Answer
$1$
Work Step by Step
f(x) = $x^{2}e^{-x}$
f'(x) = $-x^{2}e^{-x}+2xe^{-x}$
f(a) = $a^{2}e^{-a}$
f'(a) = $-a^{2}e^{-a}+2ae^{-a}$
y = f'(a)(x-a)+f(a)
y = $(-a^{2}e^{-a}+2ae^{-a})(x-a)+a^{2}e^{-a}$
origin (0, 0) replace x = 0 and y =0
0 = $(-a^{2}e^{-a}+2ae^{-a})(0-a)+a^{2}e^{-a}$
0 = $(a^{3}-2a^{2}+a^{2})e^{-a}$
0 = $(a^{3}-a^{2})e^{-a}$
0 = $(a-1)a^{2}e^{-a}$
a = 0, 1
a > 0 so a = 1
