Rational functions and graphs — practice questions

1. Asymptotes:
   • Vertical Asymptote: Set denominator to zero: $x-1=0 \implies x=1$. The line $x=1$ is a vertical asymptote.
   • Oblique Asymptote: The degree of the numerator (2) is one more than the denominator (1). Perform polynomial long division:

         x - 1
     x-1 | x² - 2x - 3
         -(x² -  x)
         ---------
              -x - 3
             -(-x + 1)
             --------
                  -4
     

     So, $y = x-1 - \frac{4}{x-1}$. The oblique asymptote is $y=x-1$.
2. Intercepts:
   • y-intercept (set x=0): $y = \frac{0-0-3}{0-1} = 3$. The curve crosses the y-axis at $(0, 3)$.
   • x-intercepts (set y=0): Set numerator to zero: $x^2 - 2x - 3 = 0 \implies (x-3)(x+1)=0$. So, $x=3$ and $x=-1$. The curve crosses the x-axis at $(3, 0)$ and $(-1, 0)$.
3. Sketch:
   • Draw dashed lines for the asymptotes $x=1$ and $y=x-1$.
   • Plot the intercepts $(0, 3)$, $(3, 0)$, and $(-1, 0)$.
   • Sketch the curve. For $x > 1$, the term $-\frac{4}{x-1}$ is negative, so the curve is below the oblique asymptote. For $x < 1$, the term is positive, so the curve is above the oblique asymptote. The graph consists of two branches in the regions defined by the asymptotes, passing through the calculated points.

(i) Asymptotes: - Vertical Asymptotes: Set denominator to zero: $x^2 - 4 = 0 \implies (x-2)(x+2)=0$. The vertical asymptotes are $x=2$ and $x=-2$. - Horizontal Asymptote: The degree of the numerator (2) equals the degree of the denominator (2). The asymptote is $y = \frac{\text{leading coeff of num}}{\text{leading coeff of den}} = \frac{2}{1} = 2$. The horizontal asymptote is $y=2$.

(ii) Stationary Point: Use the quotient rule: $u=2x^2+5, v=x^2-4 \implies u'=4x, v'=2x$. $\frac{dy}{dx} = \frac{v u' - u v'}{v^2} = \frac{(x^2-4)(4x) - (2x^2+5)(2x)}{(x^2-4)^2}$ $\frac{dy}{dx} = \frac{4x^3 - 16x - 4x^3 - 10x}{(x^2-4)^2} = \frac{-26x}{(x^2-4)^2}$ Set $\frac{dy}{dx}=0 \implies -26x = 0 \implies x=0$. When $x=0$, $y = \frac{2(0)^2+5}{0^2-4} = -\frac{5}{4}$. The stationary point is $(0, -1.25)$. Since the denominator of $\frac{dy}{dx}$ is always positive, the sign of the derivative is determined by $-26x$. For $x<0$, $\frac{dy}{dx}>0$ (increasing). For $x>0$, $\frac{dy}{dx}<0$ (decreasing). This confirms $(0, -1.25)$ is a local maximum.

(iii) Sketch: - Draw asymptotes $x=2$, $x=-2$, and $y=2$. - Plot the stationary point (which is also the y-intercept) at $(0, -1.25)$. - Note that the numerator $2x^2+5$ is always positive, so there are no x-intercepts. - Check behaviour: As $x \to \infty$, $y = \frac{2x^2+5}{x^2-4} > 2$, so the curve approaches $y=2$ from above. As $x \to 2^+$, $y \to +\infty$. As $x \to 2^-$, $y \to -\infty$. By symmetry (it's an even function), the behaviour around $x=-2$ is mirrored. - Sketch the three branches of the curve: one U-shaped branch between the vertical asymptotes with a maximum at $(0, -1.25)$, and two branches in the top-left and top-right quadrants defined by the asymptotes.