# Thread: Sketching graphs using partial fractions!

1. ## Sketching graphs using partial fractions!

Sketch the graph of the following (showing turning points):

$
y=\frac{2x+3}{x^2+2x+1}
$

My working out so far:

$y=\frac{2x+3}{x^2+2x+1}= \frac{2x+3}{(x+1)^2}$

$\frac{2x+3}{(x+1)^{2}}\equiv \frac{a}{x+1} + \frac{b}{(x+1)^{2}}\equiv \frac{a(x+1)+b}{{(x+1)}^2}$

$2x+3\equiv a(x+1)+b$
Letting $x = -1$
$b=1$
Letting $x=0$
$3=a +1$
$a=2$
Vertical asymptotes: $x=-1$
Horizontal asymptote: $y=0$

$\frac{2}{x+1} + \frac{1}{(x+1)^2} = 0$
$2(x+1)^2 +x+1=0$
$2x^2+4x+2+x+1=0$
$2x^2+5x+3=0$
$2x^2+3x+2x+3=0$
$x(2x+3)+1(2x+3)=0$
$(x+1)(2x+3)=0$
$x=-1$
$x=\frac{-3}{2}$

I've drawn the corresponding graph, using addition-of-ordinates.
But have I made any mistakes in my working out so far? And how do I find the turning points for this graph?

Anyway help will be welcomed with open arms and be highly appreciated!

2. Originally Posted by Joker37
Sketch the graph of the following (showing turning points):

$
y=\frac{2x+3}{x^2+2x+1}
$

My working out so far:

$y=\frac{2x+3}{x^2+2x+1}= \frac{2x+3}{(x+1)^2}$

$\frac{2x+3}{(x+1)^{2}}\equiv \frac{a}{x+1} + \frac{b}{(x+1)^{2}}\equiv \frac{a(x+1)+b}{{(x+1)}^2}$

$2x+3\equiv a(x+1)+b$
Letting $x = -1$
$b=1$
Letting $x=0$
$3=a +1$
$a=2$
Vertical asymptotes: $x=-1$
Horizontal asymptote: $y=0$

$\frac{2}{x+1} + \frac{1}{(x+1)^2} = 0$
$2(x+1)^2 +x+1=0$
That's incorrect. Multiplying on both sides by $(x+1)^2$ and you get 2(x+ 1)+ 1= 0 so that 2x+ 3= 0 and then x= -3/2. The "-1" you get is an "extraneous" root.

$2x^2+4x+2+x+1=0$
$2x^2+5x+3=0$
$2x^2+3x+2x+3=0$
$x(2x+3)+1(2x+3)=0$
$(x+1)(2x+3)=0$
$x=-1$
$x=\frac{-3}{2}$

I've drawn the corresponding graph, using addition-of-ordinates.
But have I made any mistakes in my working out so far? And how do I find the turning points for this graph?

Anyway help will be welcomed with open arms and be highly appreciated!

3. Hello, Joker37!

I don't understand the method used in this problem.
And evidently, neither do you.

Sketch the graph of the following (showing turning points):

. . $y\:=\:\dfrac{2x+3}{x^2+2x+1}$

You can find these two facts by "eyeballing" the original function.

. . $\begin{array}{ccc}\text{Vertical asymptotes:} & x\:=\:\text{-}1 \\
\text{Horizontal asymptote:} & y\:=\:0\end{array}$

Similarly, you can also find these two facts:

. . . . $\begin{array}{ccc}x\text{-intercept:} & (\text{-}\frac{3}{2},\:0) \\
\,y\text{-intercept:} & (0,3) \end{array}$

To find turning points, we solve $y' \,=\,0$

. . We have: . $y' \:=\:\dfrac{\text{-}2(x+2)}{x+1} \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}2$

Hence, the turning point is: . $(\text{-}2,\:\text{-}1)$}

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Using the process of Partial Fraction Decomposition,

. . you found that: . $\displaystyle y \;=\;\frac{2}{x+1} + \frac{1}{(x+1)^2}$

But why?

What possible purpose does it serve?

PFD is normally used facilitate integration, which is not involved here.

4. $\frac{2x+3}{x^2+2x+1}=\frac{2(x+1)+1}{(x+1)^2}=\fr ac{2}{x+1}+\frac{1}{(x+1)^2}$

Asymptotes:

$x\rightarrow\pm\infty\Rightarrow\ y\rightarrow\ 0$

$x\rightarrow\ -1\Rightarrow\ y\rightarrow\pm\infty$

Intercepts:

$\frac{2}{x+1}+\frac{1}{(x+1)^2}=0 \Rightarrow\frac{2}{x+1}=-\frac{1}{(x+1)^2}\Rightarrow\frac{2(x+1)}{(x+1)^2} =-\frac{1}{(x+1)^2}$

$\Rightarrow\ x+1=-\frac{1}{2}\Rightarrow\ x=-\frac{3}{2}$

Remember also, the function is undefined for $x=-1$

$x=0\Rightarrow\ y=2+1=3$

Turning points:

$f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}=-\left[\frac{2(x+1)+2}{(x+1)^3}\right]=0\Rightarrow\ x=-2$

$f(-2)=-2+1=-1$

5. Originally Posted by Soroban
To find turning points, we solve $y' \,=\,0$

. . We have: . $y' \:=\:\dfrac{\text{-}2(x+2)}{x+1} \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}2$
How did you get y'?

Originally Posted by Archie Meade
Turning points:

$f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}$
How did you get f'(x)?

6. Originally Posted by Soroban

Using the process of Partial Fraction Decomposition,

. . you found that: . $\displaystyle y \;=\;\frac{2}{x+1} + \frac{1}{(x+1)^2}$

But why?

What possible purpose does it serve?

PFD is normally used facilitate integration, which is not involved here.
Well, it gives you a better idea of the shape of the graph doesn't it? I still need to sketch the graph and this partial fraction is easier to analyze and graph than the more complex looking equation given originally.

7. Originally Posted by Joker37

How did you get f'(x)?
Using the Quotient Rule of differentiation on both partial fractions.
Or write the fractions using negative exponents instead...

$f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}=2(x+1)^{-1}+(x+1)^{-2}$

$\Rightarrow\ f'(x)=-2(x+1)^{-2}-2(x+1)^{-3}=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}$

8. Originally Posted by Archie Meade
$\frac{2x+3}{x^2+2x+1}=\frac{2(x+1)+1}{(x+1)^2}=\fr ac{2}{x+1}+\frac{1}{(x+1)^2}$

Asymptotes:

$x\rightarrow\pm\infty\Rightarrow\ y\rightarrow\ 0$

$x\rightarrow\ -1\Rightarrow\ y\rightarrow\pm\infty$

Intercepts:

$\frac{2}{x+1}+\frac{1}{(x+1)^2}=0 \Rightarrow\frac{2}{x+1}=-\frac{1}{(x+1)^2}\Rightarrow\frac{2(x+1)}{(x+1)^2} =-\frac{1}{(x+1)^2}$

$\Rightarrow\ x+1=-\frac{1}{2}\Rightarrow\ x=-\frac{3}{2}$

Remember also, the function is undefined for $x=-1$

$x=0\Rightarrow\ y=2+1=3$

Turning points:

$f(x)=\frac{2}{x+1}+\frac{1}{(x+1)^2}\Rightarrow\ f'(x)=-\frac{2}{(x+1)^2}-\frac{2}{(x+1)^3}=-\left[\frac{2(x+1)+2}{(x+1)^3}\right]=0\Rightarrow\ x=-2$

$f(-2)=-2+1=-1$
Was this all the complete working out that I had to show?

9. Yes.

The axis intercepts are useful to have.
The major aspects are the asymptotes and the turning points.
It's then possible to draw a rough freehand sketch of the graph.

Accuracy is improved by calculating a few more points, though the axis intercepts help a lot there.