1. Find the next three missing terms of the sequences: A) Given sequence: 35, 32, 29, 26, _______, _______, _______ - This is an arithmetic sequence where each term decreases by 3. - Formula for nth term of arithmetic sequence: $$a_n = a_1 + (n-1)d$$ where $d$ is the common difference. - Here, $d = -3$. - Next terms: - 5th term: $26 - 3 = 23$ - 6th term: $23 - 3 = 20$ - 7th term: $20 - 3 = 17$ B) Given sequence: -3, -6, -12, -24, _______, _______, _______ - This is a geometric sequence where each term is multiplied by -2. - Formula for nth term of geometric sequence: $$a_n = a_1 \times r^{n-1}$$ where $r$ is the common ratio. - Here, $r = 2$ (absolute value), but signs alternate, so $r = -2$. - Next terms: - 5th term: $-24 \times -2 = 48$ - 6th term: $48 \times -2 = -96$ - 7th term: $-96 \times -2 = 192$ 2. Difference between arithmetic and geometric patterns: - Arithmetic pattern: each term changes by adding or subtracting a constant difference. - Geometric pattern: each term changes by multiplying or dividing by a constant ratio. 3. Identify patterns: A) 3, 5, 7, ... - Difference between terms is 2, so arithmetic. B) 9, -3, -1, -1/3, ... - Ratios: $-3/9 = -1/3$, $-1/-3 = 1/3$, $-1/3 / -1 = 1/3$ (not consistent ratio), but seems geometric with ratio $-1/3$. - Actually, check carefully: - $-3/9 = -1/3$ - $-1/-3 = 1/3$ (not same as previous ratio) - Since ratio changes, not geometric. - Differences: $-3 - 9 = -12$, $-1 - (-3) = 2$, not constant. - Neither arithmetic nor geometric strictly. C) 10, 0, -10, -20, ... - Differences: $0 - 10 = -10$, $-10 - 0 = -10$, $-20 - (-10) = -10$ - Constant difference, so arithmetic. 4. Difference between domain and range: - Domain: set of all possible input values (x-values) of a relation. - Range: set of all possible output values (y-values) of a relation. 5. Difference between function and relation: - Relation: any set of ordered pairs. - Function: a relation where each input (x) has exactly one output (y). 6. Vertical line test: - A test to determine if a graph represents a function. - If any vertical line intersects the graph more than once, it is not a function. 7. Identify functions: A) $F : [0, \infty] \to [0, \infty], f(x) = x^2$ - Not injective on $[0, \infty]$ because $x^2$ is increasing and one-to-one there. - Surjective onto $[0, \infty]$ because all non-negative values are covered. - So, $f$ is injective and surjective (bijective) on this domain and codomain. B) $F : \mathbb{R} \to \mathbb{R}, f(x) = 5x + 1$ - Linear function with slope 5. - Injective (one-to-one) and surjective (onto) over all real numbers. - So, bijective. C) $F : \mathbb{R} \to [0, \infty], f(x) = 4 - x^2$ - Not injective because $f(x) = f(-x)$. - Surjective onto $[0,4]$ but not onto $[0, \infty]$ since max is 4. - So, neither injective nor surjective onto $[0, \infty]$. 