CALC. Dalco Manufacturing Profit Formula - Profit Formula: The weekly profit, P, in hundreds of dollars can be approximated by the formula: $P = -4x² + 8x + 6$ where: - $P$ = Profit in hundreds of dollars - $x$ = Number of units produced per week (in thousands). ### A. Maximum Profit Calculation - Objective: Determine the number of units (x) to produce weekly to maximize profit. - Method: Use the vertex of the parabola represented by the quadratic function. - The vertex can be found using $x = -\frac{b}{2a}$ where the function is in the form $ax^2 + bx + c$. - Given: - $a = -4$ - $b = 8$ - Maximum profit occurs when:

$x = -\frac{8}{2(-4)} = 1$ - Units to Produce: The company should produce 1,000 units (since x is in thousands). ### B. Maximum Weekly Profit Calculation - To verify the maximum profit: - Substitute $x = 1$ back into the profit formula:

$P(1) = -4(1)^2 + 8(1) + 6$

$= -4 + 8 + 6$

$= 10$ - Maximum Profit: The maximum weekly profit is $1,000 (10 hundreds). ## Pascal's Triangle for Binomial Expansion - Question: Use Pascal's Triangle to expand the following binomials: - A: 1. Expansion of $(x - 2)¹$( - According to binomial expansion, $(a + b)^n$ is given by the coefficients from Pascal's Triangle. - For $(x - 2)^1$ the expansion is:

$x - 2$ 2. Expansion of $(2x + 3y)³$: - The third row of Pascal's triangle: 1, 3, 3, 1 coefficients apply: - $1(2x)³ + 3(2x)²(3y) + 3(2x)(3y)² + 1(3y)³$ - $= 8x³ + 36x²y + 27y³$. ## Rational Zeros and Finding All Zeros - Task: Identify all possible rational zeros for the function and find all zeros: - Given function: $f(x) = 2x³ - x² - 5x - 2$ - Finding Rational Zeros: - Possible rational zeros defined by the Rational Root Theorem: - Factors of the constant term (-2): $\pm1$, $\pm2$ - Factors of the leading coefficient (2): $\pm1$, $\pm2$ - List of Possible Rational Zeros: $\pm1$, $\pm2$, $\pm1/2$. - Finding Actual Zeros: - Use synthetic division or substitution to test each possible rational zero. For example, test $x=2$ by substituting $f(2) = 2(2)^3 - (2)^2 - 5(2) - 2 = 16 - 4 - 10 - 2 = 0$. So, $x=2$ is a zero. ## Polynomial Evaluation Using Synthetic Substitution - Problem: Evaluate $f(x) = -2x + 3x + x³ - x² + 6x + 3$ at $x = -1$. - Substitute x = -1 into the polynomial: - First combine like terms:

$f(x) = x³ - x² + (3 - 2 + 6)x + 3$

$f(x) = x³ - x² + 7x + 3$ - Thus,

$f(-1) = (-1)³ - (-1)² + 7(-1) + 3$

$= -1 - 1 - 7 + 3$ - $f(-1) = -6$. ## Basic Algebraic Operations ### Subtraction: - Problem: Subtract $(4x² - 5x + 1) - (x³ + x² - 3x + 4)$ - Step 1: Distribute the negative sign:

$4x² - 5x + 1 - x³ - x² + 3x - 4$ - Step 2: Rearrange terms by degree:

$- x³ + 4x² - 1x² - 5x + 3x + 1 - 4$ - Step 3: Combine like terms:

$- x³ + (4 - 1)x² + (-5 + 3)x + (1 - 4)$ - Result: $- x³ + 3x² - 2x - 3$. ### Polynomial Product and Multiplication: - Problem (a): Find the product $(2x + 1)(x² - x - 3)$. - Step 1: Distribute each term from the first polynomial to the second:

$= 2x(x² - x - 3) + 1(x² - x - 3)$ - Step 2: Multiply each distributed term:

$= (2x \cdot x²) + (2x \cdot -x) + (2x \cdot -3) + (1 \cdot x²) + (1 \cdot -x) + (1 \cdot -3)$

$= 2x³ - 2x² - 6x + x² - x - 3$ - Step 3: Combine like terms:

$= 2x³ + (-2x² + x²) + (-6x - x) - 3$ - Expansion yields $2x³ - x² - 7x - 3$. - Problem (b): Simplify $(x + 2)(x + 2)(x + 2)$ (Cubic Power): - Use cubic expansion to yield:

