Question 835733
Q:
Let f(x) = 3x^2 - 4x. Find the constant k such that  f(x) = f(k - x) for all real numbers x.
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A:
f(x) = f(k - x)
{{{3x^2 - 4x}}} = {{{3(k - x)^2 - 4(k - x)}}}
{{{3x^2 - 4x}}} = {{{3(k^2 - 2kx + x^2) - 4k + 4x}}}
{{{3x^2 - 4x}}} = {{{3x^2 +(4 - 6k)x + (3k^2 - 4k)}}}
Therefore, 4 - 6k = -4 
                 k = {{{highlight(4/3)}}}