The function provided is $h(x) = \frac{-2x}{x + 1}$.

Let's calculate the values for each part:

(a) $h(5)$

$h(5) = \frac{-2(5)}{5 + 1} = \frac{-10}{6} = \frac{-5}{3}$ So, $h(5) = \frac{-5}{3}$.

(b) $h(3)$

$h(3) = \frac{-2(3)}{3 + 1} = \frac{-6}{4} = \frac{-3}{2}$ So, $h(3) = \frac{-3}{2}$.

(c) $h(0)$

$h(0) = \frac{-2(0)}{0 + 1} = \frac{0}{1} = 0$ So, $h(0) = 0$.

(d) $h\left(\frac{1}{2}\right)$

$h\left(\frac{1}{2}\right) = \frac{-2\left(\frac{1}{2}\right)}{\frac{1}{2} + 1} = \frac{-1}{\frac{3}{2}} = \frac{-1 \times 2}{3} = \frac{-2}{3}$ So, $h\left(\frac{1}{2}\right) = \frac{-2}{3}$.

Summary:

• (a) $h(5) = \frac{-5}{3}$
• (b) $h(3) = \frac{-3}{2}$
• (c) $h(0) = 0$
• (d) $h\left(\frac{1}{2}\right) = \frac{-2}{3}$

Would you like further explanation on any part of the process?

Here are 5 related questions to explore:

1. What is the domain of the function $h(x) = \frac{-2x}{x+1}$?
2. What is the range of $h(x)$ based on its expression?
3. Can you find the vertical asymptote of the function $h(x)$?
4. How would you calculate $h(x)$ when $x = -1$?
5. How does $h(x)$ behave as $x$ approaches infinity?

Tip: Always simplify fractions by dividing both the numerator and denominator by their greatest common factor!