Answer
Evaluate $h(x) = \frac{x^{2}+4}{5}$:
$h(2) = 1.6 $
$h(-2) = 1.6$
$h(a) = \frac{a^{2}+4}{5}$
$h(-x) = \frac{x^{2}+4}{5}$
$h(a-2) = \frac{a^{2}-4a+ 8}{5}$
$h(\sqrt x) = \frac{x+4}{5}$
Work Step by Step
x = 2
$h(2) = \frac{2^{2}+4}{5}$ ... square 2
$= \frac{4+4}{5}$ ... add numerator
$= \frac{8}{5}$ ... divide
$=1.6$
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x = -2
$h(-2) = \frac{(-2)^{2}+4}{5}$ ... square -2
$= \frac{4+4}{5}$ ... add numerator
$= \frac{8}{5}$ ... divide
$=1.6$
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x = a
$h(a) = \frac{a^{2}+4}{5}$
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x = -x
$h(-x) = \frac{(-x)^{2}+4}{5}$ ... square -x
$= \frac{x^{2}+4}{5}$
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x = a-2
$h(a-2) = \frac{(a-2)^{2}+4}{5}$ ... square (a-2)
$= \frac{(a^{2}-4a+4)+4}{5}$ ... add numerator
$= \frac{a^{2}-4a+ 8}{5}$
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x = $\sqrt x$
$h(\sqrt x) = \frac{(\sqrt x)^{2}+4}{5}$ ... square $\sqrt x$
$= \frac{x+4}{5}$
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