Question

Help me solve this rational problem?

How do I solve this

Answer 1

See explanation.

Explanation:

Given: `(x^2-16)/(x^2+x-20)`

There are several things to note.

`color(brown)("Consider the numerator:")`

`x^2-16 -> a^2-b^2=(a-b)(a+b)`

Thus we have `(x-4)(x+4)`

From this it MAY be the case that the question designer intended one of them to cancel out in the denominator.

`color(brown)("Consider the denominator:")`

As it is possible that there MAY be a cancelling out it is reasonable to explore `x=+-4` as a factor

Note that `4xx5=20 and 5-4=1` so lets look at

`x^2+x-20color(white)("ddd")->color(white)("ddd")(x+5)(x-4)`

`color(white)("dddddddddddd")->color(white)("ddd") x^2+5x-4x-20 larr" Works"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Putting it all back together")`

`(x^2-16)/(x^2+x-20) -> (cancel((x-4))(x+4))/((x+5)cancel((x-4))) =(x+4)/(x+5) ->f(x)`

As the equation is undefined when the denominator becomes 0 we have a vertical asymptote at `x=-5`

`lim_(x->+oo) f(x)->k=+1`

`lim_(x->-oo) f(x)->k==+1`

Thus there is a horizontal asymptote at `y=1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Any excluded value")`

Although the `(x-4)` in the denominator cancels 'away' it still forms part of the original expression. Thus it has to be taken into account. This is done by setting `(x-4)=0` and declaring `x=4` as the excluded value.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("The general behaviour of the plot")`

I will let you investigate this. You will need to shoe your logic in your solution.

We already know that there is a horizontal asymptote at `y=1`

Test out what happens very close to the vertical asymptote.

color(white)("d")

Answer 2

`"see explanation"`

Explanation:

`"let "f(x)=(x^2-16)/(x^2+x-20)`

`"the numerator is a "color(blue)"difference of squares"`

`x^2-16=(x-4)(x+4)`

`"denominator"`

`"the factors of "-20" which sum to "+1`
`"are "+5" and "-4`

`x^2+x-20=(x+5)(x-4)`

`f(x)=(cancel((x-4))(x+4))/((x+5)cancel((x-4)))=(x+4)/(x+5)`

`"the removal of the factor "(x-4)" from the numerator"`

`" and denominator indicates a hole at"`

`x-4=0rArrx=4toy=(4+4)/(4+5)=8/9`

`"hole at "(4,8/9)`

`"the graph of "f(x)=(x+4)/(x+5)" is the same as"`

`(x^2-16)/(x^2+x-20)" but without the hole"`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

`"solve "x+5=0rArrx=-5" is the asymptote"`

`"Horizontal asymptotes occur as"`

`lim_(xto+-oo),f(x)toc" ( a constant)"`

`"divide terms on numerator/denominator by "x`

`f(x)=(x/x+4/x)/(x/x+5/x)=(1+4/x)/(1+5/x)`

`"as "xto+-oo,f(x)to(1+0)/(1+0)`

`y=1" is the asymptote"`

`"there is a discontinuity at "x=-5`

`"domain is "x in(-oo,-5)uu(-5,oo)`

`"there is a discontinuity at "y=1`

`"range is "y in(-oo,1)uu(1,oo)`

`"For Intercepts"`

`x=0rArry=4/5larrcolor(red)"y-intercept"`

`x+4=0rArrx=-4larrcolor(red)"x-intercept"`
graph{(x+4)/(x+5) [-10, 10, -5, 5]}