Sketching Graphs (VCE SSCE Mathematical Methods): Quizzes

Numerator degree is one more than denominator

Where the denominator equals zero

$y = x - 1$

$(0, 3)$

No, because $x^2 + 3 \neq 0$ for real $x$

$1 - \frac{4}{(x + 1)^2}$

$x = 1$ and $x = -3$

Local minimum at $(1, 2)$

Vertical asymptote at $x = -1$ splits domain

$(-3, -6)$