What is the vertex form of $y=x^2+4x+16$ ?

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Answer 1 · 2016-01-09T15:58:56

$y = (x + 2 )^2 + 12$

The standard form of a quadratic equation is:

$y = ax^2 + bx + c$

The vertex form is : $y = (x - h )^2 + k$ where (h , k ) are the coordinates of the vertex.

For the given function $a = 1$ , $b = 4$ , and $c = 16$ .

The x-coordinate of the vertex (h) $= -b/(2a) = - 4/2 = - 2$

and the corresponding y-coordinate is found by substituting x =- 2 into the equation :

$rArr y = (- 2 )^2+4(- 2 ) + 16 = 4 - 8 + 16 = 12$

the coordinates of the vertex are (- 2 , 12 ) = (h , k )

the vertex form of $y = x^2 + 4x + 16$ is then :

$y = (x + 2 )^2 + 12$

check:

$(x + 2 )^2 + 12 = x^2 + 4x +16$

What is the vertex form of y=x^2+4x+16y=x^2+4x+16?