Parabolas

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Write the equation of the *parabola* $x=y^{2}+2y+6$ 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.

**Answer**

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

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

$x=\left(y^{2}+2y+1\right)-1+6$

$x=\left(y+1\right)^2+5$

$x-5= \left(y+1\right)^2$

The *parabola* opens to the right.

The *vertex* is given by $V\left(5,-1\right)$

The *focus* is given by $F\left(\frac{21}{4},-1\right)$

The *directrix* is given by $x= \frac{19}{4}$

The *axis of symmetry* is given by $y=-1$