# How do you prove that the function: T(x) = 1 / (abs(x-2)-x^2) is continuous between [1.5,8]?

Dec 10, 2016

Only ( infinite ) discontinuities at $x = - 2 \mathmr{and} x = 1$. The vertical asymptotes are $x = 1 \mathmr{and} x = - 2$. The horizontal asymptote is x = 0. y-intercept is $\frac{1}{2}$.

#### Explanation:

Besides the graph, I have provided some data for clarity. Further, the

given function is representing the couple

$T = \frac{1}{x - 2 - {x}^{2}} , x \ge 2$ and

$T = \frac{1}{2 - x - {x}^{2}} , x \le 2.$

graph{y(x^2-|x-2|)+1=0 [-10, 10, -5, 5]}