To determine where the graph of the function [tex]\( f(x)=(x-1)\left(x^2+x-20\right) \)[/tex] crosses the [tex]\( x \)[/tex]-axis, we need to find the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
When the function crosses the [tex]\( x \)[/tex]-axis, the [tex]\( y \)[/tex]-coordinate (i.e., [tex]\( f(x) \)[/tex]) is 0. So, we set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex].
The function given is:
[tex]\[ f(x) = (x-1)(x^2+x-20) \][/tex]
First, we set the function equal to zero:
[tex]\[ (x-1)(x^2 + x - 20) = 0 \][/tex]
For this product to be zero, at least one of the factors must be zero. We will solve for [tex]\( x \)[/tex] by setting each factor to zero separately.
1. [tex]\( x - 1 = 0 \)[/tex]
[tex]\[ x = 1 \][/tex]
2. [tex]\( x^2 + x - 20 = 0 \)[/tex]
We need to solve the quadratic equation [tex]\( x^2 + x - 20 = 0 \)[/tex]. This can be factored as follows:
[tex]\[ x^2 + x - 20 = (x + 5)(x - 4) \][/tex]
By setting each factor equal to zero, we have:
[tex]\[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \][/tex]
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
So, the solutions to the equation [tex]\( (x-1)(x^2+x-20) = 0 \)[/tex] are:
[tex]\[ x = 1, \, x = -5, \, x = 4 \][/tex]
Thus, the ordered pairs where the graph of [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis (where [tex]\( y = 0 \)[/tex]) are:
[tex]\[ (1, 0), \, (-5, 0), \, (4, 0) \][/tex]
Out of the provided options, the correct ordered pair is:
[tex]\[ (-5, 0) \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{(-5,0)} \][/tex]
