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Graphing Quadratic Functions

Parabolas, vertex, axis of symmetry

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Graphing Quadratic Functions

What is a Quadratic Function?

A quadratic function has the form:

f(x) = ax² + bx + c or y = ax² + bx + c

where a ≠ 0.

The graph of a quadratic function is a parabola - a U-shaped curve.

Key features:

  • Opens up (if a > 0) or down (if a < 0)
  • Has a vertex (highest or lowest point)
  • Has an axis of symmetry (vertical line through vertex)
  • May have 0, 1, or 2 x-intercepts

The Standard Form

Standard Form: y = ax² + bx + c

What each coefficient tells us:

a (leading coefficient):

  • If a > 0: parabola opens UP (∪)
  • If a < 0: parabola opens DOWN (∩)
  • |a| > 1: narrower parabola (steeper)
  • |a| < 1: wider parabola (flatter)

c (constant term):

  • The y-intercept (where graph crosses y-axis)
  • The point (0, c)

b (middle coefficient):

  • Affects the position of the vertex horizontally
  • Used to find axis of symmetry

Parent Function: y = x²

The simplest quadratic is y = x².

Key points:

  • Vertex at origin (0, 0)
  • Opens upward
  • Axis of symmetry: x = 0 (y-axis)
  • Symmetric about y-axis

Table of values:

xy = x²
-24
-11
00
11
24

The parabola is symmetric: same y-values for ±x

The Vertex

The vertex is the turning point of the parabola.

If a > 0: Vertex is the MINIMUM (lowest point)

If a < 0: Vertex is the MAXIMUM (highest point)

Finding the vertex from y = ax² + bx + c:

x-coordinate: x = -b/(2a)

y-coordinate: Substitute x-value into equation

Example: Find vertex of y = x² - 4x + 3

x = -(-4)/(2(1)) = 4/2 = 2

y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1

Vertex: (2, -1)

Axis of Symmetry

The axis of symmetry is a vertical line through the vertex.

Equation: x = -b/(2a)

This is the same as the x-coordinate of the vertex!

Example: y = 2x² + 8x + 5

Axis of symmetry: x = -8/(2·2) = -8/4 = -2

The parabola is symmetric about the line x = -2.

Finding the Y-Intercept

The y-intercept is where the graph crosses the y-axis (when x = 0).

Simply evaluate f(0) = c

Example: y = 2x² - 5x + 3 Y-intercept: (0, 3)

Finding X-Intercepts (Zeros/Roots)

The x-intercepts are where the graph crosses the x-axis (when y = 0).

Set y = 0 and solve: ax² + bx + c = 0

Use factoring, quadratic formula, or other methods.

Example: Find x-intercepts of y = x² - 5x + 6

Set y = 0: x² - 5x + 6 = 0 Factor: (x - 2)(x - 3) = 0 x = 2 or x = 3

X-intercepts: (2, 0) and (3, 0)

Number of X-Intercepts

Use the discriminant b² - 4ac:

b² - 4ac > 0: Two x-intercepts (parabola crosses x-axis twice)

b² - 4ac = 0: One x-intercept (parabola touches x-axis at vertex)

b² - 4ac < 0: No x-intercepts (parabola doesn't reach x-axis)

Example 1: y = x² - 4 Discriminant: 0² - 4(1)(-4) = 16 > 0 Two x-intercepts: x = ±2

Example 2: y = x² + 1 Discriminant: 0² - 4(1)(1) = -4 < 0 No x-intercepts

Graphing Using Key Points

Step-by-step process:

Step 1: Determine if parabola opens up or down (check sign of a)

Step 2: Find vertex using x = -b/(2a), then find y

Step 3: Draw axis of symmetry (vertical line through vertex)

Step 4: Find y-intercept (0, c)

Step 5: Find x-intercepts if they exist (solve ax² + bx + c = 0)

Step 6: Plot additional points if needed

Step 7: Draw smooth U-shaped curve through points

Example 1: Complete Graphing

Graph: y = x² - 4x + 3

Step 1: a = 1 > 0, opens UP

Step 2: Find vertex x = -(-4)/(2·1) = 2 y = (2)² - 4(2) + 3 = -1 Vertex: (2, -1)

Step 3: Axis of symmetry: x = 2

Step 4: Y-intercept: (0, 3)

Step 5: X-intercepts: x² - 4x + 3 = 0 (x - 1)(x - 3) = 0 x = 1 or x = 3 Points: (1, 0) and (3, 0)

Step 6: Use symmetry If (0, 3) is on the graph, then (4, 3) is also on it (symmetric about x = 2)

Step 7: Draw parabola through: (1,0), (2,-1), (3,0), (0,3), (4,3)

Example 2: Parabola Opening Down

Graph: y = -x² + 2x + 3

Step 1: a = -1 < 0, opens DOWN

Step 2: Vertex x = -2/(2·(-1)) = 1 y = -(1)² + 2(1) + 3 = 4 Vertex: (1, 4) - this is the maximum!

