Question

What are the coordinates of the vertex?What is the equation of the axis of symmetry?What is/are the x-intercept(s)?What is the y-intercept?

Answer 1

We will find the vertex of the parabola as follows:

*Given a quadratic function of the form:

$y=ax^2+bx+c$

We will have that the vertex is given by:

$(-\frac{b}{2a},\frac{b^2-4ac}{4a})$

So, for the function given we will have the following vertex:

$(-\frac{(-8)}{2(-1)},\frac{(-8)^2-4(-1)(0)}{4(-1)})\to(-4,16)$

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:

$x=-4$

So, the axis of symmetry is x = -4.

We will determine the x-intercepts as follows:

$-x^2-8x=0\Rightarrow x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(-1)(0)}}{2(-1)}$$\Rightarrow\begin{cases}x=-8 \\ \\ x=0\end{cases}$

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.