Question

y=-x^2-4x-8Identify the vertex, the axis of symmetry, the maximum or minimum value, and the range of the parabola.

Answer 1

Here we have the following parabola:

`y=-x^2-4x-8`

To find the vertex, we could use the following formula:

`V(x,y)=V((-b)/(2a),(-b^2)/(4a)+c)`

Where a, b and c are the coefficients of the quadratic function:

`y=ax^2+bx+c`

As you can see, in this problem a = -1 , b = -4 and c = -8. Thus,

`V(x,y)=V((-(-4))/(2(-1)),(-(-4)^2)/(4(-1))-8)`

This is:

`V(-2,-4)`

Then, the vertex of the parabola is (-2,-4)

The axis of symmetry of the parabola is the line x=-2. Since the vertex is situated at the coordinates (-2,-4), that means that the parabola is symmetrical around this line.

The vertex is maximum point of the parabola.

The range, is defined as all the values that the y-axis could take. If we notice, that is:

`(-\infty,-4\rbrack`

I'm going to upload a picture of the parabola: