Melody

You sound surprised.

You asked me privately to give you a detailed answer to help you understand graphing lines.

I have done that.

I hope you have looked at my answer properly. 

Do you understand that my answer and your answer are the same only mine is slightly better presented?

Anyway, I am glad you understand the question.

How do you know kanisan tried?

Maybe 4 was given to them by someone else (likely a dummy answer here).

Kanison did not state how the 4 was derived.

EP was right to ask.

And it is quite annoying when our 'teaching' answers are overridden.

Especially when the original person has not even responded to our response question yet.

\(x^2 + y^2 = 2x - 6y + 6\\ (x^2 -2x)+( y^2 -6y)= 6\\ (x^2 -2x+1)+( y^2 -6y+9)= 6+1+9\\ (x-1)^2+(y-3)^2=16 \)

This is a circle with a radius of 3

you can find the area from there.

The purple line is the asymptote.

I treated it as a graphing exercise.

Hi  ilorty,

For this specific question your way is MUCH better than mine.

I have only been introduced to your method very recently, perhaps by you, and it will take me a while to integrate it into my own memory and use.

Thanks very much for reminding me   

This was my approach

\(y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\ y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\ y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\ \text{I note that the denominator is positive for all real values of x}\\ \text{As x tends to }\pm\infty \text{ y tends to +1}\\ y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\ y'=0 \quad when \quad \\ (x^2+2x+3)(2)-(2x+2)(2x+1)=0\\ 2x^2+4x+6-(4x^2+6x+2)=0\\ -2x^2-2x+4=0\\ x^2+x-2=0\\ (x+2)(x-1)=0\\ x=1\;\;or \;\;x=-2\\ \)

When x=1 y=4

When x=-2, y=-5

I now have enough info to know that the graph looks something like this:

\(\bf\text{So the range of this function is   }   -5 \le x\le 4\)

LaTex:

y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\
y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\
y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\
\text{I note that the denominator is positive for all real values of x}\\
\text{As x tends to }\pm\infty \text{  y tends to +1}\\
y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\
y'=0 \quad when \quad \\
(x^2+2x+3)(2)-(2x+2)(2x+1)=0\\
2x^2+4x+6-(4x^2+6x+2)=0\\
-2x^2-2x+4=0\\
x^2+x-2=0\\
(x+2)(x-1)=0\\
x=1\;\;or \;\;x=-2\\

6*5*3*67 = 6030

So you will need at least 6030 m^3 of storage.  But you will probably need more because the boxes have to fit without being crushed.

If they were in a straight snake that was 67 boxes long then it would be exactly 6030m^3

WOW!

That is really cool !

Thanks EP.  I will just add my 2c worth

The line appears to go through   (1,0)  and (-1,-1)

\(gradient=\frac{rise}{run}\\ gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\ gradient=\frac{y_2-y_1}{x_2-x_1}\\ gradient=\frac{-1-0}{-1-1}\\ gradient=\frac{-1}{-2}\\ gradient=\frac{1}{2}\\\)

now to get the equation I usually just think to myself 

 gradient =gradient

\(gradient=\frac{y_2-y_1}{x_2-x_1}\\ \text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\ \frac{1}{2}=\frac{y-0}{x-1}\\ 1(x-1)=2(y-0)\\ x-1=2y\\ x-2y-1=0 \)

(0,0) is in the wanted part of the graph so you have to sub in (0,0) and choose the correct inequality

(it is a dashed line, not a solid one, so there will be no equal sign)

\(LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\ LHS=0-0-1\\ LHS=-1\\ LHS<0\\ so\\ x-2y-1<0\)

LaTex:

gradient=\frac{rise}{run}\\
gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\
gradient=\frac{y_2-y_1}{x_2-x_1}\\
gradient=\frac{-1-0}{-1-1}\\
gradient=\frac{-1}{-2}\\
gradient=\frac{1}{2}\\

gradient=\frac{y_2-y_1}{x_2-x_1}\\
\text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\
\frac{1}{2}=\frac{y-0}{x-1}\\
1(x-1)=2(y-0)\\
x-1=2y\\
x-2y-1=0

LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\
LHS=0-0-1\\
LHS=-1\\
LHS<0\\
so\\
x-2y-1<0