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Graphing y = (-2x^2+6x-3)/(x-1)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            2          
       - 2*x  + 6*x - 3
f(x) = ----------------
            x - 1      
$$f{\left(x \right)} = \frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1}$$
f = (-2*x^2 + 6*x - 3)/(x - 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
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:
$$\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = \frac{3}{2} - \frac{\sqrt{3}}{2}$$
$$x_{2} = \frac{\sqrt{3}}{2} + \frac{3}{2}$$
Numerical solution
$$x_{1} = 2.36602540378444$$
$$x_{2} = 0.633974596215561$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (-2*x^2 + 6*x - 3)/(x - 1).
$$\frac{-3 + \left(- 2 \cdot 0^{2} + 0 \cdot 6\right)}{-1}$$
The result:
$$f{\left(0 \right)} = 3$$
The point:
(0, 3)
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
$$\frac{6 - 4 x}{x - 1} - \frac{\left(- 2 x^{2} + 6 x\right) - 3}{\left(x - 1\right)^{2}} = 0$$
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
$$\frac{2 \left(-2 + \frac{2 \left(2 x - 3\right)}{x - 1} - \frac{2 x^{2} - 6 x + 3}{\left(x - 1\right)^{2}}\right)}{x - 1} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1}\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1}\right) = -\infty$$
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 (-2*x^2 + 6*x - 3)/(x - 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x \left(x - 1\right)}\right) = -2$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = - 2 x$$
$$\lim_{x \to \infty}\left(\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x \left(x - 1\right)}\right) = -2$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = - 2 x$$
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:
$$\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1} = \frac{- 2 x^{2} - 6 x - 3}{- x - 1}$$
- No
$$\frac{\left(- 2 x^{2} + 6 x\right) - 3}{x - 1} = - \frac{- 2 x^{2} - 6 x - 3}{- x - 1}$$
- No
so, the function
not is
neither even, nor odd