Question

Set up an equation of a tangent to the graph of the following function.
At what points does the tangent to the graph of the function $y=\frac{x+2}{x-2}$ makes an angle of ${135}^{\circ }$ with the x-axis?

Answer 1

$y=\frac{x+2}{x-2}$

For ab angle of ${135}^0$, slope $=\tan 135=-1$

$y^{\prime }=-1$

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

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

$\Rightarrow x-2=\pm 4$

$x=4$ or $x=0$

$y\left(4\right)=\frac{4+2}{4-2}=3$

$y\left(0\right)=\frac{0+2}{0-2}=-1$

$\therefore$ at point $(0,-1)and(4,3)$, the tangent makes an angle of ${135}^0$ with $O_x$ axis$.