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Parabolas, vertex, axis of symmetry
Learn step-by-step with practice exercises built right in.
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:
Standard Form: y = ax² + bx + c
What each coefficient tells us:
a (leading coefficient):
c (constant term):
Does the parabola open upward or downward?
Avoid these 3 frequent errors
See how this math is used in the real world
Solve .
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b (middle coefficient):
The simplest quadratic is y = x².
Key points:
Table of values:
| x | y = x² |
|---|---|
| -2 | 4 |
| -1 | 1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
The parabola is symmetric: same y-values for ±x
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)
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.
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)
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)
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
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
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)
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: 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
y = x² + k: Vertical shift
y = (x - h)²: Horizontal shift
y = ax²: Vertical stretch/compression
Example: y = (x - 2)² + 3
Starting from y = x²:
If a > 0 (opens up):
If a < 0 (opens down):
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: 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, ∞)
Width: Determined by |a|
Example:
When in doubt, make a table!
Example: y = x² + 2x - 3
| x | y | Point |
|---|---|---|
| -3 | 0 | (-3, 0) |
| -2 | -3 | (-2, -3) |
| -1 | -4 | (-1, -4) |
| 0 | -3 | (0, -3) |
| 1 | 0 | (1, 0) |
Vertex appears to be at (-1, -4) Verify: x = -2/(2·1) = -1 ✓
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.
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)
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!)
Quick sketch needs:
Precise graph needs:
Wrong direction Check sign of a carefully!
Forgetting negative in vertex formula x = -b/(2a), not b/(2a)
Not simplifying vertex coordinates Show actual numbers, not formulas
Assuming parabola crosses x-axis Check discriminant!
Plotting points but not drawing smooth curve Use a smooth U-shape, not line segments
Forgetting axis of symmetry Use it to find additional points easily
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:
Width:
Level 1: Simple parabolas
Level 2: Shifted parabolas
Level 3: Standard form with factoring
Level 4: Need quadratic formula
Level 5: Applications
Look at the coefficient of :
Since (negative), the parabola opens downward.
Answer: Downward
Find the vertex of
Identify: , ,
Step 1: Find the x-coordinate of the vertex
What is the axis of symmetry for ?
The axis of symmetry is the vertical line through the vertex.
Use:
Identify: ,
Step 2: Find the y-coordinate by substituting
Answer: Vertex is at
Answer: