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  • How to use it?

  • Graphing y =:
  • x^5+4x
  • (x+5)^2(x-1)+7
  • x/(4-x^2)
  • -x^4+x^2+5

  • Identical expressions

  • sqrt(x+ three)+ one /(x^ two)
  • square root of (x plus 3) plus 1 divide by (x squared )
  • square root of (x plus three) plus one divide by (x to the power of two)
  • √(x+3)+1/(x^2)
  • sqrt(x+3)+1/(x2)
  • sqrtx+3+1/x2
  • sqrt(x+3)+1/(x²)
  • sqrt(x+3)+1/(x to the power of 2)
  • sqrtx+3+1/x^2
  • sqrt(x+3)+1 divide by (x^2)

  • Similar expressions

  • sqrt(x-3)+1/(x^2)
  • sqrt(x+3)-1/(x^2)
Plotting a function graph/ sqrt(x+3)+1/(x^2)

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
         _______   1 
f(x) = \/ x + 3  + --
                    2
                   x
$$f{\left(x \right)} = \sqrt{x + 3} + \frac{1}{x^{2}}$$ 
f = sqrt(x + 3) + 1/(x^2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 0$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to sqrt(x + 3) + 1/(x^2).
$$\frac{1}{0^{2}} + \sqrt{3}$$
The result:
$$f{\left(0 \right)} = \tilde{\infty}$$
sof doesn't intersect Y
Vertical asymptotes
Have:
$$x_{1} = 0$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\sqrt{x + 3} + \frac{1}{x^{2}}\right) = \infty i$$
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
$$\lim_{x \to \infty}\left(\sqrt{x + 3} + \frac{1}{x^{2}}\right) = \infty$$
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 sqrt(x + 3) + 1/(x^2), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x + 3} + \frac{1}{x^{2}}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sqrt{x + 3} + \frac{1}{x^{2}}}{x}\right) = 0$$
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:
$$\sqrt{x + 3} + \frac{1}{x^{2}} = \sqrt{3 - x} + \frac{1}{x^{2}}$$
- No
$$\sqrt{x + 3} + \frac{1}{x^{2}} = - \sqrt{3 - x} - \frac{1}{x^{2}}$$
- No
so, the function
not is
neither even, nor odd
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