The rule of the division of differentiation is
[tex]\frac{d}{dx}(\frac{u}{v})=\frac{u^{\prime}v-uv^{\prime}}{v^2}[/tex]
The given function is
[tex]y=f(x)=\frac{x^2+3x+3}{x+2}[/tex]
a)
Let u the numerator and v the denominator
[tex]\begin{gathered} u=x^2+3x+3 \\ u^{\prime}=2x+3 \end{gathered}[/tex][tex]\begin{gathered} v=x+2 \\ v^{\prime}=1 \end{gathered}[/tex]
Substitute them in the rule above
[tex]\begin{gathered} \frac{dy}{dx}=\frac{(2x+3)(x+2)-(x^2+3x+3)(1)}{(x+2)^2} \\ \frac{dy}{dx}=\frac{2x^2+7x+6-x^2-3x-3}{(x+2)^2} \\ \frac{dy}{dx}=\frac{x^2+4x+3}{(x+2)^2} \\ \frac{dy}{dx}=\frac{(x+3)(x+1)}{(x+2)^2} \end{gathered}[/tex]
We will differentiate dy/dx again to find d^2y/dx^2
[tex]\begin{gathered} u=x^2+4x+3 \\ u^{\prime}=2x+4 \end{gathered}[/tex][tex]\begin{gathered} v=(x+2)^2=x^2+4x+4 \\ v^{\prime}=2x+4 \end{gathered}[/tex]
Then substitute them in the rule above
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{(2x+4)(x^2+4x+4)-(x^2+4x+3)(2x+4)}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)\lbrack x^2+4x+4-x^2-4x-3\rbrack}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)\lbrack1\rbrack}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)}{(x+2)^4} \\ \frac{d^2y}{dx^2}=\frac{2(x+2)}{(x+2)^4} \\ \frac{d^2y}{dx^2}=\frac{2}{(x+2)^3} \end{gathered}[/tex]
b)
The turning point is the point that has dy/dx = 0
Equate dy/dx by 0 to find the values of x
[tex]\begin{gathered} \frac{dy}{dx}=\frac{(x+3)(x+1)}{(x+2)^2} \\ \frac{dy}{dx}=0 \\ \frac{(x+3)(x+1)}{(x+2)^2}=0 \end{gathered}[/tex]
By using the cross multiplication
[tex]\begin{gathered} (x+3)(x+1)=0 \\ x+3=0,x+1=0 \\ x+3-3=0-3,x+1-1=0-1 \\ x=-3,x=-1 \end{gathered}[/tex]
Substitute x by -3 and -1 in f(x) to find y
[tex]\begin{gathered} f(-3)=\frac{(-3)^2+3(-3)+3}{-3+2} \\ f(-3)=\frac{3}{-1} \\ y=-3 \end{gathered}[/tex][tex]\begin{gathered} f(-1)=\frac{(-1)^2+3(-1)+3}{-1+2} \\ f(-1)=\frac{1}{1} \\ y=1 \end{gathered}[/tex]
The turning points are (-3, -3) and (-1, 1)
c)
To find the inflection point equate d^2y/dx^2 by 0 to find x
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{2}{(x+2)^3} \\ \frac{d^2y}{dx^2}=0 \\ \frac{2}{(x+2)^3}=0 \end{gathered}[/tex]
By using the cross multiplication
[tex]2=0[/tex]
Which is wrong 2 can not be equal to zero, then
NO inflection point for the curve
d)
Since the denominator of the curve is x + 2, then
Equate it by 0 to find the vertical asymptote
[tex]\begin{gathered} x+2=0 \\ x+2-2=0-2 \\ x=-2 \end{gathered}[/tex]
There is a vertical asymptote at x = -1
Since the greatest power of x up is 2 and the greatest power of down is 1, then there is an Oblique asymptote by dividing up and down
[tex]\begin{gathered} \frac{x^2+3x+3}{x+2}=x+1 \\ y=x+1 \end{gathered}[/tex]
The Oblique asymptote is y = x + 1
No horizontal asymptote
e)
This is the graph of y = f(x)
This is the graph of y = f(IxI)
f)
For the curve
[tex]y=\frac{x^2-3x+3}{2-x}[/tex]
Take (-) sign as a common factor down, then
[tex]\begin{gathered} y=\frac{(x^2+3x+3)}{-(-2+x)} \\ y=-\frac{(x^2-3x+3)}{(x-2)} \end{gathered}[/tex]
Since the sign of y is changed, then
[tex]y=-f(x)[/tex]
Then it is the reflection of f(x) about the y-axis we can see it from the attached graph
The red graph is f(x)
The purple graph is -f(x) which is the equation of the last part
