Answer
$x=\dfrac{1}{5}$
Work Step by Step
We have: $y=5x^2-2x+1$
We differentiate both sides with respect to $x$.
$y^{\prime}=\dfrac{d}{dx} [5x^2-2x+1] \\=10x-2$
The tangent line to the graph will be horizontal when $ \dfrac{dy}{dx} =0$
Therefore, $10x -2=0 \implies x=\dfrac{1}{5}$
