Verified answer

1)

y ≤ x^2 − 5x − 6

x = -b/(2a)

x = - -5/(2 * 1)

x = 5/2

y = (5/2)^2 - 5 * (5/2) - 6

y = -49/4 <-----> vertex at ( 5/2 , -49/4 )

zeros:

0 = x^2 − 5x − 6

0 = (x - 6)(x + 1) ----> x = -1 & 6

you can start graphing using these three points.

Shading: Testing (0 , 0)

0 ≤ 0^2 - 5 * 0 - 6

0 ≤ - 6 <----- false

shading will be in the region where it doesn't have the (0 , 0)

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2)

x^2 + 6x + 9 ≤ 0

(x + 3)(x + 3) ≤ 0

(x + 3)^2 ≤ 0

we need to check on a number line numbers before and after the -3 into the original inequality.

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. . . . . . .-3. . . . . .

x = -3 is the only solution.

{ -3 }

Set-Builder Notation:

{ x I x = -3 }

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3)

0 ≤ 5x^2 − 20x + 20

0 ≤ x^2 − 4x + 4

0 ≤ (x - 2)(x - 2)

0 ≤ (x - 2)^2

TTTTTTTTTTTTTTTTTT

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. . . . . . .2. . . . . .

Interval Notation:

( -∞ , ∞ )

Set-Builder Notation:

{ x I x is all real numbers }

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4)

0 > x^2 + 7x + 12

0 > (x + 3)(x + 4)

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. . . .-4 . . . . -3 . . . . .

Interval Notation:

( -4 , -3 )

Set-Builder Notation:

{ x I -4 < x < -3 }

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5)

x^2 + 2x − 5 ≥ 0

x = [ -b + /- √( b^2 - 4 * a * c ) ] / [ 2 * a ]

x = [ -2 + /- √( 2^2 - 4 * 1 * -5 ) ] / [ 2 * 1 ]

x = [ -2 + /- √( 4 + 20 ) ] / [ 2 ]

x = [ -2 + /- √(24) ] / [ 2 ]

x = [ -2 + /- 2√(6) ] / [ 2 ]

x = -1 + /- √(6)

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. . . .-1 - √(6) . . . . -1 + √(6) . . . . .

Interval Notation:

( -∞ , -1 - √(6) ] U [ -1 + √(6) , ∞ )

Set-Builder Notation:

{ x I x ≤ -1 - √(6) or x ≥ -1 + √(6) }

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