Parabolas
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Question

Write the equation of the parabola $y=x^{2}+4x-6$ in the standard form. Then determine the vertex, focus, directrix and the axis of symmetry of the parabola. Also, state which way the parabola opens.

$y=x^{2}+4x-6$

$y=\left(x^{2}+4x\right)-6$

$y=\left(x^{2}+4x+4\right)-4-6$

$y=\left(x+2\right)^2-10$

$y+10= \left(x+2\right)^2$

The parabola opens up words.

The vertex is given by $V\left(-2,-10\right)$

The focus is given by $F\left(-2,-\frac{39}{4}\right)$

The directrix is given by $y= -\frac{41}{4}$

The axis of symmetry is given by $x=-2$