Answer:
(1, -3)
Explanation:
The function contains a hole if there is a value of x that makes the numerator and denominator equal to 0.
The values that make the numerator equal to 0 can be calculated solving the following equation
x² + x - 2 = 0
(x + 2)(x - 1) = 0
Then
x + 2 = 0
x + 2 - 2 = 0 - 2
x = -2
or
x - 1 = 0
x - 1 + 1 = 0 + 1
x = 1
So, the values are x = -2 and x = 1
In the same way, the values that make the denominator equal to 0 are
x² - 3x + 2 = 0
(x - 2)(x - 1) = 0
Then
x - 2 = 0
x - 2 + 2 = 0 + 2
x = 2
or
x - 1 = 0
x - 1 + 1 = 0 + 1
x = 1
So, the values are x = 2 and x = 1.
It means that there is a hole in x = 1. To know the coordinates for the location of the hole, we need to simplify the initial expression as:
[tex]\begin{gathered} f(x)=\frac{x^2+x-2}{x^2-3x+2} \\ f(x)=\frac{(x+2)(x-1)}{(x-2)(x-1)} \\ f(x)=\frac{(x+2)}{(x-2)} \end{gathered}[/tex]
Now, we can replace x by 1 to get:
[tex]f(1)=\frac{1+2}{1-2}=\frac{3}{-1}=-3[/tex]
So, the coordinates for the location of the hole are (1, -3)
