Find the vertex of the parabola y=x^2+3x-(1)/(4) Simplify both coordinates and write them as proper fractions,improper fractions, or inte (square ,square )

To find the vertex of the parabola given by the equation $y=x^{2}+3x-\frac {1}{4}$, we can use the vertex formula for a parabola in the form $y=ax^{2}+bx+c$. The x-coordinate of the vertex is given by $x=-\frac{b}{2a}$, and the y-coordinate is found by substituting this x-value back into the original equation.<br /><br />In this case, $a=1$, $b=3$, and $c=-\frac{1}{4}$. Plugging these values into the vertex formula, we get:<br /><br />$x=-\frac{3}{2(1)}=-\frac{3}{2}$<br /><br />Now, substitute $x=-\frac{3}{2}$ back into the original equation to find the y-coordinate:<br /><br />$y=(-\frac{3}{2})^{2}+3(-\frac{3}{2})-\frac{1}{4}=\frac{9}{4}-\frac{9}{2}-\frac{1}{4}=\frac{9}{4}-\frac{18}{4}-\frac{1}{4}=-\frac{10}{4}=-\frac{5}{2}$<br /><br />Therefore, the vertex of the parabola is $\left(-\frac{3}{2}, -\frac{5}{2}\right)$.