Worked Examples: ACT Algebra & Functions

Example 1: Quadratic in Context (Medium)

A ball is thrown upward from a height of 4 feet with an initial velocity of 32 feet per second. Its height $h$ in feet after $t$ seconds is given by $h = -16t^2 + 32t + 4$. After how many seconds does the ball reach its maximum height?

Solution: The maximum of a downward-opening parabola occurs at the vertex: $t = -\frac{b}{2a} = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1$ second.

The maximum height is $h = -16(1) + 32(1) + 4 = 20$ feet.

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Example 2: Logarithm Equation (Medium-Hard)

If $\log_5(2x - 1) = 2$, what is the value of $x$?

Solution: Convert to exponential form: $5^2 = 2x - 1$, so $25 = 2x - 1$, giving $2x = 26$ and $x = 13$.

Verify: $\log_5(2(13) - 1) = \log_5(25) = \log_5(5^2) = 2$. Confirmed.

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Example 3: Geometric Sequence (Medium)

The first term of a geometric sequence is 3 and the common ratio is 4. What is the 5th term?

Solution: $a_n = a_1 \cdot r^{n-1} = 3 \cdot 4^{5-1} = 3 \cdot 4^4 = 3 \cdot 256 = 768$.

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Example 4: Function Composition (Medium)

If $f(x) = 2x + 3$ and $g(x) = x^2 - 1$, what is $g(f(2))$?

Solution: Work from the inside out.

• $f(2) = 2(2) + 3 = 7$
• $g(7) = 7^2 - 1 = 49 - 1 = 48$

Common trap: Computing $f(g(2))$ instead of $g(f(2))$. Order matters. $f(g(2)) = f(3) = 9$, which would be a wrong answer choice.

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Example 5: System of Equations from Word Problem (Hard)

A store sells two types of gift baskets. Basket A contains 3 candles and 2 soaps. Basket B contains 1 candle and 4 soaps. The store used 27 candles and 38 soaps to fill all baskets. How many of Basket A were made?

Setup: Let $a$ = number of Basket A, $b$ = number of Basket B. $3a + b = 27 \quad \text{(candles)}$ $2a + 4b = 38 \quad \text{(soaps)}$

Solve: From the first equation: $b = 27 - 3a$. Substitute into the second: $2a + 4(27 - 3a) = 38$ $2a + 108 - 12a = 38$ $-10a = -70$ $a = 7$

Answer: 7 Basket A's were made.

Verify: $b = 27 - 3(7) = 6$. Candles: $3(7) + 1(6) = 27$. Soaps: $2(7) + 4(6) = 14 + 24 = 38$. Confirmed.

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Example 6: Arithmetic Sequence Sum (Hard)

An auditorium has 20 rows. The first row has 15 seats, and each subsequent row has 3 more seats than the row before it. How many total seats are in the auditorium?

Solution: This is an arithmetic sequence with $a_1 = 15$, $d = 3$, $n = 20$.

Last row: $a_{20} = 15 + (20-1)(3) = 15 + 57 = 72$.

Total: $S_{20} = \frac{20}{2}(15 + 72) = 10(87) = 870$ seats.