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Write the vertex form of the following equation. y= x^2+6x-5

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Answer:


\boxed{\bold{y=(x+3)^2-14}}

Explanation:

y= x²+6x-5

completing the square: a²+2ab+b² = (a+b)² where a²=x² (so a=x) and 2ab=6x (so b would be 3):


y=x^2+6x-5\\\\y=\underline{x^2+2\cdot x\cdot3+3^2}-3^2-5\\\\y=(x+3)^2-9-5\\\\\underline{y=(x+3)^2-14}\\\\\\vertex:\ \ (-3,\,-14)

User Jesufer Vn
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