Find the vertex of the parabola y=x^2+8x+(44)/(5) Simplify both coordinates and write them as proper fractions,improper fractions, or integer:

To find the vertex of the parabola given by the equation $y=x^{2}+8x+\frac {44}{5}$, we can use the vertex formula for a parabola in the form $y=ax^{2}+bx+c$, which is given by:<br /><br />$x_{vertex} = -\frac{b}{2a}$<br /><br />$y_{vertex} = c - \frac{b^2}{4a}$<br /><br />In this case, $a=1$, $b=8$, and $c=\frac{44}{5}$.<br /><br />First, let's find the x-coordinate of the vertex:<br /><br />$x_{vertex} = -\frac{8}{2(1)} = -4$<br /><br />Now, let's find the y-coordinate of the vertex:<br /><br />$y_{vertex} = \frac{44}{5} - \frac{8^2}{4(1)} = \frac{44}{5} - \frac{64}{4} = \frac{44}{5} - 16 = \frac{44 - 80}{5} = -\frac{36}{5}$<br /><br />Therefore, the vertex of the parabola is at the point $(-4, -\frac{36}{5})$.