Have out: U1D6 Bellwork: a) x2 + 10x + 24 b) x3 – 4x2 – 45x Yesterday’s assignment, pencil, red pen, highlighter, calculator U1D6 Bellwork: Factor completely. Look out for GCFs! total: a) x2 + 10x + 24 b) x3 – 4x2 – 45x 24 Product –45 Product +2 x(x2 – 4x – 45) 4 6 (x + 4)(x + 6) –9 5 x(x + 5)(x – 9) 10 Sum –4 Sum +3 c) 3x2 + 7x + 2 d) 2x2 – 3x – 5 x + 2 2x – 5 (3) (2) (2) (–5) 6 –10 3x 3x2 + 6x x 2x2 –5x 6 1 –5 2 + 1 + x + 2 + 1 + 2x –5 7 –3 (3x + 1)(x + 2) (2x – 5)(x + 1) +2 +2

Investigate the equation y = x2 + 6x + 5. a) Complete the table for -7 ≤ x ≤ 1. x y –7 –6 –5 –4 –3 –2 –1 1 b) Graph the parabola and line of symmetry. 12 x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = –3 5 y = x2 + 6x + 5 –3 –4 –3 5 12

b) Algebraically verify the x–intercepts. Steps: Set y = ___ Factor _________ Use the _________________ Write as ______ x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = –3 completely y = x2 + 6x + 5 Zero Product Property (x, 0) x2 + 6x + 5 = 0 (x + 5)(x + 1) = 0 x + 5 = 0 or x + 1 = 0 x = –5 x = –1 (–5, 0) (–1, 0)

c) Algebraically solve for the line of symmetry. Steps: Average the ___________ x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = –3 x–intercepts y = x2 + 6x + 5 (–5, 0) (–1, 0)

d) Verify the vertex by substituting the line of symmetry into the equation. Write the answer as (x, y). x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = –3 Line of Symmetry: x = –3 y = x2 + 6x + 5 y = x2 + 6x + 5 y = (–3)2 + 6(–3) + 5 y = 9 – 18 + 5 y = –4 (–3, –4)

e) Algebraically verify the y–intercept. x y 10 2 4 6 8 –10 –2 –4 –6 –8 Steps:  Set x = ___  Solve for y.  Write as _______ x = –3 y = x2 + 6x + 5 (0, y) y = x2 + 6x + 5 y = (0)2 + 6(0) + 5 y = 5 (0, 5)

Practice: For each parabola, algebraically solve for the x–intercept(s), y–intercept, line of symmetry, and vertex. Then graph the parabola and line of symmetry without making a table.

1) y = x2 – 6x + 8 a) x–intercept(s): x2 – 6x + 8 = 0 10 2 4 6 8 –10 –2 –4 –6 –8 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x – 2 = 0 or x – 4 = 0 x = 2 x = 4 (2, 0) (4, 0) b) y–intercept: y = x2 – 6x + 8 y = (0)2 – 6(0) + 8 y = 8 (0, 8)

1) y = x2 – 6x + 8 c) line of symmetry: (2, 0) (4, 0) d) vertex: 10 2 4 6 8 –10 –2 –4 –6 –8 x = 3 (2, 0) (4, 0) y = x2 – 6x + 8 d) vertex: y = x2 – 6x + 8 y = (3)2 – 6(3) + 8 y = 9 – 18 + 8 Since parabolas are symmetric, we can place the “mirror image” of the y–intercept here. y = –1 (3, –1)

2) y = x2 – 4x – 5 a) x–intercept(s): x2 – 4x – 5 = 0 10 2 4 6 8 –10 –2 –4 –6 –8 x2 – 4x – 5 = 0 (x + 1)(x – 5) = 0 x + 1 = 0 or x – 5 = 0 x = –1 x = 5 (–1, 0) (5, 0) b) y–intercept: y = x2 – 4x – 5 y = (0)2 – 5(0) – 5 y = –5 (0, –5)

y – intercept “mirror image” y = –9 (2, –9) 2) y = x2 – 4x – 5 c) line of symmetry: x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = 2 (–1, 0) (5, 0) y = x2 – 4x – 5 d) vertex: y = x2 – 4x – 5 y = (2)2 – 4(2) – 5 y = 4 – 8 – 5 y – intercept “mirror image” y = –9 (2, –9)

3) y = x2 + 2x – 3 a) x–intercept(s): x2 + 2x – 3 = 0 10 2 4 6 8 –10 –2 –4 –6 –8 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 or x – 1 = 0 x = –3 x = 1 (–3, 0) (1, 0) b) y–intercept: y = x2 + 2x – 3 y = (0)2 + 2(0) – 3 y = –3 (0, –3)

y – intercept “mirror image” y = –4 (–1, –4) 3) y = x2 + 2x – 3 c) line of symmetry: x y 10 2 4 6 8 –10 –2 –4 –6 –8 x = –1 (–3, 0) (1, 0) f(x) = x2 + 2x – 3 d) vertex: y = x2 + 2x – 3 y = (–1)2 + 2(–1) – 3 y = 1 – 2 – 3 y – intercept “mirror image” y = –4 (–1, –4)

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