Question

Investigate the behaviour of the following and construct its graph :

$y=\frac{x^2-3x+4}{x^2+3x+4}$

• A. Has maximum at $x=4$ and minimum at $x=-4$.
• B. Has maximum at $x=-4$ and minimum at $x=4$.
• C. Has maximum at $x=2$ and minimum at $x=-2$.
• D. Has maximum at $x=-2$ and minimum at $x=2$.

Answer 1

The correct option is D Has maximum at $x=-2$ and minimum at $x=2$.

Given, $y=\frac{x^2-3x+4}{x^2+3x+4}$
Now At $x=0$, we get $y=1$
Hence, $y=f\left(x\right)$ cuts the y axis at y=1.
Now $y=0$ implies $\frac{x^2-3x+4}{x^2+3x+4}=0$
$\therefore x^2-3x+4=0$
$\therefore D$ $=9-16$ $=-7$
Hence, $D<0$, complex, non-real roots.
Hence, $y=f\left(x\right)$ does not cut the $x$-axis.
Now $y^{\prime }$ $=\frac{(2x+3)(x^2-3x+4)-(2x-3)(x^2+3x+4)}{(x^2+3x+4{)}^2}$
$=\frac{-6x^2+24}{(x^2+3x+4{)}^2}$
$=0$ ... for critical points.
Hence, $-6x^2+24=0$
$\Rightarrow x^2=4$
$\Rightarrow x=\pm 2$
Now $f\left(2\right)=\frac{1}{7}$ and $f(-2)=7$
Hence, it has a maximum at $x=-2$ and minimum at $x=2$.