Notice that ##x^2+4x = (x+2)^2 – 4## and take ##abs(x+2)## outside the square root to find two slant :

##y = x+2##

and

##y = -x-2##

Let ##f(x) = y = sqrt(x^2+4x) = sqrt(x(x+4))##

As a Real valued function, this has ##(-oo, -4] uu [0, oo)##, since ##x^2+4x >= 0## if and only if ##x in (-oo, -4] uu [0, oo).##

##sqrt(x^2+4x)##

##=sqrt(x^2+4x+4-4)##

##=sqrt((x+2)^2-4)##

##=sqrt((x+2)^2(1 – 4/((x+2)^2))##

##=abs(x+2) sqrt(1-4/(x+2)^2)##

As ##x->+-oo## we find that ##4/(x+2)^2 -> 0##, so ##f(x)## is asymptotic to ##abs(x+2)##

This results in two slant asymptotes:

##y = x+2## as ##x->+oo##

and

##y = -x-2## as ##x->-oo##

graph{(y-sqrt(x^2+4x))(y – x – 2)(y + x + 2) = 0 [-11.01, 8.99, -1.08, 8.92]}

 

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A slight variant on method

##color(blue)(“Vertex” ->(x,y)->(2,-1)##

Given:##” “2x^2-8x+7##

Write as:##” ” 2(x^2-color(red)(8/2)x)+7##

Consider the ##color(red)(-8/2)” from “-8/2x##

##color(blue)(x_(“vertex”)=(-1/2)xx(color(red)(-8/2)) = +8/4 = 2)##

‘~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute ##x=color(magenta)(2)## in the original equation to find ##y_(“vertex”)##

##y=2x^2-8x+7″ “->” “y_(“vertex”)=2(color(magenta)(2))^2-8(color(magenta)(2))+7##

##color(blue)(y_(“vertex”)= 8-16+7=-1)##

‘~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~##color(blue)(“Vertex” ->(x,y)->(2,-1)##

 

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Vertical: x=3; no Horizontal asymptote; Oblique: y=x+5

1) The vertical depend on the ; the domain is obtained by solving the following:

##x^2-5x+6!=0##

that is solved by :

##x=(-b+-sqrt(b^2-4ac))/2a##

where a=1; b=-5; c=6

then

##x=(5+-sqrt(5^2-4*1*6))/(2*1)##

##=(5+-sqrt(25-24))/2##

##=(5+-1)/2##

##x_1=2;x_2=3##

The domain of the given function is:

##x!=2 and x!=3##

Now let’s calculate

##lim_(x->2) (x^3-8)/(x^2-5x+6)=lim_(x->2)((cancel(x-2))(x^2+2x+4))/((cancel(x-2))(x-3))=-12##

and

##lim_(x->3) (x^3-8)/(x^2-5x+6) =lim_(x->3)((cancel(x-2))(x^2+2x+4))/((cancel(x-2))(x-3))=oo##

Then the vertical asymptote is the line x=3

2) Let’s calculate

##lim_(x->oo) (x^3-8)/(x^2-5x+6)=oo##

then there are no horizontal asymptote

3) Let’s calculate

##m=lim_(x->oo) (x^3-8)/(x^2-5x+6)*1/x=(x^3-8)/(x^3-5x^2+6x)=1##

that’s the slope of the oblique asymptote.

Let’s calculate the intercept

##n=lim_(x->oo) (x^3-8)/(x^2-5x+6)-mx=lim_(x->oo) (x^3-8)/(x^2-5x+6)-x##

##n=lim_(x->oo) (cancelx^3-8-cancelx^3+5x^2-6x)/(x^2-5x+6)=5##

Then the oblique asymptote is the line

##y=x+5##

graph{(x^3-8)/(x^2-5x+6) [-20, 10, -15, 5]}

 

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Vertical asymptote: ##x=-3##Horizontal asymptote: noneOblique asymptote: ##y=-2x##

How to find vertical asymptote:

Set denominator to equal zero and solve for ##x##

##3x+9## ##rarr## ##3x=-9## ##rarr## ##x=-9/3## ##rarr## ##x=-3##

How to find horizontal asymptote:

The degree in the numerator is greater than the degree of the denominator, therefore there is no horizontal asymptote.

How to find oblique asymptote:

The degree in the numerator is one more than the degree in the denominator, therefore there is a slant/oblique asymptote present. To find the slant asymptote, you must make sure that the numerator is in quadratic form. In this case, since the numerator is already in quadratic form, we leave it as it is. From here, we then perform synthesis division to find the slant asymptote.

We don’t have to worry about the remainder column in this situation. This makes the slant asymptote: ##y=-2x##We also know that there is a slant asymptote because there is also a vertical asymptote present.

 

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