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find the vertex of F(x)=2(x+10)^2+20

Sagot :

Since3122015
[tex]f(x)=2(x+10)^2+20\\f(x)=2(x^2+20x+100)+20\\f(x)=2x^2+40x+200+20\\f(x)=2x^2+40x+220\\f'(x)=4x+40=0\\4x=-40\\x=-10\\\\f(-10)=2((-10)+10)^2+20\\f(-10)=2(0)^2+20\\f(-10)=20\\(-10,20)[/tex]

The vertex of f(x) is at (-10, 20).
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