Solved Exercises on Standard Forms of Parabola in Grade 1

Master standard forms of parabola: vertex form, intercept form, and transformations through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Vertex form conversion
Exercise 1
Convert to vertex form and find vertex:
\(f(x) = x^2 - 6x + 5\)
Definition:

Vertex form: \(f(x) = a(x-h)^2 + k\) where (h,k) is the vertex

Completing the square method:
  1. Factor out coefficient of x² if not 1
  2. Take coefficient of x, divide by 2, square it
  3. Add and subtract this value inside the expression
  4. Factor the perfect square trinomial
  5. Simplify to vertex form
Original
\(f(x) = x^2 - 6x + 5\)
Complete square
\(f(x) = (x-3)^2 - 4\)
Vertex
(3, -4)
Step 1: Identify coefficients

For f(x) = x² - 6x + 5: a = 1, b = -6, c = 5

Step 2: Complete the square

Take half of -6: (-6)/2 = -3, then (-3)² = 9

f(x) = x² - 6x + 9 - 9 + 5

Step 3: Factor perfect square trinomial

f(x) = (x² - 6x + 9) - 4 = (x - 3)² - 4

Step 4: Identify vertex

In vertex form f(x) = a(x-h)² + k, vertex is (h,k)

So vertex is (3, -4)

Vertex form: \(f(x) = (x-3)^2 - 4\), Vertex: (3, -4)
Final answer:

Vertex form: \(f(x) = (x-3)^2 - 4\), Vertex: (3, -4)

Applied rules:

• Completing the square: Add and subtract \((\frac{b}{2})^2\)

• Perfect square trinomial: x² + bx + \((\frac{b}{2})^2\) = (x + \(\frac{b}{2}\))²

• Vertex identification: From vertex form f(x) = a(x-h)² + k, vertex is (h,k)

2 Direction and width analysis
Exercise 2
Analyze the parabola:
\(f(x) = -2(x+1)^2 + 8\)
Definition:

Parabola orientation: If a > 0, opens upward; if a < 0, opens downward

Function
\(f(x) = -2(x+1)^2 + 8\)
Parameter a
a = -2
Vertex
(-1, 8)
Step 1: Identify the coefficient a

From vertex form f(x) = a(x-h)² + k, we have a = -2

Step 2: Determine direction

Since a = -2 < 0, the parabola opens downward

Step 3: Determine width

Since |a| = |-2| = 2 > 1, the parabola is narrower than y = x²

Step 4: Identify vertex and other features

Vertex: (h, k) = (-1, 8)

y-intercept: f(0) = -2(1) + 8 = 6, so (0, 6)

Opens downward, narrower than y=x², vertex: (-1, 8)
Final answer:

Parabola opens downward, is narrower than y = x², vertex at (-1, 8)

Applied rules:

• Direction rule: a > 0 → opens upward, a < 0 → opens downward

• Width rule: |a| > 1 → narrower, |a| < 1 → wider than y = x²

• Vertex identification: From f(x) = a(x-h)² + k, vertex is (h,k)

3 Intercepts calculation
Exercise 3
Find x and y intercepts of:
\(f(x) = x^2 - 4x - 5\)
Definition:

X-intercepts: Values where f(x) = 0, also called zeros or roots

Function
\(f(x) = x^2 - 4x - 5\)
Factor
\(f(x) = (x-5)(x+1)\)
X-intercepts
x = 5, x = -1
Step 1: Find y-intercept

Set x = 0: f(0) = 0² - 4(0) - 5 = -5

Y-intercept: (0, -5)

Step 2: Find x-intercepts

Set f(x) = 0: x² - 4x - 5 = 0

Step 3: Factor the quadratic

We need two numbers that multiply to -5 and add to -4

Those numbers are -5 and 1: (x - 5)(x + 1) = 0

Step 4: Solve for x

Either x - 5 = 0 or x + 1 = 0

So x = 5 or x = -1

X-intercepts: (5, 0) and (-1, 0)

Y-intercept: (0, -5), X-intercepts: (5, 0), (-1, 0)
Final answer:

Y-intercept: (0, -5), X-intercepts: (5, 0) and (-1, 0)

Applied rules:

