We will find the vertex of the parabola as follows:
*Given a quadratic function of the form:
[tex]y=ax^2+bx+c[/tex]
We will have that the vertex is given by:
[tex](-\frac{b}{2a},\frac{b^2-4ac}{4a})[/tex]
So, for the function given we will have the following vertex:
[tex](-\frac{(-8)}{2(-1)},\frac{(-8)^2-4(-1)(0)}{4(-1)})\to(-4,16)[/tex]
So, the vertex for the function is given by the ordered pair (-4, 16).
We will have that the equation of the axis of symmetry is given by:
[tex]x=-4[/tex]
So, the axis of symmetry is x = -4.
We will determine the x-intercepts as follows:
[tex]-x^2-8x=0\Rightarrow x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(-1)(0)}}{2(-1)}[/tex][tex]\Rightarrow\begin{cases}x=-8 \\ \\ x=0\end{cases}[/tex]
So, we will have that the x-intercepts are located at x = -8 & x = 0.
From the previous point we can see that the point (0, 0) belongs to the parabola, thus the y-intercept is y = 0.
