I am tasked with sketching the following rational function$$\frac{\left(x+1\right)\left(3x+2\right)}{\left(x+1\right)^{2}}$$
I have found that the two asymptotes are: $y=3$ and $x=-1$ and there are no turning points for this function.
To find the $x$ intercept: $$0=\frac{\left(x+1\right)\left(3x+2\right)}{\left(x+1\right)^{2}}$$
Hence$$3x^2-x-2=0$$ $$x_1=1\space \space \space x_2=-\frac{2}{3}$$
Simularly to find the $y$ intercept I did the following:$$\frac{\left(0+1\right)\left(3(0)+2\right)}{\left(0+1\right)^{2}}$$ $$y=-2$$
However plotting this into a graphing software shows that the $y$ intercept is $y=2$ and $x$ intercept is only $x=-\frac{2}{3}$ and not $x=1$.
I am extremely confused, why are the intercepts incorrect?
Note that$$\frac{(x+1)(3x+2)}{(x+1)^2}=\frac{3x+2}{x+1}\tag1$$and that therefore the only zero of your function is located at $-\frac23$. On the other hand, if $x=0$, then $(1)$ is equal to $2$.
Also, using your approach, there is no need to solve a quadratic equation using the quadratic formula, since $(3x+2)(x+1)=0$ if and only if $x=-\frac23$ or $x=-1$. And $-1$ does not belong to the domain of your function.