Melody

 Lim-->4 (x*sqrt(x)-8)/(x-4)  

\(\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\\ =\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\)

At this point I have learned from our guest.  Thankyou :)  

\(=\displaystyle\lim_{x\rightarrow4} \frac{ (x\sqrt{x}-8)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\~\\ \qquad \mbox{I need to get rid of the }\sqrt{x}-2\;\;\mbox{in the denominator}\\ \qquad \mbox{Now I used algebraic division to find that}\\ \qquad (x\sqrt{x}-8)\div (\sqrt{x}-2)=x+2\sqrt{x}+4\\\qquad so\\ \qquad(x\sqrt{x}-8)=(x+2\sqrt{x}+4) (\sqrt{x}-2)\\~\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) (\sqrt{x}-2)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) }{( \sqrt{x}+2) }\\~\\ =\frac{4+2*2+4}{2+2}\\ =3\)

Here it the graph - it has a hole in it at (4,3)

Yea that is pretty cool Chris - DS would have to see it it worked as a little icon. 

That is just trial and error.

Here is another but it is not as coold as chris's.

Thanks Coldplay and guest :))

Hi Juriemagic,

Admin knows that you and others have log in problems.  He does know how to fix it.

Have you tried using a different browser?  

Do you completely understand my answer?

You should thank guest answerer too.  He/she put in a lot of work on your behalf.   :/

Here are some more examples to practic on if you want

Hi Eloise,

I have missed you  :((   

I am glad you are back :))

Hi Juriemagic,

I see you still cannot log on    

\(\frac {(p^2q)^3(2p^3q^{-4})^2 } {p^2q^3 * (4p^{-2}q^5)^3}\\~\\ =\frac {p^6q^3(2^2p^6q^{-8}) } {p^2q^3 * (4^3p^{-6}q^{15})}\\~\\ =\frac {p^6(4p^6q^{-8}) } {p^2 * (64p^{-6}q^{15})}\\~\\ =\frac {p^4(4p^6q^{-8}) } { (64p^{-6}q^{15})}\\~\\ =\frac {4p^4p^6q^{-8} } { 64p^{-6}q^{15}}\\~\\ =\frac {p^{10}q^{-8} } { 16p^{-6}q^{15}}\\~\\ =\frac {p^{10}p^{+6} } { 16q^{15}q^{+8}} \\~\\ =\frac {p^{16} } { 16q^{23}} \\ \)

If you do not understand or you want a particular line discussed then please ask :)

Hi Solveit,

y=-x^2+k^2

Vertex is (0,k^2)

Find roots of the parabola  y=0

0=-x^2+k^2

x= +/- k

Area = 1/2 * 2k  * k^2 = 64

k^3=64

k=4

This is not a complete answer - it is just a good outline.

Let me know if you need more.

You should check it anyway :)

Thanks Solveit, Chis, Alan, Heureka and Bertie, :)

This one should go up in our Hall of Fame !

Lots to study and digest here.   

I always like that you are in the wings Bertie,  Thank you  

(I assume the other guest post was also you )

Hi Solveit,

I am glad you are persisting.  Persistence is how we learn difficult things. :)

This was one of the constraints of your question

The function has a line of symmetry at x=-3. 

Your graph is not correct because the axis of symmetry is x=0  not  x=-3

g(x)=ax^2+bx-6a

for the axis to be x=-3

-b/2a must equal -3

-b/2a=-3

b=6a

so t he graph becomes

g(x)=ax^2+6ax-6a

Here is the correct graph

This is just a troll kids, for whom you are providing great amusement.