$= (x + 2)³ = x³ + 6x² + 12x + 8$. ## Solve Quadratic Equations ### Solve 4(t + 6)² = 160: - Step 1: Divide both sides by 4:

$\frac{4(t + 6)²}{4} = \frac{160}{4}$

$(t + 6)² = 40$ - Step 2: Take the square root of both sides:

$\sqrt{(t + 6)²} = \pm\sqrt{40}$

$t + 6 = \pm\sqrt{40}$ - Step 3: Isolate t:

$t = -6 \pm\sqrt{40}$ - The solutions are $t = -6 + \sqrt{40}$ and $t = -6 - \sqrt{40}$. ### Solve 9x² + 27 = 0: - Step 1: Subtract 27 from both sides:

$9x² = -27$ - Step 2: Divide by 9:

$\frac{9x²}{9} = \frac{-27}{9}$

$x² = -3$ - Step 3: Take the square root of both sides:

$\sqrt{x²} = \pm\sqrt{-3}$ - Step 4: Express the imaginary solution:

$x = \pm i\sqrt{3}$ - Therefore, the two solutions are $x = i\sqrt{3}$ and $x = -i\sqrt{3}$. ## Number and Type of Zeros of Functions ### Determine Zeros: - Function: $f(x) = 5x - 29x³ + 55x² - 28x$ - Approach: - Factorization or numerical approximation to find roots. ### Factor the Polynomial: 1. Polynomial: $20x³ - 5x$. - Step 1: Factor out the common monomial factor:

$5x(4x² - 1)$ - Step 2: Recognize the difference of squares pattern ($a² - b² = (a - b)(a + b)$):

$5x((2x)² - 1²)$ - Factor: $5x(2x + 1)(2x - 1)$. 2. Polynomial: $3x² - 15x + x - 5$. - Step 1: Combine like terms:

$3x² - 14x - 5$ - Step 2: Apply Quadratic Formula (or factoring by grouping if applicable): $x = \frac{-(-14) \pm \sqrt{(-14)² - 4(3)(-5)}}{2(3)}$. - Step 3: Simplify the expression under the square root:

$x = \frac{14 \pm \sqrt{196 - (-60)}}{6}$

$x = \frac{14 \pm \sqrt{256}}{6}$

$x = \frac{14 \pm 16}{6}$ - Step 4: Calculate the two solutions:

$x_1 = \frac{14 + 16}{6} = \frac{30}{6} = 5$

$x<em>2 = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3}$ 3. Polynomial: $10x² - 13x - 3$. - Using the quadratic formula (or factoring by trial and error): - Identify $a=10$, $b=-13$, $c=-3$. - $x = \frac{-(-13) \pm \sqrt{(-13)² - 4(10)(-3)}}{2(10)}$ - Simplify: $x = \frac{13 \pm \sqrt{169 + 120}}{20} = \frac{13 \pm \sqrt{289}}{20} = \frac{13 \pm 17}{20}$ - Solutions: $x</em>1 = \frac{13 + 17}{20} = \frac{30}{20} = \frac{3}{2}$ and $x_2 = \frac{13 - 17}{20} = \frac{-4}{20} = -\frac{1}{5}$. 4. Polynomial: $8x³ + 125$. - Recognizes sum of cubes factorization: $(a³ + b³ = (a + b)(a² - ab + b²))$:

$(2x)³ + (5)³$ - Factor: $(2x + 5)((2x)² - (2x)(5) + 5²)$

$(2x + 5)(4x² - 10x + 25)$ which can be solved accordingly (one real root from $2x+5=0$ and two complex conjugate roots from the quadratic factor). 5. Polynomial: $x + 4x³ - 32x$. - Rearranging gives: $4x³ - 31x$ can be factored out for finding roots. - It seems there was a typo in the original prompt about $x$ being $x^4$. Assuming it was $x^4 + 4x^3 - 32x^2$. Let's correct it by assuming the first term should have been $x^2$. - Corrected assumption: $x² + 4x³ - 32x$ (not standard order) or perhaps $x^4 + 4x^3 - 32x^2$. Let's use the given as is for factoring: $4x³ + x - 32x$. - Combining: $4x³ - 31x$ - Factor out $x$: $x(4x² - 31)$ - Set factors to zero: - $x = 0$ - $4x² - 31 = 0 \Rightarrow 4x² = 31 \Rightarrow x² = \frac{31}{4} \Rightarrow x = \pm \frac{\sqrt{31}}{2}$ - Thus, three real roots: $0$, $\frac{\sqrt{31}}{2}$, and $- \frac{\sqrt{31}}{2}$. ### Determine Number and Type of Roots: - Analysis of polynomial expressions based on counts of degree and possible rational zeros reveals roots types (real vs. complex). ## Projectile Motion and Vertex Calculations ### Ball Thrown into the Air: - Initial Velocity: 24 ft/s, height function:

