curve sketching

G

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y= (-5x^2 + 3x) / (2x^2-5)

find
1. vertical asymptotes
2.maximum and minimum values
3.points of inflection
please help me
 
Given \(\displaystyle \L
\begin{array}{l}
y &=& \frac{{ - 5x^2 + 3}}{{2x^2 - 5}} \\
y' &=& \frac{{ - 6x^2 + 50x - 15}}{{\left( {2x^2 - 5} \right)^2 }} \\
y'' &=& \frac{{2\left( {12x^3 - 150x^2 + 90x - 125} \right)}}{{\left( {2x^2 - 5} \right)^3 }} \\
\end{array}\)
you find where y’=0 & y’’=0.
 
-6x^2+50x-15=0
x=-1.3,x=6.98 this answer righ or wrong?
ican't find y''
 
Hello, batticaloa!

Sorry, but I have to say this . . .
    \displaystyle \;\;You're in Calculus, but you can't solve a quadratic?


\(\displaystyle \L y\;=\;\frac{-5x^2\,+\,3x}{2x^2\,-\,5}\)

\(\displaystyle \L y'\;=\;\frac{-6x^2\,+\,50x\,-\,15}{(2x\,-\,5)^2}\)

We have:  −6x2 + 50x − 15 = 0    ⇒    6x2 − 50x + 15 = 0\displaystyle \,-6x^2\,+\,50x\,-\,15\:=\:0\;\;\Rightarrow\;\;6x^2\,-\,50x\,+\,15\:=\:0

                a = 6,  b = −50,  c = 15\displaystyle \;\;\;\;\;\;\;\;a\,=\,6,\;b\,=\,-50,\;c\,=\,15

Quadratic Formula:   x  =  −(−50) ± (−50)2 − 4(6)(15)2(6)  =  50 ± 214012  =  50 ± 253512\displaystyle \;x\;=\;\frac{-(-50)\,\pm\,\sqrt{(-50)^2\,-\,4(6)(15)}}{2(6)} \;=\;\frac{50\,\pm\,\sqrt{2140}}{12}\;=\;\frac{50\,\pm\,2\sqrt{535}}{12}

And we have:  x1 = 25 + 5356 = 8.021677835x2 = 25 − 5356 = 0.311655498    \displaystyle \:\begin{array}{cc}x_1\:=\:\frac{25\,+\,\sqrt{535}}{6}\:=\:8.021677835 \\ x_2\:=\:\frac{25\,-\,\sqrt{535}}{6}\:=\:0.311655498\end{array}\;\; . . . critical values


\(\displaystyle \;\;\;y''\;=\;\frac{(2x\,-\,5)^2(-12x\.+\,50)\,-\,(-6x^2\,+\,50x\,-\,15)\cdot2(2x\,-\,5)\cdot 2}{(2x\,-\,5)^4}\)

Factor:  (2x − 5) (2x − 5)(−12x + 50) − 4(−6x2 + 50x − 15)(2x − 5)4\displaystyle \,(2x\,-\,5)\,\frac{(2x\,-\,5)(-12x\,+\,50)\,-\,4(-6x^2\,+\,50x\,-\,15)}{(2x\,-\,5)^4}

      =  −24x2 + 100x + 60x − 250 + 24x2 − 200x + 60(2x − 5)3  =  −40x − 190(2x − 5)3\displaystyle \;\;\;=\;\frac{-24x^2\,+\,100x\,+\,60x\,-\,250\,+\,24x^2\,-\,200x\,+\,60}{(2x\,-\,5)^3} \;= \;\frac{-40x\,-\,190}{(2x\,-\,5)^3}

Hence: \(\displaystyle \L\,y''\;=\;-10\,\frac{4x\,+\,19}{(2x\,-\,5)^3}\;\;\) . . . etc.
 
Next time when you are asked, after having been provided assistance, to kindly reply showing some effort of your own, please consider at least attempting the exercise yourself, rather than just reposting.

Thank you.

Eliz.
 
stapel said:
Next time when you are asked, after having been provided assistance, to kindly reply showing some effort of your own, please consider at least attempting the exercise yourself, rather than just reposting.

Thank you.

Eliz.
Click to expand...

Yes, and I guess it "worked" for this person, because Soroban just gave them the answer and saved them from doing any work. :roll:
 
y''=2(12x^3-175x^2+90x-125)/(2x^2-5)^3
y''=0
it's too hard solve this equation?
 
batticaloa said:
y''=2(12x^3-175x^2+90x-125)/(2x^2-5)^3
y''=0
it's too hard solve this equation?
Click to expand...
Please post new exercises as new threads, not as replies to old threads.

Thank you.

Eliz.
 
jolly said:
batticaloa, are you a thief?
Click to expand...
Looks more like just lazy, to me. Too bad the process is so frequently profitable. :roll:

Eliz.
 
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