# Let f(x) \frac{2x^2-4}{x^2-3}. a. State the y intercept b. State the x intercept(s) c. State the...

## Question:

Let {eq}f(x) \frac{2x^2-4}{x^2-3}. {/eq}

a. State the {eq}y {/eq} intercept

b. State the {eq}x {/eq} intercept(s)

c. State the vertical asymptote(s)

d. State the horizontal asymptote

## Rational functions

This example illustrates many of the properties of a rational function, in which the numerator and denominator are each polynomials.

The example shows how to determine horizontal and vertical asymptotes, and also the intercepts with the axes.

Part a

The y intercept occurs when x = 0, so:

{eq}f(0) = \dfrac{2(0)^2 - 4}{0^2 - 3} = \dfrac{4}{3}. {/eq}

Hence the y intercept is at {eq}(0, \dfrac{4}{3}). {/eq}

Part b

The x intercepts occur when y = 0, giving:

{eq}0 = \dfrac{2x^2 - 4}{x^2 - 3} \\ \Rightarrow 2x^2 - 4 = 0 \\ \Rightarrow x^2 = 2 \\ \Rightarrow x = \pm \sqrt{2} {/eq}

Hence the x intercepts are at {eq}(\pm \sqrt{2}, 0). {/eq}

Part c

The vertical asymptotes of a rational function occur when the denominator is zero; hence:

{eq}x^2 - 3 = 0 \\ \Rightarrow x = \pm \sqrt{3}. {/eq}

So these are the vertical asymptotes.

Part d

For a rational function in which the numerator and denominator are both polynomials, the horizontal asymptote can be determined by considering just the highest powers of x in the numerator and denominator. Hence:

{eq}f(x) = \dfrac{2x^2 - 4}{x^2 - 3} \approx \dfrac{2x^2}{x^2} = 2. {/eq}

this gives the horizontal asymptote as {eq}y = 2. {/eq}