• Y-intercept: Set x = 0 and solve for f(0)

• X-intercepts: Set f(x) = 0 and solve for x

• Factoring: Find two numbers that multiply to c and add to b

Parabola Rules and Methods
\(f(x) = a(x-h)^2 + k\)
Vertex Form
Vertex Form
\(f(x) = a(x-h)^2 + k\)
Vertex: (h,k)
Standard Form
\(f(x) = ax^2 + bx + c\)
Y-intercept: (0,c)
Intercept Form
\(f(x) = a(x-r_1)(x-r_2)\)
X-intercepts: r₁,r₂
Key definitions:

Parabola: U-shaped curve that is the graph of a quadratic function

Vertex: Highest or lowest point on the parabola, axis of symmetry passes through it

Axis of symmetry: Vertical line that divides the parabola into two mirror images

X-intercepts: Points where the parabola crosses the x-axis (zeros of the function)

Y-intercept: Point where the parabola crosses the y-axis

Complete methodology:
  1. Identify the form: Determine if given in standard, vertex, or intercept form
  2. Extract parameters: Find a, h, k, or x-intercepts as needed
  3. Analyze characteristics: Direction, width, vertex, intercepts
  4. Transform if needed: Convert between forms using algebraic methods
Tip 1: The coefficient 'a' controls direction and width of the parabola.
Tip 2: Vertex form immediately reveals the vertex coordinates.
Tip 3: Intercept form immediately reveals the x-intercepts.
Tip 4: Axis of symmetry is always x = h in vertex form.
Common errors: Forgetting to change signs when identifying vertex from vertex form, incorrect factoring, miscalculating intercepts.
Exam preparation: Practice converting between forms, memorize vertex formulas, understand the role of each parameter.
Formulas to know by heart:

• Vertex form: \(f(x) = a(x-h)^2 + k\), vertex at (h,k)

• Standard form: \(f(x) = ax^2 + bx + c\), y-intercept at (0,c)

• Intercept form: \(f(x) = a(x-r_1)(x-r_2)\), x-intercepts at (r₁,0), (r₂,0)

• Axis of symmetry: x = h in vertex form, x = -b/(2a) in standard form

• Vertex x-coordinate: x = -b/(2a) when in standard form

Solution: Exercises 4 to 5
4 Transformation analysis
Exercise 4
Describe transformations from y = x²:
\(f(x) = 3(x-2)^2 - 4\)
Definition:

Transformation: Changes to parent function y = x² through shifts, stretches, and reflections

Function
\(f(x) = 3(x-2)^2 - 4\)
Transformations
Vertical stretch by 3, right 2, down 4
New vertex
(2, -4)
Step 1: Compare to parent function

Parent: y = x², Given: f(x) = 3(x-2)² - 4

Step 2: Identify each transformation

• Coefficient a = 3: Vertical stretch by factor of 3

• (x-2): Horizontal shift right by 2 units

• -4: Vertical shift down by 4 units

Step 3: Determine the effect on the graph

• The parabola opens upward (a > 0) but is narrower (|a| > 1)

• Vertex moves from (0,0) to (2,-4)

Step 4: Verify the vertex

From vertex form f(x) = a(x-h)² + k, vertex is (h,k) = (2,-4)

Vertical stretch by 3, horizontal shift right 2, vertical shift down 4
Final answer:

The parabola is vertically stretched by factor 3, shifted right 2 units, and shifted down 4 units

Applied rules:

• Vertical stretch: Coefficient a multiplies function values

• Horizontal shift: (x-h) shifts right by h, (x+h) shifts left by h

• Vertical shift: +k shifts up by k, -k shifts down by k

5 Application problem
Exercise 5
A ball is thrown upward with height:
\(h(t) = -16t^2 + 64t + 5\) (t in seconds, h in feet)
Definition:

Projectile motion: Height modeled by quadratic function with negative leading coefficient

