Module 1 Practice Problems

Solution:

(a) 0: Whole, Integer, Rational, Real

Explanation: 0 is not a natural number (natural numbers start at 1), but it IS a whole number, integer, can be written as 0/1 (rational), and is on the number line (real).

(b) -7: Integer, Rational, Real

Explanation: -7 is negative, so it's NOT natural or whole. But it IS an integer (whole numbers + negatives), rational (-7 = -7/1), and real.

(c) √5: Irrational, Real

Explanation: √5 ≈ 2.236... cannot be written as a fraction (the decimal never repeats or terminates). It's irrational and real.

(d) 3/4: Rational, Real

Explanation: 3/4 is already written as a fraction of two integers, so it's rational. It equals 0.75 (terminates), which confirms it's rational. Also real.

(e) π: Irrational, Real

Explanation: π ≈ 3.14159... never repeats or terminates and cannot be written as a fraction. It's irrational and real.

Quick Reminder: Natural ⊂ Whole ⊂ Integers ⊂ Rational ⊂ Real. Irrational numbers are also part of Real numbers but NOT rational.

Solution:

(a) (3, 5): Quadrant I

Reasoning: Both x and y are positive (x = 3, y = 5). Quadrant I has (+, +).

(b) (-2, 4): Quadrant II

Reasoning: x is negative (-2), y is positive (4). Quadrant II has (−, +).

(c) (-6, -3): Quadrant III

Reasoning: Both x and y are negative. Quadrant III has (−, −).

(d) (4, -7): Quadrant IV

Reasoning: x is positive (4), y is negative (-7). Quadrant IV has (+, −).

(e) (0, -5): On the y-axis (not in any quadrant)

Reasoning: When x = 0, the point is on the y-axis. Since y = -5 (negative), it's on the negative y-axis.

Quadrant Sign Memory:

• I: (+, +) - upper right
• II: (−, +) - upper left
• III: (−, −) - lower left
• IV: (+, −) - lower right

Solution:

Given points: (-2, 3) and (4, -5)

Let (x₁, y₁) = (-2, 3) and (x₂, y₂) = (4, -5)

Step 1: Apply the distance formula

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Step 2: Substitute values

d = √[(4 - (-2))² + (-5 - 3)²]

d = √[(4 + 2)² + (-8)²]

d = √[(6)² + (-8)²]

Step 3: Simplify

d = √[36 + 64]

d = √100

d = 10

Answer: The distance is 10 units.

Verification: This makes sense! The horizontal distance is |4 - (-2)| = 6, and the vertical distance is |-5 - 3| = 8. We have a 6-8-10 right triangle (which is a multiple of the 3-4-5 Pythagorean triple).

Solution:

Given points: (1, 7) and (9, -3)

Let (x₁, y₁) = (1, 7) and (x₂, y₂) = (9, -3)

Step 1: Apply the midpoint formula

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Step 2: Substitute values

M = ((1 + 9)/2, (7 + (-3))/2)

Step 3: Simplify

M = (10/2, 4/2)

M = (5, 2)

Answer: The midpoint is (5, 2).

Interpretation: The midpoint is exactly halfway between the two points. Notice that x goes from 1 to 9 (average = 5), and y goes from 7 to -3 (average = 2). The midpoint averages the x-coordinates and y-coordinates separately!

Solution:

(a) {(1, 2), (3, 4), (5, 6)}: YES, this IS a function

Reasoning: Each input (x-value) appears only once: 1→2, 3→4, 5→6. Every input has exactly ONE output, which satisfies the definition of a function.

(b) {(2, 3), (3, 4), (2, 5)}: NO, this is NOT a function

Reasoning: The input x = 2 appears twice with different outputs: 2→3 and 2→5. Since the input 2 maps to TWO different outputs, this violates the function rule. A function requires each input to have exactly ONE output.

