Question

How do you find the asymptotes for `h(x) = sqrt(x^2-3x) - x`?

Answer 1

`2y+4x-3=0` and `y=-3/2`

Explanation:

`h(x)=y=sqrt(x^2-3x)-x`
`y+x=sqrt(x^2-3x`
Square both side
`(y+x)^2=x^2-3x`
`y^2+2xy+x^2=x^2-3x`
`y^2+2xy+3x=0`

Let equation of Asymptote is `y=mx+c`

Put x=1 and y=m in the terms of highest power (2) of equation of Curve

`phi_2(m)=m^2+2m`
Put x=1 and y=m in the terms of power 1
`phi_1(m)=3`
differentiate with respect to m of `phi_2(m)`
`phi'_2(m)=2m+2`

To find values of m
`phi_2(m)=0`
`impliesm^2+2m=0`
`impliesm(m+2)=0`
either `m=0 or m=-2`

Value of `c=-(phi_1(m))/(phi'_2(m))=-3/(2m+2)`

for `m=0`, `c=-3/2`
Asymptotes are
`y=0x+(-3/2)` or `y=-3/2`

and for `m=-2`, `c=3/2`

Asymptotes are
`y=-2x+3/2` or `2y+4x-3=0`