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Mathematics 21 Online
OpenStudy (anonymous):

What are the coordinates of the vertex for f(x) = x2 + 4x + 10?

OpenStudy (sidsiddhartha):

try to write it as\[f(x)=(x-h)^2+k\]

OpenStudy (sidsiddhartha):

try to make a square first

OpenStudy (anonymous):

x^2 + 16x^2 + 100?

OpenStudy (sidsiddhartha):

no , we can write it as\[f(x)=x^2+2.x.2+2^2+6=(x+2)^2+6\] ok?

OpenStudy (anonymous):

not ok, lol I don't get it?

OpenStudy (sidsiddhartha):

i've just simplified it by making a square

OpenStudy (sidsiddhartha):

\[x^2+4x+10=(x^2+2.2.x+4)+6=(x+2)^2+6\\using,a^2+2ab+b^2=(a+b)^2\]

OpenStudy (sidsiddhartha):

so we've now \[f(x)=(x+2)^2+6\\comparing~with\\f(x)=(x-h)^2+k\\we'\ll ~have\\vertex=(h,k)=(-2,6)\]

OpenStudy (anonymous):

square rooting?

OpenStudy (sidsiddhartha):

no not square rooting , making a whole square because then we can compare with the standard equation

OpenStudy (anonymous):

so basically plugging in again?

OpenStudy (sidsiddhartha):

no just making it comparable to the standard equation|dw:1416065435103:dw|

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