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\(x^2 - 16x + 3 = 0\\ \color{blue} x\in \{0,16\}\)
The length of the section on the x-axis is 16.
\(f(x)=x^2 - 16x + 3 \\ f'(x)=2x-16=0\\ \color{blue} x=8\\ 8^2 - 16\cdot 8 + 3 =\color{blue} -61 \)
The length of the section on the axis of symmetry is 61.
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