(c) {(0, 1), (1, 1), (2, 1)}: YES, this IS a function

Reasoning: Each input appears only once: 0→1, 1→1, 2→1. Even though all three inputs map to the SAME output (1), that's perfectly fine! A function can have multiple inputs that produce the same output. What matters is that each input has only ONE output.

Function Rule: Each input (x) must map to EXACTLY ONE output (y). Multiple x's can share the same y, but one x cannot have multiple y's!

Solution:

(a) Parabola opening upward: IS a function

Vertical Line Test: If you draw any vertical line (like x = 2), it will only intersect the parabola at ONE point. Since no vertical line hits the graph more than once, it passes the vertical line test and IS a function.

(b) Circle: NOT a function

Vertical Line Test: If you draw a vertical line through the middle of a circle, it will intersect the circle at TWO points (top and bottom). Since at least one vertical line hits the graph more than once, it fails the vertical line test and is NOT a function.

(c) Straight line with slope 2: IS a function

Vertical Line Test: A non-vertical straight line (like y = 2x + 1) will only be intersected by any vertical line at ONE point. It passes the test and IS a function. (Note: A VERTICAL line like x = 3 would NOT be a function!)

(d) Sideways parabola: NOT a function

Vertical Line Test: For a sideways parabola like x = y², a vertical line will intersect the graph at TWO points (for example, at x = 4, you get y = 2 and y = -2). It fails the test and is NOT a function.

Remember: Vertical line test = If ANY vertical line hits the graph MORE THAN ONCE, it's NOT a function!

Solution:

(a) Find f(3):

Substitute x = 3 into f(x) = x² - 4x + 1:

f(3) = (3)² - 4(3) + 1

f(3) = 9 - 12 + 1

f(3) = -2

(b) Find f(-2):

Substitute x = -2 into f(x) = x² - 4x + 1:

f(-2) = (-2)² - 4(-2) + 1

f(-2) = 4 + 8 + 1

f(-2) = 13

(c) Find f(a + 1):

Substitute x = (a + 1) into f(x) = x² - 4x + 1:

f(a + 1) = (a + 1)² - 4(a + 1) + 1

Expand (a + 1)² = a² + 2a + 1:

f(a + 1) = (a² + 2a + 1) - 4a - 4 + 1

Combine like terms:

f(a + 1) = a² + 2a - 4a + 1 - 4 + 1

f(a + 1) = a² - 2a - 2

Key Concept: To evaluate f(anything), replace every x in the function with "anything" and simplify!

Solution:

(a) f(x) = x² - 3

Domain: All real numbers (−∞, ∞)

Reasoning: You can square any real number, so there are no restrictions on x.

Range: [−3, ∞)

Reasoning: x² ≥ 0 for all x, so x² - 3 ≥ -3. The minimum value is -3 (when x = 0), and it goes up to infinity.

(b) g(x) = 1/(x - 2)

Domain: All real numbers except x = 2, or (−∞, 2) ∪ (2, ∞)

Reasoning: Cannot divide by zero! Set denominator ≠ 0: x - 2 ≠ 0, so x ≠ 2.

Range: All real numbers except y = 0, or (−∞, 0) ∪ (0, ∞)

Reasoning: A fraction 1/(x-2) can never equal zero (numerator is 1, not 0), but it can get arbitrarily close to 0 and can be any other real number.

(c) h(x) = √(x + 4)

Domain: [−4, ∞)

Reasoning: Square root requires non-negative input: x + 4 ≥ 0, so x ≥ -4.

Range: [0, ∞)

Reasoning: Square root always produces non-negative outputs. Minimum is √0 = 0 (when x = -4), maximum is ∞.

(d) k(x) = |x| + 1

Domain: All real numbers (−∞, ∞)

Reasoning: Absolute value works for any real number.

Range: [1, ∞)

Reasoning: |x| ≥ 0 for all x, so |x| + 1 ≥ 1. Minimum value is 1 (when x = 0), and it increases without bound.

