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February 1, 2015

February 1, 2015

Posted by **Algebra** on Thursday, April 3, 2008 at 10:45pm.

minimum or maximum

f(x)=2x^2+8x-2

a=2

x=(-b)/(2a)=(-8)/(2)(2)=-4

x=-4

y=2(-4)^2+8(-4)-2

2(16)-32-2

32-32-2

y=-2

(-4,-2)

axis of symmetry is x=-4

range[-2,oo]

minimum value -2

- John -
**bobpursley**, Thursday, April 3, 2008 at 10:49pmYou have the correct min.

- John -
**Anonymous**, Thursday, April 3, 2008 at 10:55pmIsn't this the same question as this one?

w-ww-.-jiskha.-c-om/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 11:07pmYour 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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