Answer
The $x$-intercept: at $(0,0)$
The $y$-intercept: $(0,0)$
The vertical asymptote: $x=-3$ and $x=1$
The horizontal asymptote: $y=0$
See graph
Work Step by Step
Equate the numerator to $0$ to find the $x$-intercept:
$$2x=0$$ $$x=0$$
Thus, the $x$-intercept is at $(0,0)$.
Set $f(x)=0$ to find the $x$-intercept:
$$0=\frac{2x}{x^2+2x-3}$$ $$x=0$$
Thus, the $y$-intercept is also at $(0,0)$.
The function is undefined when the denominator is $0$:
$$x^2+2x-3=0$$ $$(x+3)(x-1)=0$$ $$x+3=0$$ $$x=-3$$ $$x-1=0$$ $$x=1$$
The vertical asymptote are $x=-3$ and $x=1$.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is:
$$y=0$$
The sketch of the graph of the function is as shown.
