Graph the function $f(x) = \dfrac{x^2+2x}{x^2-4x+3}$, labelling all intercepts and asymptotes.
I found the vertical and horizonal assymptotes myself.I plugged the problem into an online graph calculator to see how it would look. Why in the top two quartiles and not the bottom two and in between the vertical asymptotes going up but not down?
My work:
Vertical Asymptotes: The process here is
- Factor
- Cancel
- Set denominator $= 0$
$$\frac{x(x+2)}{(x-1)(x-3)}=0 \\ \implies \text{VA's:}\quad x= 1,\quad x=3$$
Horizontal Asymptotes:
$$\frac{x^2}{x^2}=1 \\ \implies \text{HA:}\quad y=1$$
Here's the graph:
If the formula for a function can be expressed as a ratio of two polynomials, then the function can change from being positive to negative or vice versa only at a vertical asymptote or at a place where it crosses the $x$-axis.
In your example, the asymptotes occur at $x = 1$ and $x = 3$. The zeros occur at $x = 0$ and $x = -2$ (look at the numerator). So on each of these intervals, the graph of your function must lie above the $x$-axis for the whole interval, or below the axis for the whole interval: $(−∞, -2)$, $(-2,0)$, $(0,1)$, $(1, 3)$, and $(3, +∞).$ We can pick any convenient value in each interval to see where it lies. This allows us to sketch in some more detail.
In addition, we need to look at the limits at infinity as indicated in the comments to see whether the graph approaches the horizontal asymptote from above or from below.
We can also set the expression equal to 1 and solve for x to find out where the graph crosses the horizontal asymptote.