Answer: 4) (-2, 6)
5) (-∞, -3) U (-3, ∞)
6) (-∞, -1) U (8/5, ∞)
Step-by-step explanation:
Find the zeros.
When the a-value is positive, the curve will be positive to the left of the leftmost zero and to the right of the rightmost zero. + - +
←---|----|--→
When the a-value is negative, the curve will be positive between the zeros.
- + -
←---|----|--→
4) y = -x² + 4x + 12
y = -(x² - 4x - 12)
y = -(x + 2)(x - 6)
0 = -(x + 2)(x - 6)
0 = x + 2 0 = x - 6 -- + --
x = -2 x = 6 ←------|-----------|--------→
-2 6
Positive Interval: (-2, 6)
5) y = 2x² + 12x + 18
y = 2(x² + 6x + 9)
y = 2(x + 3)(x + 3)
0 = 2(x + 3)(x + 3)
0 = x + 3 0 = x + 3 + +
x = -3 x = -3 ←---------|----------→
-3
Positive Interval: (-∞, -3) U (-3, ∞)
6) y = 5x² - 3x - 8
y = 5x² + 5x - 8x - 8
y = 5x(x + 1) - 8(x + 1)
y = (5x - 8)(x + 1)
0 = (5x - 8)(x + 1)
0 = 5x - 8 0 = x + 1 + -- +
x = 8/5 x = -1 ←------|-----------|--------→
-1 8/5
Positive Interval: (-∞, -1) U (8/5, ∞)
