Home › ACSEE (Form 6) › Advanced Mathematics › Functions › For $f(x)=\dfrac{x^2-2x-3}{x^2-4}$, the $x$-inte…
For $f(x)=\dfrac{x^2-2x-3}{x^2-4}$, the $x$-intercepts and the $y$-intercept are:
A$-1$ and $3$; $y=-\tfrac34$
B$1$ and $-3$; $y=\tfrac34$
C$-1$ and $3$; $y=\tfrac34$
D$-2$ and $2$; $y=\tfrac34$
Answer & Solution
Correct answer: C. $-1$ and $3$; $y=\tfrac34$
1. $x$-intercepts need $f(x)=0$, so set the numerator to zero.
2. $x^2-2x-3=(x-3)(x+1)=0$ gives $x=3$ and $x=-1$.
3. The $y$-intercept is $f(0)=\dfrac{0-0-3}{0-4}=\dfrac{-3}{-4}=\dfrac34$.
4. Both negatives cancel, so the $y$-intercept is positive.
5. $-2$ and $2$ are the asymptotes, not intercepts; the sign-flipped pair mis-factorises the numerator.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 6: Functions_
Related questions
A rule counting sign changes to bound the number of positive real zeros belongs to:A single point where a rational graph is undefined, shown by an open circle, is a hole or A horizontal line that a graph approaches as inputs grow without bound is a horizontal:A vertical line that a graph rushes towards but never crosses is a vertical:A polynomial with real coefficients has 2 plus 3i as a zero, so it must also have as a zerFor a rational zero, numerators come from factors of the constant term and denominators frA polynomial takes a negative value at 3 and a positive value at 4, so between them it musA graph crosses straight through the x-axis at a zero. That zero has multiplicity that is: