# How do you find vertical, horizontal and oblique asymptotes for (2x-1)/(x^2-7x+3)?

Feb 25, 2017

"vertical asymptotes at "x≈6.54,x≈0.46
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{let } f \left(x\right) = \frac{2 x - 1}{{x}^{2} - 7 x + 3}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} - 7 x + 3 = 0$

This quadratic does not factorise so use the $\textcolor{b l u e}{\text{quadratic formula}}$ to find the roots.

$\text{here "a=1,b=-7" and } c = 3$

• x=(-b+-sqrt(b^2-4ac))/(2a)

$x = \frac{7 \pm \sqrt{49 - 12}}{2} = \frac{7 \pm \sqrt{37}}{2}$

$\Rightarrow x = 6.54 \text{ or "x=0.46" to 2 decimal places}$

$\Rightarrow x = 6.54 \text{ and "x=0.46" are the asymptotes}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = \frac{\frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{7 x}{x} ^ 2 + \frac{3}{x} ^ 2} = \frac{\frac{2}{x} - \frac{1}{x} ^ 2}{1 - \frac{7}{x} + \frac{3}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0 - 0}{1 - 0 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator.This is not the case here (numerator-degree 1, denominator-degree 2) Hence there are no oblique asymptotes.
graph{(2x-1)/(x^2-7x+3) [-10, 10, -5, 5]}