Given three clues, can we find the equation of the curve?

Given three clues, can we find the equation of the curve?

Question

The gradient of a curve is $ax + b$ at all points, where $a$ and $b$ are constants. Find the equation of the curve, given that it passes through the points $(0,4)$ and $(1,3)$, and that the tangent at $(1,3)$ is parallel to the line $y = x$.

Solution

Let’s say the equation of the curve is $y = f(x)$, so $y' = ax + b$. This means that $y= \int ax + b \:dx = \frac{ax^2}{2} + bx + c$ for some constant of integration, $c$. (This is a consequence of the fundamental theorem of calculus.)

As the curve passes through the point $(0,4)$ we know that $\frac{a \times 0^2}{2} + b \times 0 + c = 4 \implies c = 4,$ and as it passes through the point $(1,3)$, we also have that $\frac{a \times 1^2}{2} + b \times 1 + c = 3 \implies \frac{a}{2} + b + 4 = 3 \implies a + 2b = -2.$ Now the tangent to the curve at $(1,3)$ is parallel to the line $y = x$, which has gradient $1$, so $a + b = 4$ (since $y' = ax + b$).

We therefore have simultaneous equations in

$a$

and

$b$

, which are

$$\begin{align*} a + 2b &= -2, \\ a + b &= 1. \end{align*}$$

Subtracting, we have $b = -3$, and hence $a = 4$. The equation of the curve is therefore $y = 2x^2 - 3x + 4$.