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# Find the vertex of parabola $x^2+4x+4-y=0$

$(a)\;(-2,0)\qquad(b)\;(-3,0)\qquad(c)\;(-4,0)\qquad(d)\;(-5,0)$

Can you answer this question?

$x^2+4x+4=y$
$x^2+2\times 2x+(2)^2=y$
$(x+2)^2=y$
Hence vertex $y=0,x+2\Rightarrow x=-2$
Vertex$(-2,0)$
Hence (a) is the correct answer.
answered Feb 5, 2014