Solution:

(a) (f + g)(x) = f(x) + g(x)

(f + g)(x) = (2x + 1) + (x² - 3)

(f + g)(x) = x² + 2x + 1 - 3

(f + g)(x) = x² + 2x - 2

(b) (f - g)(x) = f(x) - g(x)

(f - g)(x) = (2x + 1) - (x² - 3)

Be careful with the subtraction! Distribute the negative:

(f - g)(x) = 2x + 1 - x² + 3

(f - g)(x) = -x² + 2x + 4

(c) (f · g)(x) = f(x) · g(x)

(f · g)(x) = (2x + 1)(x² - 3)

Use FOIL/distribution:

(f · g)(x) = 2x(x²) + 2x(-3) + 1(x²) + 1(-3)

(f · g)(x) = 2x³ - 6x + x² - 3

(f · g)(x) = 2x³ + x² - 6x - 3

Order matters! Notice (f - g)(x) ≠ (g - f)(x). Addition and multiplication are commutative, but subtraction is not!

Solution:

Step 1: Find (f/g)(x)

(f/g)(x) = f(x) / g(x) = (x² - 4) / (x - 2)

Step 2: Factor the numerator

Notice that x² - 4 is a difference of squares: x² - 4 = (x + 2)(x - 2)

(f/g)(x) = [(x + 2)(x - 2)] / (x - 2)

Step 3: Simplify (with restriction!)

For x ≠ 2, we can cancel (x - 2):

(f/g)(x) = x + 2, where x ≠ 2

Step 4: State the domain

Domain: All real numbers except x = 2, or (−∞, 2) ∪ (2, ∞)

Important Note: Even though (f/g)(x) simplifies to x + 2 (which has domain all real numbers), we must exclude x = 2 because g(2) = 0 would make the original expression undefined. The domain restriction stays even after simplification!

Graph interpretation: The graph of (f/g)(x) looks like the line y = x + 2, but with a "hole" at the point (2, 4) where x = 2 is excluded.

Solution:

(a) (f + g)(2)

Method 1 (Add functions first):

(f + g)(x) = 3x + (x² - 1) = x² + 3x - 1

(f + g)(2) = (2)² + 3(2) - 1 = 4 + 6 - 1 = 9

Method 2 (Evaluate separately then add):

f(2) = 3(2) = 6, g(2) = (2)² - 1 = 3

(f + g)(2) = f(2) + g(2) = 6 + 3 = 9

(b) (f · g)(-1)

Method 2 is easier here - evaluate separately then multiply:

f(-1) = 3(-1) = -3

g(-1) = (-1)² - 1 = 1 - 1 = 0

(f · g)(-1) = f(-1) · g(-1) = (-3)(0) = 0

(c) (f / g)(4)

Evaluate separately then divide:

f(4) = 3(4) = 12

g(4) = (4)² - 1 = 16 - 1 = 15

(f / g)(4) = f(4) / g(4) = 12/15 = 4/5

Pro Tip: For evaluating combined functions at specific numbers, it's often easier to evaluate each function separately and then combine the results!

Solution:

(a) U-shaped parabola: f(x) = x² (Quadratic)

Description: Classic parabola opening upward with vertex at (0, 0).

(b) V-shaped graph: f(x) = |x| (Absolute Value)

Description: Two rays meeting at origin forming a V-shape.

(c) Hyperbola in Quadrants I and III: f(x) = 1/x (Reciprocal)

Description: Two curves that approach but never touch the axes (asymptotes at x = 0 and y = 0).

(d) Diagonal line through origin: f(x) = x (Linear)

Description: Straight line with slope 1 passing through (0, 0).

(e) S-curve through origin: f(x) = x³ (Cubic)

Description: Curve that increases from bottom-left to top-right, passing through origin with an S-shape.

(f) Half-parabola starting at origin, going right: f(x) = √x (Square Root)

Description: Starts at (0, 0) and curves upward to the right, like half a sideways parabola.

