Question 258558
Lets take each one separately:
1. d(x)=(2x-5)(x+4)
set d(x) = 0 to get
(2x-5)(x+4) = 0
set each parenthesis = 0 and solve for x to get
(2x-5)=0
2x=5
x = 5/2
and then
x+4=0
x=-4
The minimum occurs at (-3/4, -21.125)
graph is
{{{ graph( 300, 200, -10, 10, -30, 10, (2x-5)(x+4)) }}}
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2. f(x)=x^2+5x-6
factoring, we get
(x+6)(x-1)
Set f(x) = 0 to get
(x+6)(x-1) = 0
Set each parenthesis = 0 and solve for x to get
(x+6)=0
x = -6
and then
x-1=0
x=1
The minimum occurs at (-5/2, 12.75)
graph is
{{{ graph( 300, 200, -10, 10, -20, 20, x^2+5x-6) }}}
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3. g(x)=x^2-12x
factoring, we get
(x)(x-12)
Set f(x) = 0 to get
(x)(x-12) = 0
Set each parenthesis = 0 and solve for x to get
x=0
and then
x-12=0
x=12
The minimum occurs at (6, -36)
graph is
{{{ graph( 300, 200, -10, 10, -40, 20, x^2-12x) }}}
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4. h(x)=-x^2+7x+18 
factoring, we get
-1(x+2)(x-9)
Set f(x) = 0 to get
(x+2)(x-9) = 0
Set each parenthesis = 0 and solve for x to get
x+2=0
x = -2
and then
x-9=0
x=9
The maximum occurs at (7/2, 30.25)
graph is
{{{ graph( 300, 200, -10, 10, -20, 40, -x^2+7x+18) }}}