Math IH 2nd Semester Prep Study Guide

Geometric Constructions and Fundamentals

The study of geometry involves the precise construction of figures using a compass and a straightedge. Question 15 presents a diagram requiring identification among four options: the construction of a segment bisector, the construction of a perpendicular line through point $A$, the construction of a parallel line through point $A$, or the construction of a median. Correct identification depends on recognizing the arcs and intersection points characteristic of each specific procedure. Similarly, questions 16 and 17 require identifying the specific kind of construction being made based on visual diagrams involving arcs from points $A$, $B$, and $P$. In question 54, the objective is to use a compass and a straightedge to construct two parallel lines, a process typically involving the copy of an angle or the construction of two lines perpendicular to the same transversal.

Linear Equations, Slope, and Coordinate Geometry

Linear relationships are defined by their slope ($m$) and intercepts. Question 18 explores the relationship between lines that are parallel or perpendicular to the line $y = x + 5$. For parallel lines, the slope remains identical to the original line ($m = 1$), whereas for perpendicular lines, the slope is the negative reciprocal ($m = -1$). Question 19 tasks us with finding the equation of a line perpendicular to $y = -3x - 9$ that passes through the specific point $(9, -10)$. Since the original slope is $-3$, the perpendicular slope is $\frac{1}{3}$. Using the point-slope form, the equation is derived as $y - (-10) = \frac{1}{3}(x - 9)$, which simplifies to $y = \frac{1}{3}x - 13$. Question 53 asks for a similar derivation for a line perpendicular to $y = x - 9$ passing through $(-9, 10)$, where the perpendicular slope is $-1$.

In question 20, multiple pairs of lines are analyzed to determine if they are perpendicular. Perpendicularity is confirmed if the product of the slopes equals $-1$. The options include pairs such as $y = x$ and $y = \frac{1}{2}x$ (not perpendicular), $y = -2x$ and $y = \frac{1}{2}x$ (perpendicular because $-2 \times \frac{1}{2} = -1$), $y = -3x$ and $y = -\frac{4}{3}x$ (not perpendicular), and $y = x$ and $y = 4x$ (not perpendicular).

Question 21 requires calculating both the distance and the slope between the points $(4, 3)$ and $(6, 10)$. The slope is calculated as $m = \frac{10 - 3}{6 - 4} = \frac{7}{2}$. The distance ($d$) is found using the distance formula: d = \root { 2 } { (6 - 4)^2 + (10 - 3)^2 } = \root { 2 } { 2^2 + 7^2 } = \root { 2 } { 4 + 49 } = \root { 2 } { 53 }.

Geometric Applications of Algebra

Algebraic expressions represent physical dimensions in geometric problems. Question 22 involves solving for $x$ given that the perimeter of a triangle is $36$. The side lengths are provided as $x$, $x + 5$, and $x + 7$. The perimeter equation is $x + (x + 5) + (x + 7) = 36$, which simplifies to $3x + 12 = 36$. Subtracting $12$ from both sides yields $3x = 24$, and dividing by $3$ gives $x = 8$.

Question 25 requires finding the area of a rectangle with side lengths of $2x + 2$ and $4x + 3$. The area is found by multiplying these binomials: $(2x + 2)(4x + 3) = 8x^2 + 6x + 8x + 6 = 8x^2 + 14x + 6$. In question 26, Bobo builds a square armadillo fence with a side length of $5x - 3$ meters. The total amount of meters needed (the perimeter) is $4(5x - 3) = 20x - 12$. Question 28 asks for the shaded area within a larger rectangle of dimensions $3x$ and $3x + 10$, with a smaller rectangle removed. The total area is $3x(3x + 10) = 9x^2 + 30x$, and the subtracted area consists of a rectangle with dimensions such as $(x + 4)$ and an undisclosed side (visual interpretation needed).

