[tex]\begin{gathered} D=(-\infty,-4)U(-4,5)U(5,\infty) \\ \frac{x^2+2x}{x^2-9x+20} \end{gathered}[/tex]
c) For this expression, let's start by rewriting it as a rational function:
[tex]\begin{gathered} y=\frac{x}{x+4}\cdot\frac{x+2}{x-5} \\ y=\frac{x(x+2)_{}}{(x+4)(x-5)} \end{gathered}[/tex]
So to find out the Domain, we need to determine for which values do this function is defined. So looking at the denominator we can write out:
[tex]\begin{gathered} x+4\ne0 \\ x\ne-4 \\ x-5\ne0 \\ x\ne5 \\ D=(-\infty,-4)U(-4,5)U(5,\infty) \end{gathered}[/tex]
Note that if we plug into the denominator x=5 or x=-4 the function becomes undefined for the denominator would become equal to 0. For all values but -4, and 5 this function is defined.
We could draw the diagram to check the intersections.
2) Now, let's perform the product of that ratios, as indicated.
[tex]\frac{x}{x-4}\cdot\frac{x+2}{x-5}=\frac{x(x+2)}{(x-4)(x-5)}=\frac{x^2+2x}{x^2-5x-4x+20}=\frac{x^2+2x}{x^2-9x+20}[/tex]
Note that when we multiply ratios we multiply both numerator and denominator simultaneously. We can leave it in its factored form as well.
3) Hence, the answer is:
[tex]\begin{gathered} D=(-\infty,-4)U(-4,5)U(5,\infty) \\ \frac{x^2+2x}{x^2-9x+20} \end{gathered}[/tex]
