What is the vertex form of $y=2x^2 + 13x + 6$ ?

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Answer 1 · 2016-02-08T09:51:43

$1/2(y-(-121/8))=(x--13/4)^2$

From the given
$y=2x^2+13x+6$

factor out 2 first

$y=2(x^2+(13x)/2)+6$

$y=2(x^2+(13x)/2+169/16-169/16)+6$

$y=2((x+13/4)^2-169/16)+6$

$y=2(x+13/4)^2-169/8+6$

$y=2(x+13/4)^2-169/8+48/8$

$y=2(x+13/4)^2-121/8$

$1/2(y-(-121/8))=(x--13/4)^2$

have a nice day from the Philippines!

What is the vertex form of y=2x^2 + 13x + 6y=2x^2 + 13x + 6?