Step 3: Axis of symmetry: x = 1

Step 4: Y-intercept: (0, 3)

Step 5: X-intercepts: -x² + 2x + 3 = 0 Multiply by -1: x² - 2x - 3 = 0 (x - 3)(x + 1) = 0 Points: (-1, 0) and (3, 0)

Step 6: Plot and draw downward parabola

Vertex Form

Vertex form: y = a(x - h)² + k

where (h, k) is the vertex.

This form makes graphing easier!

Converting from vertex form to standard form: Expand the squared binomial

Example: y = 2(x - 3)² + 1

Vertex: (3, 1) Opens up (a = 2 > 0) Narrower than y = x² (a = 2 > 1)

Expand to standard form: y = 2(x² - 6x + 9) + 1 y = 2x² - 12x + 18 + 1 y = 2x² - 12x + 19

Transformations from y = x²

y = x² + k: Vertical shift

  • k > 0: shift UP k units
  • k < 0: shift DOWN |k| units

y = (x - h)²: Horizontal shift

  • h > 0: shift RIGHT h units
  • h < 0: shift LEFT |h| units

y = ax²: Vertical stretch/compression

  • a > 1: vertical stretch (narrower)
  • 0 < a < 1: vertical compression (wider)
  • a < 0: reflection over x-axis

Example: y = (x - 2)² + 3

Starting from y = x²:

  • Shift right 2 units
  • Shift up 3 units
  • Vertex moves from (0,0) to (2,3)

Maximum and Minimum Values

If a > 0 (opens up):

  • Minimum value at vertex
  • No maximum value

If a < 0 (opens down):

  • Maximum value at vertex
  • No minimum value

Example: y = -2x² + 8x - 3

Vertex: x = -8/(2·(-2)) = 2 y = -2(2)² + 8(2) - 3 = -8 + 16 - 3 = 5

Since a < 0, maximum value is 5 (at x = 2)

Domain and Range

Domain: All quadratic functions have domain: all real numbers (-∞, ∞) or {x | x ∈ ℝ}

Range: Depends on vertex and direction

If a > 0 (opens up): Range: [k, ∞) where k is y-coordinate of vertex

If a < 0 (opens down): Range: (-∞, k] where k is y-coordinate of vertex

Example: y = x² - 4x + 3 Vertex: (2, -1), opens up Range: [-1, ∞)

Comparing Parabolas

Width: Determined by |a|

  • Larger |a|: narrower parabola
  • Smaller |a|: wider parabola

Example:

  • y = 3x² is narrower than y = x²
  • y = (1/2)x² is wider than y = x²
  • y = -x² and y = x² have same width (|a| = 1)

Using a Table of Values

When in doubt, make a table!

Example: y = x² + 2x - 3

xyPoint
-30(-3, 0)
-2-3(-2, -3)
-1-4(-1, -4)
0-3(0, -3)
10(1, 0)

Vertex appears to be at (-1, -4) Verify: x = -2/(2·1) = -1 ✓

Applications: Projectile Motion

Height of projectile: h(t) = -16t² + v₀t + h₀

The graph is a parabola opening down.

Example: Ball thrown at 48 ft/s from ground level h(t) = -16t² + 48t

Maximum height at vertex: t = -48/(2·(-16)) = 1.5 seconds h = -16(1.5)² + 48(1.5) = 36 feet

The ball reaches 36 feet at 1.5 seconds.

Applications: Business

Profit/Revenue functions often quadratic.

Example: Profit = -2x² + 40x - 50 where x is number of items (in hundreds)

Maximum profit at vertex: x = -40/(2·(-2)) = 10 P = -2(10)² + 40(10) - 50 = 150

Maximum profit is $15,000 (when x = 10 hundreds = 1000 items)

Applications: Area

Example: You have 100 feet of fence. What dimensions maximize the rectangular area?

