# f(x)= ((x^3)+(3x^2)+(3x)+1)/((x^2)+(2x)+0) is defined on the interval         [-16,17] A. The function f(x) has vertical asympototes at ? B. f(x) is concave down on the region

You need to remember that if you need to find the vertical asmptotes to the graph of rational function, you should look for the roots of denominator such that:

`x^2 + 2x = 0`

You need to factor out x such that:

`x(x+2) = 0`

`x = 0 or x + 2 = 0 =gt x = -2`

Hence, the vertical lines x = 0 and x = -2 are the vertical asymptotes of function.

b) You need to remember that the graph of function is concave down for f''(x)<0 such that:

`f'(x) = ((x^3+3x^2+3x+1)'*(x^2+2x) - (x^3+3x^2+3x+1)*(x^2+2x)')/((x^2+2x)^2)`

`f'(x) = ((3x^2 + 6x + 3)*(x^2+2x) - (x^3+3x^2+3x+1)*(2x+2))/((x^2+2x)^2)`

`f'(x) = (3x^4 + 6x^3 + 6x^3 + 12x^2 + 3x^2 + 6x - 2x^4 - 2x^3 - 6x^3 - 6x^2 - 6x^2 - 6x - 2x - 2)/((x^2+2x)^2)`

`f'(x) = (x^4 + 4x^3 + 3x^2 - 2x - 2)/((x^2+2x)^2)`

You need to evaluate the second derivative such that:

`f''(x) = ((4x^3 + 12x^2 + 6x - 2)(x^2+2x)^2 - 4(x^2+2x)(x^4 + 4x^3 + 3x^2 - 2x - 2)(x+1))/((x^2+2x)^4)`

`f''(x) = ((4x^3 + 12x^2 + 6x - 2)(x^2+2x) - 4(x^4 + 4x^3 + 3x^2 - 2x - 2)(x+1))/((x^2+2x)^3)`

`4x^5 + 8x^4 + 12x^4 + 24x^3 + 6x^3 + 12x^2 - 2x^2 - 4x - 4x^5 - 4x^4 - 16x^4 - 16x^3 -12x^3 - 12x^2 + 8x^2 + 8x + 8x + 8 = 0`

`24x^3 + 6x^3 + 12x^2 - 2x^2 - 4x - 16x^3 -12x^3 - 12x^2 + 8x^2 + 8x + 8x + 8 = 0`

`-4x^3 + 6x^2 + 12x + 8 = 0`

`2x^3 - 3x^2 - 6x - 4 = 0`

If `x in [-16,17] =gt f''(x)lt0` , hence the graph of the function is concave down.