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Plotting a function graph/ (6e^(x-2)-3x^2+4x+2(x-2))+16(3cos^2(x-2)sin^2(x-2)+cos^4(x-2))/(cos^6(x-2))-2

Graphing y = (6e^(x-2)-3x^2+4x+2(x-2))+16(3cos^2(x-2)sin^2(x-2)+cos^4(x-2))/(cos^6(x-2))-2

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
                                              /     2           2             4       \    
          x - 2      2                     16*\3*cos (x - 2)*sin (x - 2) + cos (x - 2)/    
f(x) = 6*E      - 3*x  + 4*x + 2*(x - 2) + -------------------------------------------- - 2
                                                              6                            
                                                           cos (x - 2)
$$f{\left(x \right)} = \left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2$$ 
f = (16*(sin(x - 2)^2*(3*cos(x - 2)^2) + cos(x - 2)^4))/cos(x - 2)^6 + 2*(x - 2) + 4*x + 6*E^(x - 2) - 3*x^2 - 2
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 3.5707963267949$$
$$x_{2} = 6.71238898038469$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 6*E^(x - 2) - 3*x^2 + 4*x + 2*(x - 2) + (16*((3*cos(x - 2)^2)*sin(x - 2)^2 + cos(x - 2)^4))/cos(x - 2)^6 - 2.
$$-2 + \left(\left(\left(-2\right) 2 + \left(0 \cdot 4 + \left(- 3 \cdot 0^{2} + \frac{6}{e^{2}}\right)\right)\right) + \frac{16 \left(\cos^{4}{\left(-2 \right)} + \sin^{2}{\left(-2 \right)} 3 \cos^{2}{\left(-2 \right)}\right)}{\cos^{6}{\left(-2 \right)}}\right)$$
The result:
$$f{\left(0 \right)} = -6 + \frac{6}{e^{2}} + \frac{16 \cos^{4}{\left(2 \right)} + 48 \sin^{2}{\left(2 \right)} \cos^{2}{\left(2 \right)}}{\cos^{6}{\left(2 \right)}}$$
The point:
(0, -6 + 6*exp(-2) + (16*cos(2)^4 + 48*cos(2)^2*sin(2)^2)/cos(2)^6)
Vertical asymptotes
Have:
$$x_{1} = 3.5707963267949$$
$$x_{2} = 6.71238898038469$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
True

Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty}\left(\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2\right)$$
$$\lim_{x \to \infty}\left(\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 6*E^(x - 2) - 3*x^2 + 4*x + 2*(x - 2) + (16*((3*cos(x - 2)^2)*sin(x - 2)^2 + cos(x - 2)^4))/cos(x - 2)^6 - 2, divided by x at x->+oo and x ->-oo
True

Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2}{x}\right)$$
$$\lim_{x \to \infty}\left(\frac{\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesn’t exist
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:
$$\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2 = - 3 x^{2} - 6 x + \frac{48 \sin^{2}{\left(x + 2 \right)} \cos^{2}{\left(x + 2 \right)} + 16 \cos^{4}{\left(x + 2 \right)}}{\cos^{6}{\left(x + 2 \right)}} + 6 e^{- x - 2} - 6$$
- No
$$\left(\frac{16 \left(\sin^{2}{\left(x - 2 \right)} 3 \cos^{2}{\left(x - 2 \right)} + \cos^{4}{\left(x - 2 \right)}\right)}{\cos^{6}{\left(x - 2 \right)}} + \left(2 \left(x - 2\right) + \left(4 x + \left(6 e^{x - 2} - 3 x^{2}\right)\right)\right)\right) - 2 = 3 x^{2} + 6 x - \frac{48 \sin^{2}{\left(x + 2 \right)} \cos^{2}{\left(x + 2 \right)} + 16 \cos^{4}{\left(x + 2 \right)}}{\cos^{6}{\left(x + 2 \right)}} - 6 e^{- x - 2} + 6$$
- No
so, the function
not is
neither even, nor odd
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