Let x = width Then (100 - 2x)/2 = length = 50 - x

Area: A = x(50 - x) = -x² + 50x

Maximum at vertex: x = -50/(2·(-1)) = 25 feet Length = 25 feet

Maximum area: 625 square feet (a square!)

Sketching vs. Precise Graphing

Quick sketch needs:

  • Vertex
  • Direction (up/down)
  • Y-intercept
  • General shape

Precise graph needs:

  • Vertex
  • Axis of symmetry
  • Y-intercept
  • X-intercepts
  • Additional points
  • Scale marked clearly

Common Mistakes to Avoid

  1. Wrong direction Check sign of a carefully!

  2. Forgetting negative in vertex formula x = -b/(2a), not b/(2a)

  3. Not simplifying vertex coordinates Show actual numbers, not formulas

  4. Assuming parabola crosses x-axis Check discriminant!

  5. Plotting points but not drawing smooth curve Use a smooth U-shape, not line segments

  6. Forgetting axis of symmetry Use it to find additional points easily

Quick Reference

Standard Form: y = ax² + bx + c

Vertex: (-b/(2a), f(-b/(2a)))

Axis of Symmetry: x = -b/(2a)

Y-intercept: (0, c)

X-intercepts: Solve ax² + bx + c = 0

Direction:

  • a > 0: opens up
  • a < 0: opens down

Width:

  • |a| > 1: narrow
  • |a| < 1: wide

Practice Strategy

Level 1: Simple parabolas

  • y = x²
  • y = x² + 3
  • y = -x²

Level 2: Shifted parabolas

  • y = (x - 2)²
  • y = (x + 1)² - 4

Level 3: Standard form with factoring

  • y = x² - 4x + 3

Level 4: Need quadratic formula

  • y = x² + 4x + 1

Level 5: Applications

  • Projectile motion
  • Business optimization
  • Area problems

Tips for Success

  • Always find the vertex first
  • Use axis of symmetry to find symmetric points
  • Check if parabola opens up or down
  • Calculate discriminant to know about x-intercepts
  • Plot at least 5 points for accuracy
  • Draw smooth curves, not choppy lines
  • Label all key points clearly
  • Include arrows at ends of parabola
  • Practice vertex formula until automatic

📚 Practice Problems

1Problem 1easy

❓ Question:

Does the parabola y=−2x2+3x+1y = -2x^2 + 3x + 1 open upward or downward?

💡 Show Solution

Look at the coefficient of x2x^2:

a=−2a = -2

Since a<0a < 0 (negative), the parabola opens downward.

Answer: Downward

2Problem 2medium

❓ Question:

Find the vertex of y=x2−6x+5y = x^2 - 6x + 5

💡 Show Solution

Identify: a=1a = 1, b=−6b = -6, c=5c = 5

Step 1: Find the x-coordinate of the vertex x=−b2a=−−62(1)=62=3x = -\frac{b}{2a} = -\frac{-6}{2(1)} = \frac{6}{2} = 3

Step 2: Find the y-coordinate by substituting x=3x = 3 y=(3)2−6(3)+5y = (3)^2 - 6(3) + 5 y=9−18+5y = 9 - 18 + 5 y=−4y = -4

Answer: Vertex is at (3,−4)(3, -4)

3Problem 3medium

❓ Question:

What is the axis of symmetry for y=2x2+8x−3y = 2x^2 + 8x - 3?

💡 Show Solution

The axis of symmetry is the vertical line through the vertex.

Use: x=−b2ax = -\frac{b}{2a}

Identify: a=2a = 2, b=8b = 8

x=−82(2)=−84=−2x = -\frac{8}{2(2)} = -\frac{8}{4} = -2

Answer: x=−2x = -2

Explain using:

⚠️ Common Mistakes: Graphing Quadratic Functions

Avoid these 3 frequent errors

🌍 Real-World Applications: Graphing Quadratic Functions

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero

📌 Related Topics in Quadratic Equations

❓ Frequently Asked Questions

What is Graphing Quadratic Functions?▾
Parabolas, vertex, axis of symmetry
How can I study Graphing Quadratic Functions effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Graphing Quadratic Functions study guide free?▾
Yes — all study notes, flashcards, and practice problems for Graphing Quadratic Functions on Study Mondo are free to access. No account is needed.
What course covers Graphing Quadratic Functions?▾
Graphing Quadratic Functions is part of the Algebra 1 course on Study Mondo, specifically in the Quadratic Equations section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Graphing Quadratic Functions?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.