1. 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

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

3. 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)

4. 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)

5. 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 –
y = –4
(–3, –4)

6. 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)

7. 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.

8. 1) y = x2 – 6x + 8 a) x–intercept(s): x2 – 6x + 8 = 010 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)

9. 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 –
Since parabolas are symmetric, we can place the “mirror image” of the y–intercept here.
y = –1
(3, –1)

10. 2) y = x2 – 4x – 5 a) x–intercept(s): x2 – 4x – 5 = 010 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)

11. 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)

12. 3) y = x2 + 2x – 3 a) x–intercept(s): x2 + 2x – 3 = 010 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)

13. 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)

14. Finish the worksheet