Answer: f(3) = 3 and f(−1) = 3

1. f(3) = 3² − 2(3) = 9 − 6 = 3
2. f(−1) = (−1)² − 2(−1) = 1 + 2 = 3
3. (Both points lie at the same height because the parabola is symmetric about x = 1.)

Answer: Domain x ≥ 3; range y ≥ 0

1. The expression under the root must be non-negative: 2x − 6 ≥ 0.
2. 2x ≥ 6, so x ≥ 3. Domain: x ≥ 3.
3. The square root function gives outputs from 0 upwards, with f(3) = 0. Range: y ≥ 0.

Answer: Domain: all real x except x = 3 and x = −3; g(0) = −19

1. The denominator cannot be zero: x² − 9 ≠ 0.
2. (x − 3)(x + 3) ≠ 0, so x ≠ 3 and x ≠ −3.
3. Domain: all real x, x ≠ ±3.
4. g(0) = 0 + 10 − 9 = 1−9 = −19

Answer: (a) (4, 9) (b) (−1, 4)

1. (a) x − 3 shifts the graph 3 units right: x becomes 1 + 3 = 4.
2. The factor 2 doubles the y-value: 4 × 2 = 8. The + 1 then shifts it up: 8 + 1 = 9.
3. Image: (4, 9).
4. (b) f(−x) reflects the graph in the y-axis, so x changes sign and y is unchanged.
5. Image: (−1, 4).

Answer: (a) 2x² + 1 (b) 4x² + 4x + 1 (c) 9

1. (a) f(g(x)) = f(x²) = 2x² + 1
2. (b) g(f(x)) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1
3. (c) g(2) = 4, so f(g(2)) = f(4) = 2(4) + 1 = 9. (Check with part (a): 2(2)² + 1 = 9.)

Answer: f⁻¹(x) = 3x + 2, domain x ≠ 0

1. Write y = 3x − 2 and swap x and y: x = 3y − 2.
2. Make y the subject: x(y − 2) = 3, so y − 2 = 3x.
3. y = 3x + 2, so f⁻¹(x) = 3x + 2.
4. The domain of f⁻¹ is the range of f. f never equals 0 (its horizontal asymptote is y = 0), so the domain of f⁻¹ is x ≠ 0 — which also matches the 3x in the rule.

Answer: (a) odd (b) even (c) neither

1. A function is even if f(−x) = f(x) and odd if f(−x) = −f(x).
2. (a) f(−x) = (−x)³ − 4(−x) = −x³ + 4x = −(x³ − 4x) = −f(x), so f is odd (symmetric about the origin).
3. (b) g(−x) = (−x)⁴ + 2(−x)² = x⁴ + 2x² = g(x), so g is even (symmetric about the y-axis).
4. (c) h(−x) = x² − x. This is not equal to h(x) = x² + x, and not equal to −h(x) = −x² − x, so h is neither.

Answer: Vertex (2, −9); y-intercept −5; x-intercepts x = −1 and x = 5; range y ≥ −9

1. Vertex form y = (x − h)² + k gives the vertex (2, −9). It is a minimum because the coefficient of x² is positive.
2. y-intercept (x = 0): y = (−2)² − 9 = 4 − 9 = −5.
3. x-intercepts (y = 0): (x − 2)² = 9, so x − 2 = ±3, giving x = 5 or x = −1.
4. Range: the minimum value is −9, so y ≥ −9.

Answer: Asymptotes x = −1 and y = −3; y-intercept (0, −1); x-intercept (−13, 0)

1. The base graph y = 2x is shifted 1 unit left (x + 1) and 3 units down (− 3).
2. Vertical asymptote: x + 1 = 0, so x = −1. Horizontal asymptote: y = −3.
3. y-intercept (x = 0): y = 21 − 3 = −1, so (0, −1).
4. x-intercept (y = 0): 2x + 1 = 3, so 3(x + 1) = 2, x + 1 = 23, x = −13. Point (−13, 0).

Answer: y-intercept −7; x-intercept x = 3; asymptote y = −8; range y > −8

1. y-intercept (x = 0): y = 2⁰ − 8 = 1 − 8 = −7.
2. x-intercept (y = 0): 2^x = 8 = 2³, so x = 3.
3. As x → −∞, 2^x → 0, so y → −8 from above. Horizontal asymptote: y = −8.
4. 2^x > 0 for all x, so y = 2^x − 8 > −8. Range: y > −8.