The tangent to the curve $$y = x^2 - 5x + 5$$, parallel to the line $$2y = 4x + 1$$, also passes through the point:

First, we recall that the slope of any straight line written in the form $$ax+by+c=0$$ is given by the formula $$m=-\dfrac{a}{b}.$$

The line whose equation is $$2y=4x+1$$ can be rewritten as $$4x-2y+1=0.$$ Using the formula above, its slope comes out as $$m=-\dfrac{4}{-2}=2.$$ Hence every line parallel to $$2y=4x+1$$ must also have slope $$2.$$

Now we turn to the curve $$y=x^2-5x+5.$$ For a tangent drawn to this curve, the slope equals the derivative of $$y$$ with respect to $$x$$ at the point of contact. We compute

$$\dfrac{dy}{dx}=2x-5.$$

Because the desired tangent is parallel to the given line, its slope must be $$2.$$ Therefore we must have

$$2x-5 = 2.$$

Solving for $$x$$ step by step:

$$2x = 2 + 5,$$

$$2x = 7,$$

$$x = \dfrac{7}{2}.$$

We substitute this $$x$$-value back into the curve to find the corresponding $$y$$-coordinate:

$$y = \left(\dfrac{7}{2}\right)^2 - 5\left(\dfrac{7}{2}\right) + 5.$$

We now simplify term by term.

First square: $$\left(\dfrac{7}{2}\right)^2 = \dfrac{49}{4}.$$

Next product: $$5\left(\dfrac{7}{2}\right) = \dfrac{35}{2}.$$

Express every fraction with denominator $$4$$ so that we can combine them easily:

$$\dfrac{35}{2} = \dfrac{70}{4}, \quad 5 = \dfrac{20}{4}.$$

Hence

$$y = \dfrac{49}{4} - \dfrac{70}{4} + \dfrac{20}{4} = \dfrac{49 - 70 + 20}{4} = \dfrac{-1}{4}.$$

Thus the point of contact of the tangent and the curve is $$\left(\dfrac{7}{2}, -\dfrac{1}{4}\right).$$

Next, we write the equation of the tangent line using the point-slope form $$y - y_1 = m(x - x_1).$$ Here $$(x_1,y_1)=\left(\dfrac{7}{2}, -\dfrac{1}{4}\right)$$ and $$m=2.$$ Substituting, we get

$$y - \left(-\dfrac{1}{4}\right) = 2\left(x - \dfrac{7}{2}\right).$$

Simplifying first the left side and then the right side:

$$y + \dfrac{1}{4} = 2x - 2\cdot\dfrac{7}{2},$$

$$y + \dfrac{1}{4} = 2x - 7.$$

Subtracting $$\dfrac{1}{4}$$ from both sides gives the tangent in slope-intercept form:

$$y = 2x - 7 - \dfrac{1}{4} = 2x - \dfrac{29}{4}.$$

To discover which of the four given points lies on this line, we substitute the coordinates of each option into the equation $$y = 2x - \dfrac{29}{4}.$$

Option A: $$\left(\dfrac{1}{4}, \dfrac{7}{2}\right).$$ Put $$x=\dfrac{1}{4}$$:

Right-hand side $$=2\cdot\dfrac{1}{4}-\dfrac{29}{4}=\dfrac{1}{2}-\dfrac{29}{4}=\dfrac{2-29}{4}=-\dfrac{27}{4}.$$ Left-hand side $$=\dfrac{7}{2}=\dfrac{14}{4}.$$ Since $$\dfrac{14}{4}\neq-\dfrac{27}{4},$$ this point does not lie on the line.

Option B: $$\left(\dfrac{7}{2},\dfrac{1}{4}\right).$$ Put $$x=\dfrac{7}{2}$$:

Right-hand side $$=2\cdot\dfrac{7}{2}-\dfrac{29}{4}=7-\dfrac{29}{4}=\dfrac{28-29}{4}=-\dfrac{1}{4}.$$ Left-hand side $$=\dfrac{1}{4}.$$ Since $$\dfrac{1}{4}\neq-\dfrac{1}{4},$$ this point does not lie on the line.

Option C: $$\left(-\dfrac{1}{8},7\right).$$ Put $$x=-\dfrac{1}{8}$$:

Right-hand side $$=2\cdot\left(-\dfrac{1}{8}\right)-\dfrac{29}{4}=-\dfrac{1}{4}-\dfrac{29}{4}=-\dfrac{30}{4}=-\dfrac{15}{2}.$$ Left-hand side $$=7.$$ Since $$7\neq-\dfrac{15}{2},$$ this point does not satisfy the equation.

Option D: $$\left(\dfrac{1}{8},-7\right).$$ Put $$x=\dfrac{1}{8}$$:

Right-hand side $$=2\cdot\dfrac{1}{8}-\dfrac{29}{4}=\dfrac{1}{4}-\dfrac{29}{4}=-\dfrac{28}{4}=-7.$$ Left-hand side $$=-7.$$ Both sides are equal, so this point lies exactly on the tangent.

Only Option D satisfies the equation of the tangent line.

Hence, the correct answer is Option D.