Function
\(h(t) = -16t^2 + 64t + 5\)
Vertex form
\(h(t) = -16(t-2)^2 + 69\)
Max height
69 feet at t = 2 sec
Step 1: Convert to vertex form

h(t) = -16t² + 64t + 5

Factor out -16: h(t) = -16(t² - 4t) + 5

Complete the square: h(t) = -16(t² - 4t + 4 - 4) + 5

h(t) = -16((t-2)² - 4) + 5 = -16(t-2)² + 64 + 5

h(t) = -16(t-2)² + 69

Step 2: Find maximum height

Since a = -16 < 0, parabola opens downward, vertex is maximum

Maximum height: 69 feet at t = 2 seconds

Step 3: Find initial height

Initial height: h(0) = -16(0)² + 64(0) + 5 = 5 feet

Step 4: Find when ball hits ground

Set h(t) = 0: -16t² + 64t + 5 = 0

Using quadratic formula: t = (−64 ± √(64² - 4(-16)(5))) / (2(-16))

t ≈ 4.08 seconds (positive solution)

Max height: 69 ft at 2 sec, Initial height: 5 ft, Ground impact: ~4.08 sec
Final answer:

Maximum height of 69 feet occurs at 2 seconds; ball starts at 5 feet and hits ground at approximately 4.08 seconds

Applied rules:

• Projectile motion: Height follows quadratic model with negative coefficient

• Maximum point: Vertex represents maximum when a < 0

• Quadratic formula: x = (-b ± √(b² - 4ac)) / (2a) for ax² + bx + c = 0

Parabola Theory and Applications
\(f(x) = ax^2 + bx + c\)
Standard Quadratic Form
Key definitions:

Quadratic function: Polynomial of degree 2, f(x) = ax² + bx + c where a ≠ 0

Parabola: The U-shaped graph of a quadratic function

Vertex: The turning point of the parabola, minimum if a > 0, maximum if a < 0

Axis of symmetry: Vertical line through the vertex, x = -b/(2a)

Discriminant: b² - 4ac determines number of real roots

Complete methodology:
  1. Form identification: Recognize which form the function is in
  2. Parameter extraction: Identify key values (a, h, k, intercepts)
  3. Feature analysis: Determine direction, vertex, intercepts, width
  4. Transformation mapping: Understand how parameters affect the graph
Tip 1: Remember that the vertex x-coordinate is always -b/(2a) regardless of form.
Tip 2: The sign of 'a' determines whether vertex is minimum (a > 0) or maximum (a < 0).
Tip 3: Completing the square is the key technique for converting standard to vertex form.
Tip 4: Discriminant (b² - 4ac) tells you how many x-intercepts exist.
Real-world applications: Projectile motion, optimization problems, business profit models, engineering structures.
Key relationships: Vertex form reveals transformations, intercept form reveals roots, standard form reveals y-intercept.
Essential formulas:

• Vertex x-coordinate: x = -b/(2a)

• Vertex form: f(x) = a(x-h)² + k

• Standard form: f(x) = ax² + bx + c

• Intercept form: f(x) = a(x-r₁)(x-r₂)

• Quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

• Discriminant: Δ = b² - 4ac (Δ > 0: 2 roots, Δ = 0: 1 root, Δ < 0: 0 real roots)

Exercise with Visualization: Parabola Families
Exercise 6: Parabola Comparison
Compare different parabolas:
\(P_1: f(x) = x^2\) (parent function)
\(P_2: f(x) = 2x^2\) (vertical stretch)
\(P_3: f(x) = -(x-1)^2 + 3\) (reflected and shifted)

Analysis: The visualization shows how different parameters affect parabola shape and position.

  • \(P_1\): Basic parabola opening upward with vertex at origin
  • \(P_2\): Narrower parabola due to vertical stretch by factor 2
  • \(P_3\): Opens downward, shifted right 1 unit and up 3 units

Questions & Answers

Question: I'm confused about the vertex form f(x) = a(x-h)² + k. Why is the vertex at (h,k) when the x-term is (x-h)? Shouldn't it be at (-h,k)?

Answer: This is a very common source of confusion! The key is to think about what value of x makes the squared term equal zero:

In f(x) = a(x-h)² + k, the squared term (x-h)² is minimized when (x-h) = 0

Solving x - h = 0 gives us x = h

At x = h, f(h) = a(h-h)² + k = a(0)² + k = k

So the vertex is at (h, k), not (-h, k).

