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Graphing y = (x^2-16)/(x^2-4x-32)

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The graph:

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Intersection points:

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The solution

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           2        
          x  - 16   
f(x) = -------------
        2           
       x  - 4*x - 32
$$f{\left(x \right)} = \frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32}$$
f = (x^2 - 16)/(x^2 - 4*x - 32)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -4$$
$$x_{2} = 8$$
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 4$$
Numerical solution
$$x_{1} = 4$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^2 - 16)/(x^2 - 4*x - 32).
$$\frac{-16 + 0^{2}}{-32 + \left(0^{2} - 0\right)}$$
The result:
$$f{\left(0 \right)} = \frac{1}{2}$$
The point:
(0, 1/2)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\frac{2 x}{\left(x^{2} - 4 x\right) - 32} + \frac{\left(4 - 2 x\right) \left(x^{2} - 16\right)}{\left(\left(x^{2} - 4 x\right) - 32\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$- \frac{2 \left(\frac{4 x \left(x - 2\right)}{- x^{2} + 4 x + 32} + \frac{\left(x^{2} - 16\right) \left(\frac{4 \left(x - 2\right)^{2}}{- x^{2} + 4 x + 32} + 1\right)}{- x^{2} + 4 x + 32} + 1\right)}{- x^{2} + 4 x + 32} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = -4$$
$$x_{2} = 8$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x^2 - 16)/(x^2 - 4*x - 32), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{x^{2} - 16}{x \left(\left(x^{2} - 4 x\right) - 32\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{x^{2} - 16}{x \left(\left(x^{2} - 4 x\right) - 32\right)}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32} = \frac{x^{2} - 16}{x^{2} + 4 x - 32}$$
- No
$$\frac{x^{2} - 16}{\left(x^{2} - 4 x\right) - 32} = - \frac{x^{2} - 16}{x^{2} + 4 x - 32}$$
- No
so, the function
not is
neither even, nor odd