7. Find the coordinates of the hole of $$ f(x)=\frac{(x+4)(x-2)}{(x-2)(x-1)} $$

Question

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To find the coordinates of the hole for the function:

$$ f(x) = \frac{(x+4)(x-2)}{(x-2)(x-1)} $$

**Step 1: Simplify the Function**

Notice that the factor $$(x-2)$$ appears in both the numerator and the denominator. We can simplify the function by canceling this common factor:

$$ f(x) = \frac{(x+4)\cancel{(x-2)}}{\cancel{(x-2)}(x-1)} = \frac{x + 4}{x - 1} \quad \text{(for } x \neq 2 \text{)} $$

**Step 2: Identify the Hole**

A hole occurs at the $$x$$-value that makes the canceled factor zero. In this case, the canceled factor is $$(x - 2)$$, so:

$$ x = 2 $$

**Step 3: Find the $$y$$-Coordinate of the Hole**

To find the corresponding $$y$$-value, substitute $$x = 2$$ into the simplified function:

$$ f(2) = \frac{2 + 4}{2 - 1} = \frac{6}{1} = 6 $$

**Conclusion**

The hole of the function is located at the point $$(2, 6)$$.

**Answer:**  
After simplifying, the hole is at the point with coordinates ( 2, 6 ).
The hole is at the point (2, 6).

7. Find the coordinates of the hole of \( f(x)=\frac{(x+4)(x-2)}{(x-2)(x-1)} \)