Memory Tip: Learn to recognize these six parent functions by their distinctive shapes - all other functions in algebra are transformations of these!

Solution:

Given: g(x) = -2(x + 3)² + 1, starting from f(x) = x²

Step-by-step transformations (in correct order):

1. Horizontal Shift LEFT 3 units

From x² to (x + 3)²

Note: Remember, (x + 3) means shift LEFT (opposite of the sign!)

2. Vertical Stretch by factor of 2

From (x + 3)² to 2(x + 3)²

Note: The coefficient 2 makes the parabola narrower/steeper.

3. Reflection over x-axis

From 2(x + 3)² to -2(x + 3)²

Note: The negative sign flips the parabola upside down (now opens downward).

4. Vertical Shift UP 1 unit

From -2(x + 3)² to -2(x + 3)² + 1

Note: Adding 1 outside shifts the entire graph up.

Summary of transformations:

• Shift LEFT 3 units
• Stretch vertically by factor of 2
• Reflect over x-axis (flip upside down)
• Shift UP 1 unit

Key Concept - Order Matters!

1. Horizontal shifts (inside parentheses)

2. Stretches and reflections (coefficient)

3. Vertical shifts (outside addition/subtraction)

Final Graph Description: An upside-down parabola with vertex at (-3, 1), opening downward, narrower than the parent function.

Solution:

Starting function: f(x) = |x|

Step 1: Shift RIGHT 4 units

To shift right, we replace x with (x - 4):

After shift right: |x - 4|

Remember: Right shifts use MINUS (opposite of what you might expect!)

Step 2: Shift DOWN 2 units

To shift down, we subtract 2 from the entire function:

After shift down: |x - 4| - 2

Step 3: Reflect over x-axis

To reflect over x-axis, we multiply the entire function by -1:

After reflection: -(|x - 4| - 2)

Simplify:

g(x) = -(|x - 4| - 2)

g(x) = -|x - 4| + 2

Final Answer: g(x) = -|x - 4| + 2

Verification:

• (x - 4) creates the right shift by 4
• +2 outside creates the up shift by 2 (we went down 2, then reflected, which flips down→up)
• Negative sign creates the reflection over x-axis (V flips upside down)

Graph Description: An upside-down V with vertex at (4, 2).

Solution:

Parent function: f(x) = x² with vertex at (0, 0)

Transformation 1: Horizontal Shift

Apply (x - 1)²

What happens: Shift RIGHT 1 unit

Vertex moves: (0, 0) → (1, 0)

Graph at this step: Parabola opening upward with vertex at (1, 0)

Transformation 2: Reflection

Apply -(x - 1)²

What happens: Reflect over the x-axis (flip upside down)

Vertex stays at: (1, 0) - reflection doesn't move the vertex horizontally or vertically, just flips direction

Graph at this step: Parabola opening DOWNWARD with vertex at (1, 0)

Transformation 3: Vertical Shift

Apply -(x - 1)² + 3

What happens: Shift UP 3 units

Vertex moves: (1, 0) → (1, 3)

Final graph: Parabola opening DOWNWARD with vertex at (1, 3)

Complete Transformation Summary:

1. Start with f(x) = x², vertex (0, 0), opens up
2. Shift right 1 → vertex (1, 0), still opens up
3. Reflect over x-axis → vertex (1, 0), now opens DOWN
4. Shift up 3 → vertex (1, 3), opens down

Final Answer: g(x) = -(x - 1)² + 3 is an upside-down parabola with vertex at (1, 3), maximum value of 3.

Graphing Strategy:

• Identify the vertex from the equation: (h, k) = (1, 3)
• Note the direction: negative coefficient means opens DOWN
• Plot the vertex (1, 3)
• Since there's no stretch/compression (coefficient is -1), use the standard parabola shape
• Sketch symmetrically on both sides of the vertex