Can we find a curve from its gradient function?

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The first part of the question tells us that \[ \frac{dy}{dx} = 4x - 3. \] so we know that \[\begin{equation}\label{eq:indefinite-eq-for-y} y = \int \frac{dy}{dx} \:dx = 2x^2 - 3x + c \end{equation}\]

for some constant \(c\). (This is a consequence of the fundamental theorem of calculus.)

The curve is a parabola and since the coefficient of \(x^2\) is positive we know its single stationary point is a minimum. We are also given that the \(y\)-value at the minimum is \(-\dfrac{1}{8}\).

This minimum occurs where \(\dfrac{dy}{dx} = 0 \implies x = \dfrac{3}{4}\). Substituting into equation \(\eqref{eq:indefinite-eq-for-y}\), we have that \[\begin{align*} -\frac{1}{8} = 2 \left( \frac{3}{4} \right)^2 - 3 \cdot \frac{3}{4} + c &\implies c = -\frac{1}{8} - \frac{9}{8} + \frac{9}{4} \\ &\implies c = 1. \end{align*}\]

So we know that the equation of the curve is \(y = 2x^2 - 3x + 1\).

We can factorise the expression for \(y\) as \[ y = 2x^2 - 3x + 1 = (2x - 1)(x - 1). \] We thus have that the equation \(y = 0\) has two roots, at \(x = \dfrac{1}{2}\) and \(1\). The graph has a minimum at the point \(\left( \dfrac{3}{4}, -\dfrac{1}{8} \right)\), and finally, the \(y\)-intercept of the function is \(1\).