Parabolas

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Write the equation of the *parabola* $x=y^{2}-4y+5$ 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**

$x=y^{2}-4y+5$

$x=\left(y^{2}-4y\right)+5$

$x=\left(y^{2}-4y+4\right)-4+5$

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

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

The *parabola* opens to the right.

The *vertex* is given by $V\left(1,2\right)$

The *focus* is given by $F\left(\frac{5}{4},2\right)$

The *directrix* is given by $x= \frac{3}{4}$

The *axis of symmetry* is given by $y=2$