Quadratic Functions and Trajectories

Quadratic equations model real-world movement and trajectories. Question 40 describes the trajectory of Mr. C's racquet after Mrs. C threw it. The height is given by $h(x) = -x^2 + 10x + 2$, where $x$ is the distance from the house. To find the maximum height, we identify the vertex. The $x$-coordinate of the vertex is $x = \frac{-b}{2a} = \frac{-10}{2(-1)} = 5$. Substituting this back into the function, $h(5) = -(5)^2 + 10(5) + 2 = -25 + 50 + 2 = 27$. Thus, the maximum height is $27$ units. In question 41, the racquet is launched by a catapult, and the task is to calculate the average rate of change between the first and third seconds of flight ($t = 1$ to $t = 3$) using the provided graph.

Question 42 asks for the values of $x$ where $y = 0$ for the function $y = x^2 - 4x - 5$. Factoring the quadratic yields $(x - 5)(x + 1) = 0$, so the zeros are $x = 5$ and $x = -1$. Question 31 requires solving $f(x) = x^2 - x - 6$ for $x$. Factoring gives $(x - 3)(x + 2) = 0$, resulting in roots $x = 3$ and $x = -2$.

Questions 37, 38, and 39 focus on parabola features. For question 38, the axis of symmetry for $f(x) = x^2 - 8x + 12$ is found using $x = \frac{-b}{2a} = \frac{8}{2} = 4$. For question 39, finding the vertex of $f(x) = 4x^2 - 16x + 7$ involves identifying the $x$-coordinate ($x = \frac{16}{8} = 2$) and the corresponding $y$-value ($f(2) = 4(4) - 16(2) + 7 = 16 - 32 + 7 = -9$). Since the lead coefficient ($4$) is positive, the vertex ($2, -9$) is a minimum point.

Function Inverses, Transformations, and Operations

Inverses of functions swap inputs and outputs. Question 43 provides three functions for which to find the inverse: a) For $f(x) = 5x + 7$, the inverse is $f^{-1}(x) = \frac{x - 7}{5}$. b) For $f(x) = x^2 + 6$, assuming $x \ge 0$, the inverse is f^{-1}(x) = \root { 2 } { x - 6 }. c) For $f(x) = 4x^2$, the inverse is f^{-1}(x) = \frac{\root { 2 } { x }}{2}.

Transformations of a function $f(x)$ are handled in question 50. Moving a function downward is represented by $f(x) - 2$ (option c). Moving it to the right is represented by $f(x - 3)^2$ (option a). Moving it up is represented by $f(x) + 7$ (option b). Moving it to the left is represented by $f(x + 9)^2$ (option d). Question 35 and 36 require matching graphs to functions in vertex form, such as $y = (x + 1)^2 - 2$ and $y = -2(x + 2)^2 - 1$, or factored form, such as $f(x) = (x + 2)(x - 3)$.

Function operations are explored in question 61. Given $f(x) = x^2 - 4x + 5$ and $g(x) = 3x + 5$: a) $(f + g)(x) = (x^2 - 4x + 5) + (3x + 5) = x^2 - x + 10$. b) $(f \cdot g)(x) = (x^2 - 4x + 5)(3x + 5) = 3x^3 + 5x^2 - 12x^2 - 20x + 15x + 25 = 3x^3 - 7x^2 - 5x + 25$.

Exponential Growth and Decay

Exponential functions follow the form $f(x) = a(b)^x$, where $a$ is the initial value and $b$ is the growth factor $(1 + r)$ or decay factor $(1 - r)$. Question 44 identifies the function representing an exponential growth of $38 \%$ as $f(x) = 10(1.38)^x$. Question 45 classifies several functions: a) $f(x) = 10(1.2)^{3x}$ is growth because $1.2 > 1$. b) $f(x) = 5(0.6)^x$ is decay because $0.6 < 1$. c) $f(x) = -9(1.1)^{2x}$ is growth because the factor $1.1 > 1$ (though it is reflected). d) $f(x) = 7(0.97)^{4x}$ is decay because $0.97 < 1$. e) $f(x) = 100(1.61)^x$ is growth because $1.61 > 1$.

Radical and Piecewise Functions

Radical functions require careful graphing and simplification. Question 48 asks to graph f(x) = 2\root { 2 } { x - 5 } + 4, which has a starting point at $(5, 4)$ and is stretched by a factor of $2$. Question 49 is a graph of f(x) = \root { 2 } { x } + 10. Rational exponent representation is addressed in question 57, where (\root { 3 } { 7 })^1 is converted to $7^{\frac{1}{3}}$. Question 58 involves simplifying x^{\frac{1}{2}} \cdot \root { 2 } { x^3 }, which is $x^{\frac{1}{2}} \cdot x^{\frac{3}{2}} = x^{\frac{4}{2}} = x^2$.

