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4 f(x) = ------------ 2 3 + 2*x - x

$$f{\left(x \right)} = \frac{4}{- x^{2} + 2 x + 3}$$

f = 4/(-x^2 + 2*x + 3)

The graph of the function

The domain of the function

The points at which the function is not precisely defined:

$$x_{1} = -1$$

$$x_{2} = 3$$

$$x_{1} = -1$$

$$x_{2} = 3$$

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:

$$\frac{4}{- x^{2} + 2 x + 3} = 0$$

Solve this equation

Solution is not found,

it's possible that the graph doesn't intersect the axis X

so we need to solve the equation:

$$\frac{4}{- x^{2} + 2 x + 3} = 0$$

Solve this equation

Solution is not found,

it's possible that the graph doesn't intersect the axis X

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

$$\frac{4 \cdot \left(2 x - 2\right)}{\left(- x^{2} + 2 x + 3\right)^{2}} = 0$$

Solve this equation

The roots of this equation

$$x_{1} = 1$$

The values of the extrema at the points:

**Intervals of increase and decrease of the function:**

Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:

Minima of the function at points:

$$x_{1} = 1$$

The function has no maxima

Decreasing at intervals

$$\left[1, \infty\right)$$

Increasing at intervals

$$\left(-\infty, 1\right]$$

$$\frac{d}{d x} f{\left(x \right)} = 0$$

(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

$$\frac{4 \cdot \left(2 x - 2\right)}{\left(- x^{2} + 2 x + 3\right)^{2}} = 0$$

Solve this equation

The roots of this equation

$$x_{1} = 1$$

The values of the extrema at the points:

(1, 1)

Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:

Minima of the function at points:

$$x_{1} = 1$$

The function has no maxima

Decreasing at intervals

$$\left[1, \infty\right)$$

Increasing at intervals

$$\left(-\infty, 1\right]$$

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

$$\frac{8 \cdot \left(\frac{4 \left(x - 1\right)^{2}}{- x^{2} + 2 x + 3} + 1\right)}{\left(- x^{2} + 2 x + 3\right)^{2}} = 0$$

Solve this equation

Solutions are not found,

maybe, the function has no inflections

$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$

(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

$$\frac{8 \cdot \left(\frac{4 \left(x - 1\right)^{2}}{- x^{2} + 2 x + 3} + 1\right)}{\left(- x^{2} + 2 x + 3\right)^{2}} = 0$$

Solve this equation

Solutions are not found,

maybe, the function has no inflections

Vertical asymptotes

Have:

$$x_{1} = -1$$

$$x_{2} = 3$$

$$x_{1} = -1$$

$$x_{2} = 3$$

Horizontal asymptotes

Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo

$$\lim_{x \to -\infty}\left(\frac{4}{- x^{2} + 2 x + 3}\right) = 0$$

Let's take the limit

so,

equation of the horizontal asymptote on the left:

$$y = 0$$

$$\lim_{x \to \infty}\left(\frac{4}{- x^{2} + 2 x + 3}\right) = 0$$

Let's take the limit

so,

equation of the horizontal asymptote on the right:

$$y = 0$$

$$\lim_{x \to -\infty}\left(\frac{4}{- x^{2} + 2 x + 3}\right) = 0$$

Let's take the limit

so,

equation of the horizontal asymptote on the left:

$$y = 0$$

$$\lim_{x \to \infty}\left(\frac{4}{- x^{2} + 2 x + 3}\right) = 0$$

Let's take the limit

so,

equation of the horizontal asymptote on the right:

$$y = 0$$

Inclined asymptotes

Inclined asymptote can be found by calculating the limit of 4/(3 + 2*x - x^2), divided by x at x->+oo and x ->-oo

$$\lim_{x \to -\infty}\left(\frac{4}{x \left(- x^{2} + 2 x + 3\right)}\right) = 0$$

Let's take the limit

so,

inclined coincides with the horizontal asymptote on the right

$$\lim_{x \to \infty}\left(\frac{4}{x \left(- x^{2} + 2 x + 3\right)}\right) = 0$$

Let's take the limit

so,

inclined coincides with the horizontal asymptote on the left

$$\lim_{x \to -\infty}\left(\frac{4}{x \left(- x^{2} + 2 x + 3\right)}\right) = 0$$

Let's take the limit

so,

inclined coincides with the horizontal asymptote on the right

$$\lim_{x \to \infty}\left(\frac{4}{x \left(- x^{2} + 2 x + 3\right)}\right) = 0$$

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:

$$\frac{4}{- x^{2} + 2 x + 3} = \frac{4}{- x^{2} - 2 x + 3}$$

- No

$$\frac{4}{- x^{2} + 2 x + 3} = - \frac{4}{- x^{2} - 2 x + 3}$$

- No

so, the function

not is

neither even, nor odd

So, check:

$$\frac{4}{- x^{2} + 2 x + 3} = \frac{4}{- x^{2} - 2 x + 3}$$

- No

$$\frac{4}{- x^{2} + 2 x + 3} = - \frac{4}{- x^{2} - 2 x + 3}$$

- No

so, the function

not is

neither even, nor odd

The graph