Posted by Algebra on Thursday, April 3, 2008 at 10:45pm.
min/max point
minimum or maximum
f(x)=2x^2+8x2
a=2
x=(b)/(2a)=(8)/(2)(2)=4
x=4
y=2(4)^2+8(4)2
2(16)322
32322
y=2
(4,2)
axis of symmetry is x=4
range[2,oo]
minimum value 2

John  bobpursley, Thursday, April 3, 2008 at 10:49pm
You have the correct min.

John  Anonymous, Thursday, April 3, 2008 at 10:55pm
Isn't this the same question as this one?
www.jiskha.com/display.cgi?id=1207258851
Err exactly the same. The min is wrong >.<. The minimum value occurs at x = 2 but the minimum value is not 2.

John  Reiny, Thursday, April 3, 2008 at 11:07pm
Your parabola in standard form is
y = 2(x+2)^2  10 , (check it by expanding this)
so you have a parabola opening upwards with vertex at (2,10)
Minimum value is 10, when x=2
axis of symmetry is x=2
domain: any real number for x
range: y ≥ 10, y any real number
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