Answer:
The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)
Step-by-step explanation:
A rational equation is a equation where
[tex] \frac{p(x)}{q(x)} [/tex]
where both are polynomials and q(x) can't equal zero.
1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of
[tex](x - 2)[/tex]
in our denomiator.
So right now we have
[tex] \frac{p(x)}{(x - 2)} [/tex]
2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.
We can model -2.5 as
[tex](2x + 5)[/tex]
So we have as of right now.
[tex] \frac{(2x + 5)}{(x - 2)(2x + 5)} [/tex]
Now let see if this passes throught point (6,-3).
[tex] \frac{(2x + 5)}{(x - 2)(2x + 5)} = y[/tex]
[tex] \frac{(17)}{68} = \frac{1}{4} [/tex]
So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.
If we have a variable r, in the numerator that will make this applicable, we would get
[tex] \frac{(2x + 5)r}{(2x + 5)(x - 2)} = - 3[/tex]
Plug in 6 for the x values.
[tex] \frac{17r}{4(17)} = - 3[/tex]
[tex] \frac{r}{4} = - 3[/tex]
[tex]r = - 12[/tex]
So our rational equation will be
[tex] \frac{ - 12(2x + 5)}{(2x + 5)(x - 2)} [/tex]
or
[tex] \frac{ - 24x - 60}{2 {x}^{2} + x - 10} [/tex]
We can prove this by graphing
