To determine whether the graph of the function [tex]\( g(x) = \frac{x+10}{x^3 + 5x^2 - 50x} \)[/tex] has any holes, we need to follow several steps to simplify the rational function and identify any factors that would introduce holes.
1. Factor the numerator and the denominator:
The numerator of [tex]\( g(x) \)[/tex] is [tex]\( x + 10 \)[/tex].
The denominator of [tex]\( g(x) \)[/tex] is [tex]\( x^3 + 5x^2 - 50x \)[/tex].
First, factor the denominator:
[tex]\[
x^3 + 5x^2 - 50x = x(x^2 + 5x - 50)
\][/tex]
Next, further factor [tex]\( x^2 + 5x - 50 \)[/tex]. We look for two numbers that multiply to [tex]\(-50\)[/tex] and add to [tex]\(5\)[/tex]:
[tex]\[
x^2 + 5x - 50 = (x+10)(x-5)
\][/tex]
Thus, the fully factored form of the denominator is:
[tex]\[
x^3 + 5x^2 - 50x = x(x + 10)(x - 5)
\][/tex]
2. Simplify the rational function:
Now the function [tex]\( g(x) \)[/tex] can be written as:
[tex]\[
g(x) = \frac{x+10}{x(x+10)(x-5)}
\][/tex]
We see that the numerator [tex]\( x + 10 \)[/tex] and the denominator [tex]\( x + 10 \)[/tex] share a common factor. We can cancel out these terms, but we need to be careful with the values at which this factor is zero, as they indicate potential "holes."
After canceling the common factor, we get:
[tex]\[
g(x) = \frac{1}{x(x-5)}, \quad x \neq -10
\][/tex]
The condition [tex]\( x \neq -10 \)[/tex] indicates a hole at [tex]\( x = -10 \)[/tex].
3. Identifying the hole:
The function is already showing a simplified form that excludes [tex]\( x = -10 \)[/tex]. Thus, the original function has a hole where [tex]\( x + 10 = 0 \)[/tex].
4. Conclusion:
Therefore, the value of [tex]\( x \)[/tex] at which the graph of [tex]\( g(x) = \frac{x+10}{x^3 + 5x^2 - 50x} \)[/tex] has a hole is:
[tex]\[
x = -10
\][/tex]
So, the correct answer is:
[tex]\[
x = -10
\][/tex]
