Which graph best represents the function y = \frac{2x^2}{1-x^2} ? A - HSC - SSCE Mathematics Extension 1 - Question 5 - 2017 - Paper 1

Vertical asymptotes occur where the denominator is zero. So, we set the denominator equal to zero: [ 1 - x^2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 ] Thus, vertical asymptotes are at ( x = -1 ) and ( x = 1 ).

To find horizontal asymptotes, we examine the limit of ( y ) as ( x ) approaches infinity: [ \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{2x^2}{1-x^2} = \lim_{x \to \infty} \frac{2}{-1} = -2 ] This indicates a horizontal asymptote at ( y = -2 ).

To find the y-intercept, we evaluate the function at ( x = 0 ): [ y(0) = \frac{2(0)^2}{1-(0)^2} = 0 ] Thus, the y-intercept is at ( (0,0) ). For x-intercepts, set ( y = 0 ): [ \frac{2x^2}{1-x^2} = 0 \Rightarrow 2x^2 = 0 \Rightarrow x = 0 ] The x-intercept is also at ( (0,0) ).

As ( x ) approaches the vertical asymptotes ( x = -1 ) and ( x = 1 ), the function will approach ( +\infty ) or ( -\infty ). The behavior near these points will help determine which graph correctly represents the function.

Based on the population of asymptotes and intercepts:

• The vertical asymptotes at ( x = -1 ) and ( x = 1 ) suggest the graph should diverge at these x-values.
• The horizontal asymptote at ( y = -2 ) suggests the function approaches this value as ( x ) approaches infinity.
• The only graph that meets these criteria is D, as it shows the correct asymptotic behavior.

The graph that best represents the function is D.