## Turning Points of a Quotient of Quadratics

What are the turning points, if any of the graph
$y= \frac{x^2+5x+4}{x^2-5x+4}$
?
We can find the turning points by solving
$\frac{dy}{dx}=0$
.
Differentiating
$\frac{dy}{dx}= \frac{(x^2-5x+4)(2x-5)}-(x^2+5x+4)(2x+5){(x^2-5x+4)^2}= \frac{-10x^2+40}{(x^2-5x+4)^2}$

Setting
$\frac{dy}{dx}=\frac{-10x^2+40}{(x^2-5x+4)^2}$
implies
$10x^2+40=0 \rightarrow x^2 =4 \rightarrow x = \pm 2$
.