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  • Graphing y =:
  • x^3+3x^2-9x+15
  • -x^2+6x-8
  • x^2-6x+10
  • x^2-2x-8

  • Identical expressions

  • sqrt(y^ two + one)/y
  • square root of (y squared plus 1) divide by y
  • square root of (y to the power of two plus one) divide by y
  • √(y^2+1)/y
  • sqrt(y2+1)/y
  • sqrty2+1/y
  • sqrt(y²+1)/y
  • sqrt(y to the power of 2+1)/y
  • sqrty^2+1/y
  • sqrt(y^2+1) divide by y

  • Similar expressions

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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
          ________
         /  2     
       \/  y  + 1 
f(y) = -----------
            y
f(y)=y2+1yf{\left(y \right)} = \frac{\sqrt{y^{2} + 1}}{y} 
f = sqrt(y^2 + 1)/y
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
y1=0y_{1} = 0
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis Y at f = 0
so we need to solve the equation:
y2+1y=0\frac{\sqrt{y^{2} + 1}}{y} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis Y
The points of intersection with the Y axis coordinate
The graph crosses Y axis when y equals 0:
substitute y = 0 to sqrt(y^2 + 1)/y.
02+10\frac{\sqrt{0^{2} + 1}}{0}
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
Extrema of the function
In order to find the extrema, we need to solve the equation
ddyf(y)=0\frac{d}{d y} f{\left(y \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddyf(y)=\frac{d}{d y} f{\left(y \right)} =
the first derivative
1y2+1−y2+1y2=0\frac{1}{\sqrt{y^{2} + 1}} - \frac{\sqrt{y^{2} + 1}}{y^{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
d2dy2f(y)=0\frac{d^{2}}{d y^{2}} f{\left(y \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dy2f(y)=\frac{d^{2}}{d y^{2}} f{\left(y \right)} =
the second derivative
−y2y2+1−1y2+1−2y2+1+2y2+1y2y=0\frac{- \frac{\frac{y^{2}}{y^{2} + 1} - 1}{\sqrt{y^{2} + 1}} - \frac{2}{\sqrt{y^{2} + 1}} + \frac{2 \sqrt{y^{2} + 1}}{y^{2}}}{y} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
y1=0y_{1} = 0
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at y->+oo and y->-oo
lim⁡y→−∞(y2+1y)=−1\lim_{y \to -\infty}\left(\frac{\sqrt{y^{2} + 1}}{y}\right) = -1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=−1y = -1
lim⁡y→∞(y2+1y)=1\lim_{y \to \infty}\left(\frac{\sqrt{y^{2} + 1}}{y}\right) = 1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=1y = 1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sqrt(y^2 + 1)/y, divided by y at y->+oo and y ->-oo
lim⁡y→−∞(y2+1y2)=0\lim_{y \to -\infty}\left(\frac{\sqrt{y^{2} + 1}}{y^{2}}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡y→∞(y2+1y2)=0\lim_{y \to \infty}\left(\frac{\sqrt{y^{2} + 1}}{y^{2}}\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(-y) и f = -f(-y).
So, check:
y2+1y=−y2+1y\frac{\sqrt{y^{2} + 1}}{y} = - \frac{\sqrt{y^{2} + 1}}{y}
- No
y2+1y=y2+1y\frac{\sqrt{y^{2} + 1}}{y} = \frac{\sqrt{y^{2} + 1}}{y}
- No
so, the function
not is
neither even, nor odd
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