Graphical Representation of Quadratic Expressions
Key Definitions
- Parabola: The geometric shape represented by the graph of a quadratic function f(x) = ax² + bx + c. It is symmetric about a vertical line passing through its vertex.
- Vertex: The point where the parabola reaches its maximum (if a < 0) or minimum (if a > 0) value. The coordinates are (-b/2a, -D/4a).
- Axis of Symmetry: The vertical line x = -b/2a that divides the parabola into two congruent halves.
- Concavity: The orientation of the parabola’s opening. If a > 0, it is concave upwards; if a < 0, it is concave downwards.
Theory and Concepts
Consider y = ax² + bx + c. The graph’s behavior is dictated by a and D = b² – 4ac.
1. Sign of a (Concavity)
- a > 0: Opens upwards. Minimum value is -D/4a.
- a < 0: Opens downwards. Maximum value is -D/4a.
2. Sign of D (Roots and Intersections)
3. Important Conditions for JEE
- Always Positive: ax² + bx + c > 0 for all x ∈ ℝ if a > 0 and D < 0.
- Always Negative: ax² + bx + c < 0 for all x ∈ ℝ if a < 0 and D < 0.
Maximum and Minimum Values
For a quadratic function f(x) = ax² + bx + c:
- If a > 0, the function has a minimum value at x = -b/2a, and the minimum value is f(-b/2a) = (4ac – b²)/4a.
- If a < 0, the function has a maximum value at x = -b/2a, and the maximum value is f(-b/2a) = (4ac - b²)/4a.
Graphical Transformation
f(x) = a(x – h)² + k
Where h = -b/2a and k = f(h). This is the vertex form, showing horizontal shift h, vertical shift k, and stretch factor a.
Solved Examples
Solved Example 1.1: Range of k for positive expression
Problem: Find the range of k for which x² – 2kx + 7k – 12 is always positive for all real x.
Solution:
D = (-2k)² – 4(7k – 12) < 0
⇒ 4k² – 28k + 48 < 0
Divide by 4: k² – 7k + 12 < 0
⇒ (k – 3)(k – 4) < 0
Thus, k ∈ (3, 4).
Solved Example 1.2: Graph touching x-axis
Problem: If the graph of y = x² + (a-1)x + 1 touches the x-axis, find a.
Solution:
(a-1)² – 4(1)(1) = 0
⇒ (a-1)² = 4 ⇒ a-1 = ± 2
a = 3 or a = -1.
Solved Example 1.3: Finding the Vertex
Problem: Find the vertex of the parabola y = 2x² – 4x + 5.
Solution:
y_v = f(1) = 2(1)² – 4(1) + 5 = 3.
Vertex: (1, 3).
Solved Example 1.4: Determining sign of ‘a’
Problem: If a+b+c < 0, a-b+c > 0 and c > 0, find the sign of a.
Solution:
The parabola passes through (0, c) where c > 0 and drops below the x-axis at x = 1.
This indicates the parabola opens downwards.
Thus, a < 0.
Solved Example 1.5: Maximum Value
Problem: Find the maximum value of -3x² + 6x + 10.
Solution:
y_max = -156 / -12 = 13.
Maximum Value = 13.
Solved Example 1.6: Expression always below x-axis
Problem: For what values of m is y = (m-1)x² + 2mx + (m+3) always below the x-axis?
Solution:
D = 4m² – 4(m-1)(m+3) = -8m + 12 < 0 ⇒ m > 1.5.
Intersection of (m < 1) and (m > 1.5) is an empty set (φ).
Solved Example 1.7: Signs of a, b, c
Problem: Determine signs of a, b, c if the vertex is in the 4th quadrant and it opens upwards, cutting the positive y-axis.
Solution:
2. Vertex in 4th quad ⇒ x_v > 0 ⇒ -b/2a > 0 ⇒ b < 0.
3. Cuts positive y-axis ⇒ f(0) > 0 ⇒ c > 0.
Solved Example 1.8: Integral values of m
Problem: Find integral values of m for which x² + mx + 4 > 0 for all x ∈ ℝ.
