What is the vertex form of $y=x^2+2x+15$ ?

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Answer 1 · 2016-01-10T05:00:07

$y=(x+1)^2+14$

Given _

$y=x^2+2x+15$

The vertex form of the equation is -

$y=a(x-h)^2+k$

If we know the values of $a,h and k$ we can change the given equation into a vertex form.

Find the vertex $(h, k)$

$a$ is the coefficient of $x^2$
$h$ is the x-co-ordinate of the vertex
$k$ is the y-co-ordinate of the vertex

$a=1$
$h= (-b)/(2a)=(-2)/(2 xx 1)=-1$

$k=(-1)^2+2(-1)+15=1-2+15=14$

Now substitute the values of $a,h and k$ in the vertex form of the equation.

$y=(1)(x-(-1))^2+14$
$y=(x+1)^2+14$

What is the vertex form of y=x^2+2x+15y=x^2+2x+15?