# sketch the graph of the curve `f(x)=x-1/x^2-4`   find  symmetries intercepts asymptotes critical points points of inflexion

lfryerda | High School Teacher | (Level 2) Educator

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There is no y-intercept, since we have a vertical asymptote at x=0.  To find the x-intercepts, let y=0 and solve:

`x-1/x^2-4=0`   multiply by `x^2`

`x^3-4x^2-1=0`   from a graphing calculator, we see that there is a single real root at `x approx 4.061` .

To find any symmetries, we let `x-> -x` .  Then

`f(-x)=-x-1/x^2-4`  which is not equal to `f(x)` or `-f(x)` .  This is not an even or odd function.

The critical points are found by taking the derivative using the power rule:

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

`={x^3+2}/x^3`   set equal to zero and solve

`{x^3+2}/x^3=0`   multiply by `x^3`

`x^3+2=0`   rearrange

`x^3=-2`

`x=-2^{1/3} approx -1.260`

There is a single critical point at `x=-2^{1/3} approx -1.260` .

The second derivative is (again using the power rule)

`f''(x)=-6/x^4`   which is always negative and never zero

This means the critical point is a local maximum and there are no inflection points.

There is a vertical asymptote at x=0 and there is an oblique asymptote at `y=x-4` .  The oblique asymptote is determined from the initial function.

This can be combined into the following graph: