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Algebra Level 4

$f(x)=\frac{2 x^2+3 x-2}{x^2-3 x-4}\times\frac{(x-4)^2}{(x^2-4)(x-3)}\times\frac{x^2-x-2}{x-4}\div\frac{2x^2-x}{x^2-3x}$

Consider this expression above. Which of the following statements is correct?

A: A graph of $y=f(x)$ will intersect with the graph $y=x$ at the point $(1,1)$ .

B: $f(x)$ can be simplified as $Mx+N$ where $M$ and $N$ are non-zero constants and $N=1$ .

C: A graph of $y=f(x)$ has a $y$ -intercept where $y=1$ .

D: $f(x)$ is undefined for exactly 4 values of $x$ .

B C D A

1 solution

Wee Xian Bin
Jul 4, 2016

$f(x)=\frac{2 x^2+3 x-2}{x^2-3 x-4}\times\frac{(x-4)^2}{(x^2-4)(x-3)}\times\frac{x^2-x-2}{x-4}\div\frac{2x^2-x}{x^2-3x}$

$=\frac{(2x-1)(x+2)}{(x+1)(x-4)}\times\frac{(x-4)^2}{(x^2-4)(x-3)}\times\frac{(x-2)(x+1)}{x-4}\times\frac{(x-3)(x)}{(2x-1)(x)}$

$=\frac{(2x-1)(x+2) (x-4)^2 (x-2)(x+1)(x-3)(x)}{(x+1)(x-4)(x-2)(x+2)(x-3)(x-4)(2x-1)(x)}$

$=\frac{(2x-1)(x+2) (x-4)^2 (x-2)(x+1)(x-3)(x)}{(2x-1)(x+2) (x-4)^2 (x-2)(x+1)(x-3)(x)}=1$

$f(x)$ is undefined for $x=-2,-1,0,\frac{1}{2},2,3,4$ because the denominator for $f(x)$ will then become zero.

A: Since $f(1)=1$ is defined, the graph will intersect with $y=x$ .

B: $M$ will then be zero, hence false.

C: Since $f(0)$ is undefined, the graph will not intersect with the $y$ -axis.

D: False, it should be 7 instead.