Question

Find range of: (a) y=x²+2x+4; 1≤x≤3 (b) y=x²+2x+4; -3≤x≤-2 (c) y=2x²+x+1; -1≤x≤1

Answer 1

(a) Range: [7, 19]

(b) Range: [4, 7]

(c) Range: [7/8, 4]

Answer 2

For each quadratic function, complete the square (or use vertex formula) to determine whether the vertex lies in the given domain. Then, evaluate the function at the endpoints and the vertex (if in range) to get the minimum and maximum values.

(a) y = x² + 2x + 4; 1 ≤ x ≤ 3

• Write in vertex form:
  y = (x² + 2x + 1) + 3 = (x + 1)² + 3
• Vertex is at x = –1, which is outside [1, 3].
• Evaluate endpoints:
  At x = 1: y = 1² + 2·1 + 4 = 7
  At x = 3: y = 9 + 6 + 4 = 19
• Range: [7, 19]

(b) y = x² + 2x + 4; –3 ≤ x ≤ –2

• Using the same vertex form as above: y = (x + 1)² + 3, vertex at x = –1 (outside domain).
• Evaluate endpoints:
  At x = –3: y = 9 – 6 + 4 = 7
  At x = –2: y = 4 – 4 + 4 = 4
• Range: [4, 7]
  (The smaller value is 4 at x = –2 and the larger is 7 at x = –3)

(c) y = 2x² + x + 1; –1 ≤ x ≤ 1

• Find vertex:
  Derivative: dy/dx = 4x + 1
  Set 4x + 1 = 0 ⟹ x = –¼ (lies in [–1, 1])
• Compute y at vertex:
  y(–¼) = 2(1/16) + (–¼) + 1 = 1/8 – 1/4 + 1
  = 1/8 – 2/8 + 8/8 = 7/8
• Evaluate endpoints:
  At x = –1: y = 2 + (–1) + 1 = 2
  At x = 1: y = 2 + 1 + 1 = 4
• Range: [7/8, 4]

1. Complete square or differentiate to find the vertex.
2. Check if the vertex is within the domain.
3. Evaluate the function at the endpoints and vertex (if applicable).
4. The lowest value gives the minimum and the highest value the maximum.