Homework Help: John

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

min/max point
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 10:49pm
You have the correct min.
John - Anonymous, Thursday, April 3, 2008 at 10:55pm
Isn'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 11:07pm
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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