What is the the vertex of $x = {(y + 6)}^{2} - 3$ $x = (y + 6)^2 - 3$ ?

Question

1360 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-26T12:18:44

The vertex is $(- 3, - 6)$ $(-3,-6)$ .

Expand the parabola:

${(y + 6)}^{2} - 3 = y^{2} + 12 y + 36 - 3 = y^{2} + 12 y + 33$ $(y+6)^2-3 = y^2+12y+36-3=y^2+12y+33$

The vertex is the minimum of a parabola, so we can derive it and set the derivative to zero:

$2 y + 12 = 0 \Leftrightarrow y = - 6$ $2y+12=0 \iff y=-6$ .

So, the vertex has $y$ $y$ -coordinate $- 6$ $-6$ . To find the $x$ $x$ -coordinate, simply compute

$f (- 6) = (- 6 + 6) - 3 = - 3$ $f(-6)=(-6+6)-3=-3$

What is the the vertex of x=(y+6)2−3x = (y + 6)^2 - 3 ?