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What is the vertex of f(x) = x^2 + 10x + 16?

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[tex]f(x)=x^2+10x+16\\\\V(-\frac{b}{2a},-\frac{\Delta}{4a})-vertex\\\\-\frac{b}{2a}=-\frac{10}{2\cdot1}=-5\\\\\Delta=b^2-4ac\\\Delta=10^2-4\cdot1\cdot16=100-64=36\\\\-\frac{\Delta}{4a}=-\frac{36}{4\cdot1}=-9\\\\V(-5,-9)[/tex]
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