Solution:
Integral values: {-3, -2, -1, 0, 1, 2, 3}.
Total = 7.
Solved Example 1.9: Vertex in terms of roots
Problem: What is the x-coordinate of the vertex in terms of roots α, β?
Solution:
x_v = ½(α + β).
Solved Example 1.10: Sketching the Graph
Problem: Sketch y = x² – 4x + 3.
Solution:
Roots: 1, 3. Vertex: (2, -1). y-intercept: 3.
Solved Example 1.11: Range of function
Problem: Find the range of f(x) = 2x² – 6x + 5 for real x.
Solution:
h = -b/2a = 6/4 = 3/2.
f(3/2) = 2(9/4) – 6(3/2) + 5 = 4.5 – 9 + 5 = 0.5.
Range: [1/2, ∞).
Solved Example 1.12: Graph above x-axis
Problem: For what values of k does the graph of f(x) = x² – 2(k+1)x + k² + 5 lie completely above the x-axis?
Solution:
D = [-2(k+1)]² – 4(1)(k² + 5) < 0
4(k² + 2k + 1) – 4k² – 20 < 0
8k – 16 < 0 ⇒ 8k < 16
k < 2.
Solved Example 1.13: Curve fitting
Problem: The graph of y = ax² + bx + c passes through (0, 3), (1, 4), and (2, 9). Find a, b, c.
Solution:
2. f(1)=4 ⇒ a + b + 3 = 4 ⇒ a + b = 1.
3. f(2)=9 ⇒ 4a + 2b + 3 = 9 ⇒ 2a + b = 3.
Solving: (2a+b) – (a+b) = 3 – 1 ⇒ a = 2, b = -1.
y = 2x² – x + 3.
Solved Example 1.14: Maximum value and location
Problem: Find the maximum value of f(x) = -2x² + 4x + 1 and the x at which it occurs.
Solution:
x = -b/2a = -4/-4 = 1.
f(1) = -2(1)² + 4(1) + 1 = 3.
Max value is 3 at x = 1.
Solved Example 1.15: Vertex form determination
Problem: If the graph of y = ax² + bx + c has vertex at (2, 3) and passes through (0, 7), find a, b, c.
Solution:
Passes through (0,7) ⇒ 7 = a(0-2)² + 3 ⇒ 4a = 4 ⇒ a = 1.
y = (x-2)² + 3 = x² – 4x + 7.
a = 1, b = -4, c = 7.
Solved Example 1.16: Graph above a line
Problem: Find the range of values of k for which the graph of f(x) = x² – 2kx + k² + k + 1 lies entirely above the line y = 2.
Solution:
x² – 2kx + k² + k – 1 > 0.
D < 0 ⇒ 4k² - 4(k² + k - 1) < 0
-4k + 4 < 0 ⇒ k > 1.
k ∈ (1, ∞).
Solved Example 1.17: Parabola touching x-axis
Problem: If the parabola y = ax² + bx + c touches the x-axis at (2, 0) and passes through (1, 2), find a, b, c.
Solution:
Passes through (1,2) ⇒ 2 = a(1-2)² ⇒ a = 2.
y = 2(x-2)² = 2x² – 8x + 8.
a = 2, b = -8, c = 8.
Solved Example 1.18: Negative intervals
Problem: Find the values of x for which f(x) = x² – 5x + 6 is negative.
Solution:
Since a > 0, the graph is negative between the roots.
x ∈ (2, 3).
Solved Example 1.19: Non-intersection
Problem: The graph of y = x² – 2ax + a² + a – 1 does not intersect the x-axis. Find the range of a.
Solution:
-4a + 4 < 0 ⇒ a > 1.
a ∈ (1, ∞).
Solved Example 1.20: Range of k for all positive x
Problem: Find the range of k for which kx² – 2x + 3k > 0 for all real x.
Solution:
2. D < 0 ⇒ 4 - 12k² < 0 ⇒ k² > 1/3.
Taking the positive root (since k > 0):
k > 1/√3.