Question

Find coordinate point in simplest form, and find all holes of following function.

Answer 1

Given:

$f(x)=\frac{x^2+5x+4}{3x^2+3x\:}$

To find the hole and its coordinate point:

Simplifying the function we get,

$\begin{gathered} f\mleft(x\mright)=\frac{x^2+1x+4x+4}{3x^2+3x} \\ =\frac{x(x+1)+4(x+1)}{3x(x+1)} \\ =\frac{\mleft(x+1\mright)(x+4)}{3x(x+1)\: } \end{gathered}$

We can cancel (x+1) on both numerator and denominator.

So, the hole is at x = -1.

The coordinate point would be,

$\begin{gathered} f(-1)=\frac{-1+4}{3(-1)} \\ =\frac{3}{-3} \\ =-1 \end{gathered}$

Thus, the coordinate point is (-1, -1).