8. Define quadratic function and example: - A quadratic function is a polynomial function of degree 2, generally written as $f(x) = ax^2 + bx + c$ where $a \neq 0$. - Example: $f(x) = 4 - x^2$. 9. Determine domain and range: A) $R = \{(x,y): y = 3 - x, x = -2, -1, 0, 1, 2\}$ - Domain: $\{-2, -1, 0, 1, 2\}$ - Range: calculate $y$ for each $x$: - $3 - (-2) = 5$ - $3 - (-1) = 4$ - $3 - 0 = 3$ - $3 - 1 = 2$ - $3 - 2 = 1$ - Range: $\{1, 2, 3, 4, 5\}$ B) $f(x) = \frac{3}{x-1}$ - Domain: all real numbers except $x = 1$ (division by zero undefined). - Range: all real numbers except $y = 0$ (fraction never zero). C) $g(x) = \sqrt{2 - x}$ - Domain: $2 - x \geq 0 \Rightarrow x \leq 2$ - Range: $y \geq 0$ (square root outputs non-negative values). 10. Sketch graphs: A) $R = \{(x,y): y \leq 2x - 1\}$ - Region below or on the line $y = 2x - 1$. B) $Y = 2x + 4; x = -2, -1, 0, 1, 2$ - Points: - $x=-2, y=0$ - $x=-1, y=2$ - $x=0, y=4$ - $x=1, y=6$ - $x=2, y=8$ - Plot these points and draw the line. 11. Let $f(x) = 2x - 4$. i. Find: A) x-intercept: set $f(x) = 0$ - $2x - 4 = 0 \Rightarrow 2x = 4 \Rightarrow x = 2$ B) y-intercept: set $x = 0$ - $f(0) = 2(0) - 4 = -4$ C) slope: coefficient of $x$ is 2. ii. Increasing or decreasing? - Since slope $2 > 0$, function is increasing. 12. Let $f(x) = x - 2$ and $g(x) = 4x^2$. A) $(f \cdot g)(x) = f(x) \times g(x) = (x - 2)(4x^2) = 4x^3 - 8x^2$ B) $(f \cdot g)(-3) = 4(-3)^3 - 8(-3)^2 = 4(-27) - 8(9) = -108 - 72 = -180$ C) $(g/f)(x) = \frac{g(x)}{f(x)} = \frac{4x^2}{x - 2}$ D) $(g/f)(1) = \frac{4(1)^2}{1 - 2} = \frac{4}{-1} = -4$ 13. Given $f(x) = 2x^2 - 3x + 1$ and $g(x) = -x^2 + 2x + 3$. A) $(2f - g)(x) = 2f(x) - g(x) = 2(2x^2 - 3x + 1) - (-x^2 + 2x + 3)$ - $= 4x^2 - 6x + 2 + x^2 - 2x - 3 = 5x^2 - 8x - 1$ B) $(2f - g)(2) = 5(2)^2 - 8(2) - 1 = 5(4) - 16 - 1 = 20 - 16 - 1 = 3$ 14. Given $f(x) = x^2 + 7x + 12$ and $g(x) = \frac{x + 4}{x + 3}$. Find $\frac{f(x)}{g(x)} = f(x) \times \frac{1}{g(x)} = (x^2 + 7x + 12) \times \frac{x + 3}{x + 4}$ Factor $f(x)$: - $x^2 + 7x + 12 = (x + 3)(x + 4)$ So, - $\frac{f(x)}{g(x)} = (x + 3)(x + 4) \times \frac{x + 3}{x + 4} = (x + 3)^2$ Domain excludes $x = -4$ (denominator zero in $g(x)$). 15. Given $f(x) = 8x - \sqrt{8 - x}$, evaluate $f(-8)$. Calculate: - $8(-8) = -64$ - $\sqrt{8 - (-8)} = \sqrt{8 + 8} = \sqrt{16} = 4$ So, - $f(-8) = -64 - 4 = -68$ Final answers summarized: 1A) 23, 20, 17 1B) 48, -96, 192 3A) Arithmetic 3B) Neither strictly arithmetic nor geometric 3C) Arithmetic 7A) Injective and surjective (bijective) on given domain 7B) Bijective 7C) Neither injective nor surjective 9A) Domain: \{-2,-1,0,1,2\}, Range: \{1,2,3,4,5\} 9B) Domain: $x \neq 1$, Range: $y \neq 0$ 9C) Domain: $x \leq 2$, Range: $y \geq 0$ 11iA) x-intercept: 2 11iB) y-intercept: -4 11iC) slope: 2 11ii) Increasing 12A) $4x^3 - 8x^2$ 12B) -180 12C) $\frac{4x^2}{x-2}$ 12D) -4 13A) $5x^2 - 8x - 1$ 13B) 3 14) $(x + 3)^2$ 15) -68