Consider the family of functions of domain $\mathbb{R}\setminus \{a\}$, $\frac{2x^2+x-3}{x-a}$.Find a so that there are no vertical asymptotes

Question

Consider the family of functions of domain $\mathbb{R}\setminus \{a\}$ defined by $f(x) = \frac{2x^2+x-3}{x-a}$.

Determine $a$ so that $f$ has no vertical asymptotes.

I know that one way to solve this is by factoring ${2x^2+x-3}$ into $(x-a)$ times something.

I'm not very good at factoring so I messed around a bit to see if I could find the solution.

My book says the solutions are x = 1 and x = -1.5. I looked at the graph of the function an realised these were the zeros of $2x^2+x-3$.

With a little help from Wolfram I found out how to factor $2x^2+x-3$ : $(x-1)(2x+3)$.

And so $a = 1$

To get the other solution I tried this:

$${2x^2+x-3} = (x-a)(y+z)$$

$x \cdot y = 2x^2 \Leftrightarrow y = 2x$

$${2x^2+x-3} = (x-a)(2x+z) = 2x^2+xz-2xa-az = 2x^2+x(z-2a) -az$$

$az = 3$

$x(z-2a) = x \Leftrightarrow z-2a = 1$

Then I put the above two conclusions into a system and calculated them and got $a = 1 \lor a = -1.5$, no problem there.

My questions:

1. Why does $a$ take the values of the zeros of $2x^2+x-3$? Is it just a coincidence?

   2. Is there a simpler way of solving this?

Answer 1

If $a$ is a real zero of $f(x) = 2x^2+x - 3$ then it is possible to factor $f(x)=(x-a)(x-b)$ into a product of linear polynomials, whence

$$ q(x):=\frac{2x^2+x-3}{x-a} = x-b, \quad (x \in \mathbb{R} \backslash \{a\})$$

can be seen to have a finite limit as $x \to a$, and cannot have a vertical asymptote there.

However, if $a$ is not a real zero of $f(x)$, then as $x \to a$, the polynomial $f(x)$ tends to some finite nonzero value $f(a)$ while the linear polynomial $x-a$ gets arbitrarily small in absolute value. In this case, $q(x)$ will get arbitrarily large in absolute value as $x \to a$

Answer 2

Note that $$f(x) = \frac{2x^2+x-3}{x-a} = \frac{(2x+3)(x-1)}{x-a} = \frac{2(x+1.5)(x-1)}{x-a}.$$ Thus if $a=-1.5$ or $a=1$, then $f(x)$ simplifies to a linear polynomial over the domain $\mathbb{R} \setminus \{a\}$, and linear polynomials do not have vertical asymptotes.