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What is the value of ƒ(– 6) when ƒ(x) = x2 + 8x + 10?
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What is the value of ƒ(– 6) when ƒ(x) = x2 + 8x + 10?

User Bgh
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1 Answer

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f(x)= x^(2) +8x+10

f(-6) = (-6)^2+8(-6)+10

f(-6)=36-48+10 = -2

f(-6)=-2
User Phil Booth
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