How do you graph #y=(3x^2+10x-8)/(x^2+4)# using asymptotes, intercepts, end behavior?

2 Answers
Dec 8, 2016

Answer:

See calculations below

Explanation:

The domain of y is #D_y =RR#

The denominator is #x^2+4>0#, #AAx in RR#

To find the intercepts, let #x=0#

#y=-8/4=-2#

The y-intercept is #(0,-2)#

When #y=0#

#3x^2+10x-8=0#

#(3x-2)(x+4)=0#

So, #x=2/3# and #x=-4#

The x-intercepts are #(2/3,0)# and #(-4,0)#

To calculate the limits when #x->+-oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->+-oo)y=lim_(x->+-oo)(3x^2)/x^2=3#

So, the horizontal asymptote is #y=3#

When #y=3#, #=>#, #3=(3x^2+10x-8)/(x^2+4)#

#3x^2+12=3x^2+10x-8#

#10x=20#, #=>#. #x=2#

The curve cuts the horizontal asymptote at #(2,3)#

To find the maximum and minimum, you have to calculate the derivative.

#y=u/v#

#dy/dx=(u'v-uv')/v^2#

#u=3x^2+10x-8#, #=>#. #u'=6x+10#

#v=x^2+4#, #=>#, #v'=2x#

#dy/dx=((6x+10)(x^2+4)-(2x)(3x^2+10x-8))/(x^2+4)^2#

#=(6x^3+24x+10x^2+40-6x^3-20x^2+16x)/(x^2+4)^2#

#=(-10x^2+40x+40)/(x^2+4)^2#

#dy/dx=0#

#-10(x^2-4x-4)=0#

#Delta=b^2-4ac=16+16=32#

#x=(4+-sqrt32)/2=(4+-4sqrt2)/2=2+-2sqrt2#

So we have 2 points, #x=2+2sqrt2# and #x=2-2sqrt2#

We do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##2-2sqrt2##color(white)(aaaa)##2+2sqrt2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##dy/dx##color(white)(aaaaaaaa)##-##color(white)(aaaaaaaa)##+##color(white)(aaaaaaa)##-#

#color(white)(aaaa)##y##color(white)(aaaaaaaaaa)##darr##color(white)(aaaaaaaaa)##uarr##color(white)(aaaaaaa)##darr#

#x=2-2sqrt2# corresponds to a minimum

#x=2+2sqrt2# corresponds to a maximum

graph{(y-(3x^2+10x-8)/(x^2+4))(y-3)=0 [-10, 10, -5, 5]}

Dec 8, 2016

Answer:

x-intercepts #(y=0): -4 and 2/3#; y-intercept #(x=0): -2#. Horizontal asymptote: #larr y = 3 rarr. y in (-3.04, 4.04)#, nearly.

Explanation:

x-intercepts #(y=0): -4 and 2/3#; y-intercept #(x=0): -2#

By actual division,

#y =3+(10(x-2))/(x^2+4) to (2, 3)# is a point on the graph.

#y=3+(10(2-1/x))/(x(1+4/x^2)) to 3#, as x to +-oo#.

It is evident that y = 3 is the horizontal asymptote that cuts the graph

at (2, 3)..

#y'=(-10x^2+40x+40)/(x^2+1)^2= 0#, when x = 2(1+-sqrt2)=-.83 and

4.83,

nearly. These turning points are (-.93, -3.04) and (4.83, 4.04), nearly.

The graph justifies mini/max y at these points as -3.04/4.04,

nearly. y is bounded between these extremes.

graph{y(x^2+4)-3x^2-10x+8=0 [-16.29, 16.24, -8.13, 8.14]}