Question 458017: Please help I have two problems: find the vertex,line of symmetry and graph the function:
f(x)=-2x^2+2x+10
f(x)=2-x^2
Thank you
Gina
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! f(x)=-2x^2+2x+10
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The vertex occurs where x = -b/(2a) = -2/(2*-2) = 1/2
f(1/2) = -2(1/2)^2+2(1/2)+10 = -1/2 + 1 + 10 = 10 1/2 = 21/2
Vertex:::(1/2,21/2)
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f(x)=2-x^2
x = -b/(2a)
x = 0/(2*-2) = 0
f(0) = 2
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Vertex: (0,2)
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Cheers,
Stan H.
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