= Solution
Use the order of the three <Bernstein basis> functions given in this part. Endpoint evaluation gives
$$
a_1=p(1),\qquad a_3=p(0),
$$
so both endpoint coefficients have absolute value at most one. At the midpoint,
$$
p(1/2)=\frac14a_1+\frac12a_2+\frac14a_3,
\qquad
a_2=2p(1/2)-\frac12[p(0)+p(1)].
$$
The <triangle inequality> therefore gives the <quadratic Bernstein coefficient bound>
$$
\boxed{|a_1|\leq1,\qquad |a_2|\leq3,\qquad |a_3|\leq1.}
$$
All three bounds are sharp. Indeed $p(x)=8x^2-8x+1=T_2(2x-1)$ has <uniform norm> one and coefficient vector $(1,-3,1)$ in this basis. The reversed ordering of endpoint Bernstein functions does not affect these symmetric endpoint bounds.
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