Answer
$$\left( {\frac{4}{9},\frac{{20}}{9}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 9{x^2} - 8x + 4 \cr
& {\text{find the derivative of the function}} \cr
& f'\left( x \right) = {D_x}\left( {9{x^2} - 8x + 4} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& f'\left( x \right) = 9\left( {2x} \right) - 8\left( 1 \right) + 0 \cr
& f'\left( x \right) = 18x - 8 \cr
& \cr
& {\text{Find the points of }}x{\text{ where the slope of the tangent line is 0}}{\text{, set }}f'\left( x \right) = 0 \cr
& {\text{Then}} \cr
& 0 = 18x - 8 \cr
& 18x - 8 = 0 \cr
& {\text{solve for }}x \cr
& x = \frac{8}{{18}} = \frac{4}{9} \cr
& \cr
& {\text{Find the point on the graph where the slope is 0}}{\text{, evaluate }}f\left( x \right){\text{ at }}x = \frac{4}{9} \cr
& f\left( {\frac{4}{9}} \right) = 9{\left( {\frac{4}{9}} \right)^2} - 8\left( {\frac{4}{9}} \right) + 4 \cr
& f\left( {\frac{4}{9}} \right) = \frac{{16}}{9} - \frac{{32}}{9} + 4 \cr
& f\left( {\frac{4}{9}} \right) = \frac{{16 - 32 + 36}}{9} \cr
& f\left( {\frac{4}{9}} \right) = \frac{{20}}{9} \cr} $$
