Given #y={2x^2-a}/{x^2-4}# Find: i) the domain ii) the range iii) x and y intercepts iv) the asymptotes v) point of inflexion?

Given #y={2x^2-a}/(x^2-4)#
Find: i) the domain
ii) the range
iii) x and y intercepts
iv) the asymptotes
v) point of inflection

2 Answers
Dean R.
Jun 22, 2018

Some confusion about the function, let's go with

#y = (2x^2 - a)/(x^2 - 4)#

We're working over the reals presumably so the domain is all the reals except where the denominator vanishes,

#x^2-4=0#

#x= pm 2#

So the domain is #x in R, x ne pm 2,# all the reals except #2# and #-2.#

The range is tricky. Let's dispense with special case #a=8# first.

#y = (2x^2 - 8)/(x^2-4) = 2(x^2 -4)/(x^2-4)=2#

The range when #a=8# is just #y in {2}.#

When #a ne 8# there will be a parabola-like central part and a hyperbola-like outer part with the opposite sign, both wings of the hyperbolish on the same side of the x axis. Whether or not #a>8# determines whether the parabolish is open up or down but it doesn't much matter; the range will be the entire reals (we get toward #pm infty# near #x=2#) less the interval between the vertex of the parabolish and the asymptotes of the hyperbolish.

By the symmetry the stationary point must be #x=0# but let's work it out:

#0 = y' = {vdu-udv}/v^2 = {(x^2-4)(4x) - (2x^2-a)(2x) }/(x^2-4)^2#

#4x^3-16x - 4x^3 + 2ax = 0#

#2x(8-a) = 0#

We're already assuming #a ne 8# so

#x= 0#

#y = -a/-4 = a/4#

#a/4# is always in the range; it's one end of the excluded zone.

As # x to infty# we see #y to 2# so that's the other end.

The range is #x in R, x notin (a/4,2] and x notin [2,a/4)#

We write the funny conjunction to handle both #a<8# and #a>8.#

We just calculated the y intercept as #y=a/4#.

The x intercept aka zero is when the numerator is zero, #2x^2=a# or #x=pm sqrt{a/2},# with the usual caveat #a ne 8# so #x ne pm 2.#

We calculated the inflection point as the y intercept, #(0, a/4)# again excluding #a=8.#

I've had the graph crash my tab so I'll post first then add the graph.

Let's graph the cases #a=4, a=8 and a=12.#

#a=4:# graph{(2x^2 -4)/(x^2 - 4) [-10, 10, -5, 5]}

#a=8:# graph{(2x^2 -8)/(x^2 - 4) [-10, 10, -5, 5]}

#a=12:# graph{(2x^2 -12)/(x^2 - 4) [-10, 10, -5, 5]}

Narad T.
Jun 22, 2018

Please see the explanation below.

Explanation:

The function is

#y=(2x^2-a)/(x^2-4)#, #a in RR#

#(i)#

The denominator must be #!=0#

#x^2-4!=0#

#=>#, #x!=-2# and #x!=2#

The domain is # x in (-oo, -2) uu(-2,2)uu(2,+oo)#

#(ii)#

Calculate the range as follows

#y=(2x^2-a)/(x^2-4)#

#=>#, #y(x^2-4)=2x^2-a#

#=>#, #yx^2-4y=2x^2-a#

#=>#, #yx^2-2x^2=4y-a#

#=>#, #x^2(y-2)=(4y-a)#

#=>#, #x^2=(4y-a)/(y-2)#

#=>#, #x=sqrt((4y-a)/(y-2))#

Therefore,

#(4y-a)/(y-2)>=0#

#y!=2#

#=>#, #y in (-oo,2)uu(2,+oo)#

and #4y-a>=0#, #=>#, #4y>=a#

#=>#, #y>=a/4#

The range is #y in (-oo,a/4]uu(2,+oo)#

#(iii)#

The intercept are as follows:

y axis when #x=0#

#=>#, y=a/4

The point of intercept with the y-axis is #(0,a/4)#

x axis when #y=0#

#(2x^2-a)/(x^2-4)=0#

#=>#, #2x^2-a=0#

#=>#, #x^2=a/2#

#=>#, #x=+-sqrt(a/2)#

The point of intercept with the x-axis is #(-sqrt(a/2),0)# and #(sqrt(a/2),0)#

#(iv)#
The vertical asymptotes are at #x=-2# and #x=2#

The horizontal asymptotes are as follows

#lim_(x->+oo)(2x^2-a)/(x^2-4)=lim_(x->+oo)(2x^2)/x^2=2#

#lim_(x->-oo)(2x^2-a)/(x^2-4)=lim_(x->-oo)(2x^2)/x^2=2#

The horizontal asymptote is #y=2#

There are no slant asymptotes as the degree of the numerator is #=# the degree of the denominator

#(v)#

Calculate the first and second derivatives.

#dy/dx=(2(a-8)x)/(x^2-4)^2#

#(d^2y)/dx^2=-(2(a-8)(3x^2+4))/(x^2-4)^3#

As #3x^2+4>0#, #AA x in RR#

Therefore,

#(d^2y)/dx^2!=0#

So, there are no points of inflections

Graph for #a=5#

graph{(2x^2-5)/(x^2-4) [-10, 10, -5, 5]}

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