Respuesta :
Given the general quadratic expression:
[tex]y=ax^2+bx+c[/tex]we can find the x-coordinate of the vertex using the following formula:
[tex]x=-\frac{b}{2a}[/tex]Then, in this case we have the following:
[tex]\begin{gathered} y=x^2-5x-10 \\ a=1 \\ b=-5 \\ c=-10 \\ \Rightarrow x=-\frac{(-5)}{2(1)}=\frac{5}{2} \\ x=\frac{5}{2} \end{gathered}[/tex]therefore, the x-coordinate of the vertex is x=5/2. Now to find the y-coordinate, we have to evaluate x=5/2 on the original equation:
[tex]\begin{gathered} y=x^2-5x-10 \\ x=\frac{5}{2} \\ \Rightarrow y=(\frac{5}{2})^2-5(\frac{5}{2})-10=\frac{25}{4}-\frac{25}{2}-10 \\ =\frac{25-50-40}{4}=-\frac{65}{4} \\ y=-\frac{65}{4} \end{gathered}[/tex]finally, we have that the coordinates of the vertex are (5/2,-65/4)
