Mister Exam

Other calculators

Graphing y = (x^2-4x+3)/(x-1)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src]
        2          
       x  - 4*x + 3
f(x) = ------------
          x - 1    
$$f{\left(x \right)} = \frac{\left(x^{2} - 4 x\right) + 3}{x - 1}$$
f = (x^2 - 4*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:
$$\frac{\left(x^{2} - 4 x\right) + 3}{x - 1} = 0$$
Solve this equation
The points of intersection with the axis X:

Analytical solution
$$x_{1} = 3$$
Numerical solution
$$x_{1} = 3$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^2 - 4*x + 3)/(x - 1).
$$\frac{\left(0^{2} - 0\right) + 3}{-1}$$
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
$$\frac{2 x - 4}{x - 1} - \frac{\left(x^{2} - 4 x\right) + 3}{\left(x - 1\right)^{2}} = 0$$
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
$$\frac{2 \left(- \frac{2 \left(x - 2\right)}{x - 1} + 1 + \frac{x^{2} - 4 x + 3}{\left(x - 1\right)^{2}}\right)}{x - 1} = 0$$
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
$$\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 4 x\right) + 3}{x - 1}\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\frac{\left(x^{2} - 4 x\right) + 3}{x - 1}\right) = \infty$$
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 (x^2 - 4*x + 3)/(x - 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 4 x\right) + 3}{x \left(x - 1\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x$$
$$\lim_{x \to \infty}\left(\frac{\left(x^{2} - 4 x\right) + 3}{x \left(x - 1\right)}\right) = 1$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x$$
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:
$$\frac{\left(x^{2} - 4 x\right) + 3}{x - 1} = \frac{x^{2} + 4 x + 3}{- x - 1}$$
- No
$$\frac{\left(x^{2} - 4 x\right) + 3}{x - 1} = - \frac{x^{2} + 4 x + 3}{- x - 1}$$
- No
so, the function
not is
neither even, nor odd