Question

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

Answer 1

Given:

`f(x)=(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)=(x^2+1x+4x+4)/(3x^2+3x) \\ =(x(x+1)+4(x+1))/(3x(x+1)) \\ =(\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)=(-1+4)/(3(-1)) \\ =(3)/(-3) \\ =-1 \end{gathered}`

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