Let's go step-by-step and solve the two parts:

1. Finding $f(x + 8)$

Given: $f(x) = x^2 - 2x - 6$

To find $f(x + 8)$, substitute $x + 8$ into the function $f(x)$: $f(x + 8) = (x + 8)^2 - 2(x + 8) - 6$

Expand the terms:

• $(x + 8)^2 = x^2 + 16x + 64$
• $-2(x + 8) = -2x - 16$

Now, substitute these back into the expression for $f(x + 8)$: $f(x + 8) = x^2 + 16x + 64 - 2x - 16 - 6$ Simplify: $f(x + 8) = x^2 + 14x + 42$

So, $f(x + 8) = x^2 + 14x + 42$

2. Finding $g\left( \frac{x}{4} \right)$

Given: $g(x) = \frac{5x - 1}{9x - 4}$

To find $g\left( \frac{x}{4} \right)$, substitute $\frac{x}{4}$ into the function $g(x)$: $g\left( \frac{x}{4} \right) = \frac{5\left( \frac{x}{4} \right) - 1}{9\left( \frac{x}{4} \right) - 4}$

Simplify the numerator and denominator:

• Numerator: $5 \times \frac{x}{4} = \frac{5x}{4}$, so the numerator becomes $\frac{5x}{4} - 1$
• Denominator: $9 \times \frac{x}{4} = \frac{9x}{4}$, so the denominator becomes $\frac{9x}{4} - 4$

Now combine the terms: $g\left( \frac{x}{4} \right) = \frac{\frac{5x}{4} - 1}{\frac{9x}{4} - 4}$

To eliminate the fractions, multiply both the numerator and the denominator by 4: $g\left( \frac{x}{4} \right) = \frac{5x - 4}{9x - 16}$

So, $g\left( \frac{x}{4} \right) = \frac{5x - 4}{9x - 16}$

Final Answers:

1. $f(x + 8) = x^2 + 14x + 42$
2. $g\left( \frac{x}{4} \right) = \frac{5x - 4}{9x - 16}$

Would you like a detailed explanation of any part of this process or have further questions?

Here are 5 related questions:

1. How do you evaluate functions with more complex expressions like $f(x - 2)$?
2. What happens if we evaluate $g(2x)$ for the given function $g(x)$?
3. Can you explain how to simplify algebraic fractions like $\frac{5x}{4} - 1$?
4. What if you needed to compute $f(x^2)$? How would that change the approach?
5. How do you handle cases where the substitution leads to undefined values, such as $g(x)$ when the denominator becomes zero?

Tip: Always simplify the expression fully after substitution to avoid missing hidden terms that might cancel out or combine.