# How do you write an equation with a vertical asymptote of 3, slant asymptote of y=x+1, and x intercept at 2?

Jun 22, 2016

$f \left(x\right) = \frac{{x}^{2} - 2 x}{x - 3}$

#### Explanation:

Let f(x) = (n(x))/(d(x) be the choosed fractional function.

It has a vertical asymptote at $3$ so $d \left(x\right) = \left(x - 3\right)$
It has a slant symptote given by $y = x + 1$ so $n \left(x\right)$ degree must be $2$.

Now, the general polynomial of degree $2$ is

$n \left(x\right) = a {x}^{2} + b x + c$ so

$f \left(x\right) = \frac{a {x}^{2} + b x + c}{x - 3}$

We know also that $f \left(2\right) = 0$ and

$a {x}^{2} + b x + c = \left(x - 3\right) \left(x + 1\right) + d$ because

$f \left(x\right) = \frac{a {x}^{2} + b x + c}{x - 3} = \left(x + 1\right) + \frac{d}{x - 3}$ so for big values of $\left\mid x \right\mid$

$f \left(x\right) \approx \left(x + 1\right)$

then

$a {x}^{2} + b x + c = {x}^{2} - 2 x + d - 3$

so $a = 1 , b = - 2 , c = d - 3$ but from $f \left(2\right) = 0$ we extract

$0 = 4 a + 2 b + c = 4 - 4 + c \to c = 0 \to d = 3$

Finally

$f \left(x\right) = \frac{{x}^{2} - 2 x}{x - 3}$