiplication step.
Common error. Finding a solution but forgetting to check whether it makes the original denominator zero, which would make it extraneous.
▸ Practice
1. Find the vertical asymptote of f(x) = (x + 4) ⁄ (x − 6).
A) x = −4 B) x = 6 C) x = −6 D) x = 4
2. Find the horizontal asymptote of f(x) = (3x − 1) ⁄ (x + 2).
F) y = −1/2 G) y = 0 H) y = 3 J) y = 3/2
3. A(t) = 200(0.85)ᵗ. Find A(4), rounded to the nearest hundredth.
A) 104.40 B) 88.74 C) 122.83 D) 94.35
4. Evaluate log₃(81).
F) 4 G) 3 H) 5 J) 81
5. A geometric sequence has a₁ = 6 and r = 2. Find a₆.
A) 192 B) 96 C) 384 D) 32
6. P(t) = 1000(1.05)ᵗ. Find P(3), rounded to the nearest dollar.
F) 1158 G) 1150 H) 1155 J) 1200
7. Find the vertical asymptote of f(x) = (x − 2) ⁄ (x + 7).
A) x = 7 B) x = 2 C) x = −7 D) x = −2
8. A(t) = 300(1.04)ᵗ. Find A(5), rounded to the nearest hundredth.
F) 360.00 G) 365.00 H) 350.00 J) 380.00
9. Evaluate log₂(64).
A) 8 B) 6 C) 5 D) 32
10. A geometric sequence has a₁ = 5 and r = 2. Find a₆.
F) 80 G) 320 H) 40 J) 160
11. Find the horizontal asymptote of f(x) = 5 ⁄ (x + 3).
A) y = 5 B) y = 3 C) There is no horizontal asymptote D) y = 0
12. A(t) = 1000(0.85)ᵗ. Find A(4), rounded to the nearest hundredth.
F) 500.00 G) 510.00 H) 522.01 J) 515.00
13. Evaluate log₃(81).
A) 3 B) 27 C) 4 D) 5
14. f(x) = (x² − 9) ⁄ (x − 3) has a hole at x = 3. Find the y-value of that hole.
F) 0 G) 3 H) 9 J) 6
15. Find the vertical asymptote of f(x) = (x − 3) ⁄ (x + 8).
A) x = 8 B) x = −8 C) x = 3 D) x = −3
16. A(t) = 250(1.06)ᵗ. Find A(4), rounded to the nearest hundredth.
F) 315.62 G) 300.00 H) 330.75 J) 275.40
17. Evaluate log₅(125).
A) 5 B) 25 C) 125 D) 3
18. A geometric sequence has a₁ = 8 and r = 2. Find a₄.
F) 32 G) 64 H) 16 J) 128
19. Find the horizontal asymptote of f(x) = (5x + 2) ⁄ (2x − 1).
A) y = 5/2 B) y = 5 C) y = 2 D) There is no horizontal asymptote
20. A(t) = 800(0.9)ᵗ. Find A(3), rounded to the nearest hundredth.
F) 720.00 G) 583.20 H) 648.00 J) 518.40
▸ Answers and Explanations
1: B. The denominator is zero when x − 6 = 0, so x = 6. Choice A reports where the numerator equals zero instead of the denominator.
2: H. Numerator and denominator have equal degree; the horizontal asymptote is the ratio of leading coefficients, 3/1 = 3. Choice F reports the function's y-intercept (found by setting x=0) instead of its horizontal asymptote.
3: A. 200(0.85)⁴ ≈ 104.40. Choice B uses an exponent of 5 instead of 4, an off-by-one error in the number of periods.
4: F. 3 raised to the 4th power equals 81, so log₃(81) = 4. Choice G undercounts the number of times 3 must be multiplied by itself to reach 81, stopping one factor short.
5: A. a₆ = 6 · 2⁵ = 6 · 32 = 192. Choice F uses one fewer multiplication by the common ratio than required, computing a₅ instead of a₆.
6: F. 1000(1.05)³ ≈ 1157.63, which rounds to 1158. Choice G comes from a rounding or computational slip in evaluating 1.05³.
7: C. The denominator is zero at x=−7. Choice A drops the negative sign on the denominator's zero.
8: G. 300(1.04)⁵≈365.00. Choice F comes from a rounding or computational slip in evaluating 1.04⁵.
9: B. 2⁶=64, so log₂(64)=6. Choice A miscounts the number of times 2 must be multiplied by itself to reach 64.
10: J. a₆=5·2⁵=5·32=160. Choice F uses one fewer multiplication by the common ratio than required, computing a₅ instead of a₆.
11: D. The numerator's degree (0) is less than the denominator's degree (1), so the horizontal asymptote is y=0. Choice A reports the numerator's constant value directly instead of comparing the degrees of the numerator and denominator.
12: H. 1000(0.85)⁴≈522.01. Choice G comes from a rounding or computational slip in evaluating 0.85⁴.
13: C. 3⁴=81, so log₃(81)=4. Choice A undercounts the number of times 3 must be multiplied by itself to reach 81, stopping one factor short.
14: J. Factor the numerator as a difference of squares: (x−3)(x+3)⁄(x−3)=x+3 for x≠3; the hole's y-value is the limit as x approaches 3, which is 3+3=6. Choice F assumes the hole means the function value is zero there instead of finding the actual limiting value.
15: B. The denominator is zero at x=−8. Choice A drops the negative sign on the denominator's zero.
16: F. 250(1.06)⁴≈315.62. Choice G comes from a rounding or computational slip in evaluating 1.06⁴.
17: D. 5³=125, so log₅(125)=3. Choice A reports the base of the logarithm instead of the exponent that solves it.
18: G. a₄=8·2³=8·8=64. Choice F uses one fewer multiplication by the common ratio than required, computing a₃ instead of a₄.
19: A. Numerator and denominator have equal degree; the horizontal asymptote is the ratio of leading coefficients, 5/2. Choice B reports the numerator's leading coefficient alone instead of the ratio of both leading coefficients.
20: G. 800(0.9)³≈583.20. Choice F uses an exponent of 1 instead of 3, an off-by-one error stopping after only one period.