Review question

Ref: R5768

## Solution

A function, $y$, of $x$ has gradient $4x - 3$ at the point $(x,y)$ and the minimum value of the function is $-\tfrac{1}{8}$. Express $y$ in terms of $x$

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

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 $y = 2x^2 - 3x + 1$.

…and sketch the graph of $y$ against $x$.

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 $1$.