Part 1 of 4 Evaluate the function \( h(x)=x^{4}+3 x^{2}+8 \) at the given values of the independent variable and simplify. \( \begin{array}{llll}\text { a. } h(2) & \text { b. } h(-1) & \text { c. } h(-x) & \text { d. } h(3 a)\end{array} \) \( \begin{array}{l}\text { a. } h(2)=\square \text { (Simplify your answer.) }\end{array} \)

To evaluate the function \( h(x) = x^{4} + 3x^{2} + 8 \) for the given values: **a.** For \( h(2) \): \[ h(2) = (2)^{4} + 3(2)^{2} + 8 = 16 + 3 \cdot 4 + 8 = 16 + 12 + 8 = 36 \] Thus, \( h(2) = 36 \). **b.** For \( h(-1) \): \[ h(-1) = (-1)^{4} + 3(-1)^{2} + 8 = 1 + 3 \cdot 1 + 8 = 1 + 3 + 8 = 12 \] Thus, \( h(-1) = 12 \). **c.** For \( h(-x) \): \[ h(-x) = (-x)^{4} + 3(-x)^{2} + 8 = x^{4} + 3x^{2} + 8 \] Thus, \( h(-x) = h(x) \) (the function is even). **d.** For \( h(3a) \): \[ h(3a) = (3a)^{4} + 3(3a)^{2} + 8 = 81a^{4} + 3 \cdot 9a^{2} + 8 = 81a^{4} + 27a^{2} + 8 \] Thus, \( h(3a) = 81a^{4} + 27a^{2} + 8 \). Summarizing all evaluations: \[ \begin{align*} a. & \quad h(2) = 36 \\ b. & \quad h(-1) = 12 \\ c. & \quad h(-x) = h(x) \\ d. & \quad h(3a) = 81a^{4} + 27a^{2} + 8 \end{align*} \]