Question

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

Answer 1

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]}

Answer 2

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]}