Question

How do you graph `f(x)=(x^3-16x)/(-4x^2+4x+24)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

No hole, intercepts at `(-4,0)`, `(0,0)` and `(4,0)`, vertical asymptotes at `x=-2` and `x=3` and slanting asymptote at `y=-1/4x-1/4`.

Explanation:

As `f(x)=(x^3-16x)/(-4x^2+4x+24)`

= `(x(x^2-16))/(-4(x^2-x-6)`

= `(x(x-4)(x+4))/(-4(x-3)(x+2)`

As there is no common term in numerator and denominator that cancels out after factorization, there are no holes.

Observe that when `x->-2` or `x=3`, `f(x)->oo` if we approach these values from right and `f(x)->-oo` as we approach these values from left and so we have a vertical asymptotes at `x=-2` and `x=3`.

When `x=0` `f(x)=(x(x-4)(x+4))/(-4(x-3)(x+2)=0` and also when `x=+-4`, so we have a `x`-intercept as well as `y`-intercept at `(0,0)` and `x`-intercepts at `(-4,0)` and `4,0)`

In the function #f(x), we degree of numerator is one more than that of denominator, so we do not have horizontal asymptote, but we do have a slanting asymptote.

As `f(x)=(x^3-16x)/(-4x^2+4x+24)`

= `-1/4x+(-10x+x^2)/(-4x^2+4x+24)`

= `-1/4x+(-10/x+1)/(-4+4/x+24/x^2)`

when `x->+-oo` `f(x)->-1/4x-1/4` and hence we have a slanting asymptote at `y=-1/4x-1/4`.

The graph using these points and trends appears as follows:

graph{(y-(x^3-16x)/(-4x^2+4x+24))(4y+x+1)=0 [-10, 10, -5, 5]}