$h = -16t² + 24t + 4$. #### (A) Maximum Height Duration: - Use vertex formula $t = -\frac{b}{2a}$ to find when the maximum height occurs:- Substitute $b = 24$ and $a = -16$: $t = -\frac{24}{2(-16)} = -\frac{24}{-32} = 0.75$ (seconds). #### (B) Maximum Height: - Plug time back into the height equation:

$h(0.75) = -16(0.75)² + 24(0.75) + 4$

$h(0.75) = -16(0.5625) + 18 + 4$

$h(0.75) = -9 + 18 + 4 = 13$ feet. #### (C) Time When Hits Ground: - Solve for $t$ in $h = 0$ equation:

$0 = -16t² + 24t + 4$ - Use quadratic formula: $t = \frac{-24 \pm \sqrt{24² - 4(-16)(4)}}{2(-16)}$ - Simplify: $t = \frac{-24 \pm \sqrt{576 + 256}}{-32} = \frac{-24 \pm \sqrt{832}}{-32}$ - Approximate $t$ values (only positive time is physically relevant). $t \approx \frac{-24 - 28.84}{-32} \approx \frac{-52.84}{-32} \approx 1.65$ seconds. ## Completing the Square & Quadratic Formula Solve - Problem: Completing the Square for $x² + 22x + ?$: 1. The value to add is $\left(\frac{22}{2}\right)² = (11)² = 121$. ### Quadratic Formula to Solve: - 1. For the equation $-2x² + 5x = -10$: - Step 1: Rewrite the equation in standard form ($ax² + bx + c = 0$):

$-2x² + 5x + 10 = 0$ - (Alternatively, multiply by -1 to make 'a' positive):

$2x² - 5x - 10 = 0$ - Step 2: Identify a, b, and c:

$a = 2, b = -5, c = -10$ - Step 3: Apply the Quadratic Formula:

$x = \frac{-b \pm \sqrt{b² - 4ac}}{2a}$

$x = \frac{-(-5) \pm \sqrt{(-5)² - 4(2)(-10)}}{2(2)}$ - Step 4: Simplify the expression under the square root (discriminant):

$x = \frac{5 \pm \sqrt{25 - (-80)}}{4}$

$x = \frac{5 \pm \sqrt{105}}{4}$ - Resolve resulting radicand and simplify. The solutions are $x = \frac{5 + \sqrt{105}}{4}$ and $x = \frac{5 - \sqrt{105}}{4}$. - 2. For $6x² + 7x + 4 = 0$: - Step 1: Identify a, b, and c:

$a = 6, b = 7, c = 4$ - Step 2: Apply the Quadratic Formula:

$x = \frac{-7 \pm \sqrt{(7)² - 4(6)(4)}}{2(6)}$ - Step 3: Simplify the discriminant:

$x = \frac{-7 \pm \sqrt{49 - 96}}{12}$

$x = \frac{-7 \pm \sqrt{-47}}{12}$ - Step 4: Express the imaginary solutions:

$x = \frac{-7 \pm i\sqrt{47}}{12}$ - The roots are complex: $x = \frac{-7 + i\sqrt{47}}{12}$ and $x = \frac{-7 - i\sqrt{47}}{12}$. ### Solve Using Completing the Square - Problem: Solve $x² + 6x - 7 = 0$ - Step 1: Move the constant term to the right side:

$x² + 6x = 7$ - Step 2: Find the value needed to complete the square ($(\frac{b}{2})²$):

$(\frac{6}{2})² = (3)² = 9$ - Step 3: Add this value to both sides of the equation:

$x² + 6x + 9 = 7 + 9$

$x² + 6x + 9 = 16$ - Step 4: Factor the perfect square trinomial on the left side:

$(x + 3)² = 16$ - Step 5: Take the square root of both sides:

$\sqrt{(x + 3)²} = \pm\sqrt{16}$

$x + 3 = \pm4$ - Step 6: Solve for x:

$x + 3 = 4 \Rightarrow x = 4 - 3 = 1$ <br