Think of it this way: if you have f(x) = (x-3)², the parabola is shifted right by 3 units (positive direction), so the vertex moves from (0,0) to (3,0).

Similarly, f(x) = (x+3)² = (x-(-3))² shifts the parabola left by 3 units, moving the vertex from (0,0) to (-3,0).

Question: How do I remember which transformations correspond to which parts of the function? Like in f(x) = a(x-h)² + k, what does each letter do?

Answer: Here's a systematic way to remember the transformations in f(x) = a(x-h)² + k:

  • 'a' (coefficient): Controls vertical stretch/shrink and reflection
    • |a| > 1: Stretches vertically (narrower parabola)
    • |a| < 1: Shrinks vertically (wider parabola)
    • a < 0: Reflects across x-axis (opens downward)
  • 'h' (horizontal shift): Moves the parabola left/right
    • Positive h: Shifts right
    • Negative h: Shifts left
    • Remember: (x-h) means opposite of the sign of h
  • 'k' (vertical shift): Moves the parabola up/down
    • Positive k: Shifts up
    • Negative k: Shifts down
    • Same sign as k (what you see is what you get)

Memory aid: "a changes the shape, h shifts horizontally (opposite direction), k shifts vertically (same direction)"

Question: How do I convert from standard form to vertex form? The completing the square method seems tricky.

Answer: Completing the square follows a systematic process. Let's work with f(x) = ax² + bx + c:

  1. Factor out 'a' from first two terms: f(x) = a(x² + (b/a)x) + c
  2. Find the number to complete the square: Take the coefficient of x, divide by 2, square it: [(b/a)/2]² = (b/2a)²
  3. Add and subtract this number inside the parentheses: f(x) = a[x² + (b/a)x + (b/2a)² - (b/2a)²] + c
  4. Factor the perfect square trinomial: f(x) = a[(x + b/2a)² - (b/2a)²] + c
  5. Distribute and combine constants: f(x) = a(x + b/2a)² - a(b/2a)² + c
  6. Simplify to vertex form: f(x) = a(x - (-b/2a))² + (c - b²/4a)

Example with f(x) = x² - 6x + 5:

  • No need to factor 'a' since a = 1
  • Take half of -6: -3, square it: 9
  • f(x) = x² - 6x + 9 - 9 + 5
  • f(x) = (x - 3)² - 4

Question: What's the difference between the different forms of quadratic functions? When should I use each one?

Answer: Each form of a quadratic function reveals different information instantly:

  1. Standard form: f(x) = ax² + bx + c
    • Immediately shows y-intercept (0, c)
    • Easy to calculate f(0)
    • Good for general calculations and using quadratic formula
  2. Vertex form: f(x) = a(x-h)² + k
    • Immediately shows vertex (h, k)
    • Reveals transformations from parent function y = x²
    • Ideal for graphing and finding maximum/minimum values
  3. Intercept form: f(x) = a(x-r₁)(x-r₂)
    • Immediately shows x-intercepts (r₁, 0) and (r₂, 0)
    • Useful for factoring and finding zeros
    • Good for problems involving roots

Choose the form based on what information you need:

  • Need vertex/max/min? → Vertex form
  • Need x-intercepts/zeros? → Intercept form
  • Need y-intercept? → Standard form
  • Solving equations? → Standard form (quadratic formula)

Question: Are there real-world applications of parabolas? It seems like just another math concept.

Answer: Parabolas have numerous real-world applications across various fields:

  1. Physics - Projectile Motion: The path of a thrown ball, launched rocket, or water from a fountain follows a parabolic trajectory
  2. Engineering - Satellite Dishes: Parabolic reflectors focus signals to a single point (the focus)
  3. Architecture - Arches and Bridges: Parabolic shapes distribute weight efficiently
  4. Economics - Profit Optimization: Many profit functions follow quadratic models with maximum points
  5. Technology - Headlights and Flashlights: Parabolic reflectors direct light in parallel beams

Examples in daily life:

  • The arc of water from a garden hose
  • The shape of a suspension bridge cable
  • The path of a basketball through the air
  • Car headlights focusing light beams
  • Satellite dishes collecting signals

Understanding parabolas helps engineers design better structures, physicists predict motion, and economists optimize business decisions!