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:

`(-(b)/(2a),(b^2-4ac)/(4a))`

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

`(-((-8))/(2(-1)),((-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.