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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.

Answer and Explanation:

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}


Learn more about this topic:

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Rational Function: Definition, Equation & Examples

from GMAT Prep: Help and Review

Chapter 10 / Lesson 11
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