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Graphing y = (x^2+6x+3)/(x+4)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        2          
       x  + 6*x + 3
f(x) = ------------
          x + 4    
f(x)=(x2+6x)+3x+4f{\left(x \right)} = \frac{\left(x^{2} + 6 x\right) + 3}{x + 4}
f = (x^2 + 6*x + 3)/(x + 4)
The graph of the function
02468-8-6-4-2-1010-250250
The domain of the function
The points at which the function is not precisely defined:
x1=4x_{1} = -4
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:
(x2+6x)+3x+4=0\frac{\left(x^{2} + 6 x\right) + 3}{x + 4} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=36x_{1} = -3 - \sqrt{6}
x2=3+6x_{2} = -3 + \sqrt{6}
Numerical solution
x1=0.550510257216822x_{1} = -0.550510257216822
x2=5.44948974278318x_{2} = -5.44948974278318
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^2 + 6*x + 3)/(x + 4).
(02+06)+34\frac{\left(0^{2} + 0 \cdot 6\right) + 3}{4}
The result:
f(0)=34f{\left(0 \right)} = \frac{3}{4}
The point:
(0, 3/4)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\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:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
2x+6x+4(x2+6x)+3(x+4)2=0\frac{2 x + 6}{x + 4} - \frac{\left(x^{2} + 6 x\right) + 3}{\left(x + 4\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
d2dx2f(x)=0\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:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(2(x+3)x+4+1+x2+6x+3(x+4)2)x+4=0\frac{2 \left(- \frac{2 \left(x + 3\right)}{x + 4} + 1 + \frac{x^{2} + 6 x + 3}{\left(x + 4\right)^{2}}\right)}{x + 4} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=4x_{1} = -4
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
limx((x2+6x)+3x+4)=\lim_{x \to -\infty}\left(\frac{\left(x^{2} + 6 x\right) + 3}{x + 4}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
limx((x2+6x)+3x+4)=\lim_{x \to \infty}\left(\frac{\left(x^{2} + 6 x\right) + 3}{x + 4}\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 + 6*x + 3)/(x + 4), divided by x at x->+oo and x ->-oo
limx((x2+6x)+3x(x+4))=1\lim_{x \to -\infty}\left(\frac{\left(x^{2} + 6 x\right) + 3}{x \left(x + 4\right)}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the left:
y=xy = x
limx((x2+6x)+3x(x+4))=1\lim_{x \to \infty}\left(\frac{\left(x^{2} + 6 x\right) + 3}{x \left(x + 4\right)}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = 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:
(x2+6x)+3x+4=x26x+34x\frac{\left(x^{2} + 6 x\right) + 3}{x + 4} = \frac{x^{2} - 6 x + 3}{4 - x}
- No
(x2+6x)+3x+4=x26x+34x\frac{\left(x^{2} + 6 x\right) + 3}{x + 4} = - \frac{x^{2} - 6 x + 3}{4 - x}
- No
so, the function
not is
neither even, nor odd