Question 280211
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If 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(x)\ =\ x^2\ -\ 4x\ +\ 5]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(a)\ =\ (a)^2\ -\ 4(a)\ +\ 5]


So

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(4)\ =\ (4)^2\ -\ 4(4)\ +\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(-4)\ =\ (-4)^2\ -\ 4(-4)\ +\ 5]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(0)\ =\ (0)^2\ -\ 4(0)\ +\ 5]


Congratulations, you get to do your very own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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