Piecewise functions are defined differently over intervals. Question 51 provides $f(x) = 3x$ for $-2 \le x \le 2$ and $f(x) = 7$ for $2 < x < 5$. This requires graphing a segment of a line with a slope of $3$ and then a horizontal line at $y = 7$. Question 56 identifies features of $f(x) = (x - 2)^2 - 3$, including the vertex at $(2, -3)$, the minimum (extremum) at $y = -3$, the $y$-intercept at $(0, 1)$, and the axis of symmetry (AoS) at $x = 2$.

Complex Numbers and Operations

Complex numbers involve the imaginary unit i = \root { 2 } { -1 }. Question 55 provides operations: a) $(-18 + 3i) + (18 - 9i) = 0 - 6i = -6i$. b) Simple square root simplification like \root { 2 } { 40 } = 2\root { 2 } { 10 }. c) $( -12 + i )( 7 + 7i ) = -84 - 84i + 7i + 7i^2 = -84 - 77i - 7 = -91 - 77i$.

Powers of $i$ follow a cycle: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$. Question 34 requires simplifying specific powers: a) $i^{54} = (i^4)^{13} \cdot i^2 = 1 \cdot (-1) = -1$. b) $i^{13} = (i^4)^3 \cdot i = 1 \cdot i = i$. c) $i^{43} = (i^4)^{10} \cdot i^3 = 1 \cdot (-i) = -i$. d) $i^{21} = (i^4)^5 \cdot i = 1 \cdot i = i$.

Additional complex number arithmetic is found in question 32 and 33, such as $(5i - 7) + (3i + 5) = 8i - 2$ and multiplying binomials like $(6i - 3)(3i + 4) = 18i^2 + 24i - 9i - 12 = -18 + 15i - 12 = -30 + 15i$.

Statistical Analysis and Modeling

Data analysis involves interpreting historical trends and calculating fits. Question 52 tracks Dr. Sanchez's patient data for age (months) and height (inches): $(1, 20), (3, 23), (6, 27), (8, 29), (9, 31), (12, 32), (15, 34)$. A scatter plot is created to visualize the relationship. Dr. Sanchez estimates the fit using $y = x + 20$. A residual plot is then used to determine if this linear function is a good model for the data. Linear relationships are also characterized by correlation coefficients. Question 23 states a correlation coefficient of $0.488$, which indicates a weak positive linear relationship.

Polynomial Simplification and Expansion

Simplifying expressions involves combining like terms and applying properties of radicals and exponents. Question 27 requires adding and subtracting polynomials: a) $(3x^3 - 6x^2 + 7x - 11) + (-10x^3 + 3x^2 - 15x - 2) = -7x^3 - 3x^2 - 8x - 13$. b) $(-x^3 + 2x^2 - 5x + 8) - (3x^3 - 5x^2 + 7x - 2) = -4x^3 + 7x^2 - 12x + 10$.

Question 29 simplifies radical/exponent expressions: a) x^5 \cdot \root { 2 } { x^4 } = x^5 \cdot x^2 = x^7. b) \root { 3 } { a^2 } \cdot a^{\frac{1}{2}} = a^{\frac{2}{3}} \cdot a^{\frac{1}{2}} = a^{\frac{4}{6} + \frac{3}{6}} = a^{\frac{7}{6}}. c) \root { 4 } { b } \cdot \root { 4 } { b^3 } = \root { 4 } { b^4 } = b. d) $x^{\frac{1}{5}} \cdot x^{\frac{2}{7}} = x^{\frac{7}{35} + \frac{10}{35}} = x^{\frac{17}{35}}$.

Finally, question 60 asks to simplify binomial expansion: $(3v + 6m)^2 = 9v^2 + 36vm + 36m^2$. Question 33cd involves similar expansions: $(2x + y)^2 = 4x^2 + 4xy + y^2$ and $(3x - 2y)^2 = 9x^2 - 12xy + 4y^2$.