Finding tangents, computing turning points and establishing extreme values of functions all seem like problems that require calculus for us to solve. But in fact there are often techniques which involve nothing more than some geometric thinking and elementary algebra. Imagine we were in a world where calculus hadn’t yet been invented (or that you haven’t learned any calculus yet)—how would we go about maximising and minimising functions, finding tangents and so on?

As a first problem, let’s find the minimum value of $x+y$ for points on the unit circle $x^{2}+y^{2}=1$.

One way of solving this problem is to consider lines with equation $x+y=c$ for different values of the constant $c$:

Since the circle lies to the right of $x+y=-2$, we can see that $x+y$ has got to be larger than $-2$, as otherwise $(x,y)$ cannot lie on the circle. It should also be clear that we need to choose a value of $c$ so that the line is a tangent to the circle.

We can do this by observing that if we try to find the value of $x$ (or $y$) for which the line is a tangent to the curve, there will only be *one* solution to the resulting equation (corresponding to the line ‘touching’ the circle at only one point). But this corresponds to the discriminant of the quadratic resulting from solving the two equations simultaneously being zero: substituting $y=c-x$ into $x^2+y^2=1$, then doing a little rearranging gives

\[2x^{2}-2cx+(c^{2}-1)=0.\]

The discriminant of this quadratic is

\begin{align*}

\text{“}b^{2}-4ac\text{”}&=(-2c)^{2}-4\times2\times(c^{2}-1)\\&=8-4c^{2}.

\end{align*}

This is zero when $c=\pm\sqrt{2}$, and so the minimum value of $x+y$ for points on the unit circle is $-\sqrt{2}$.

## Tangents and turning points

We can use the idea of finding a single intersection point (corresponding to a line being tangent to a curve) to solve some problems where differentiation would be the standard approach: finding tangents to curves and turning points on curves.

As an example of the former, let’s suppose that we want to find the tangent to the curve $y=2x^{2}-5x+5$ at the point $(1,2)$. We know that a line with gradient $m$ through the point $(1,2)$ has equation $y-2=m(x-1)$. The diagram below shows the curve with several lines through the point $(1,2)$:

If we rearrange the equation of the general line we get $y=mx+2-m$, which we can then substitute into the equation of the curve to give

\[mx+2-m=2x^{2}-5x+5.\]

With a bit of rearranging, we can turn this into

\[2x^{2}-(5+m)x+(3+m)=0.\]

The discriminant of this quadratic is

\begin{align*}

(-(5+m))^{2}-4\times2\times(3+m)&=m^{2}+2m+1\\&=(m+1)^{2},

\end{align*}

and so the value of $m$ for which the discriminant is zero is $-1$. So the tangent to the curve at $(2,1)$ is $y=-x+3$:

As an example of finding turning points, let’s consider the curve $y=(x^{2}+3)/(x-1)$:

From the diagram, we can see that there are two turning points—shown by blue crosses—where the curve has a local maximum or minimum value. It might be worth noting at this point that we would require the quotient rule if we were to find these points by differentiation—the single intersection method here is certainly more elementary! We start by considering the graph intersecting a horizontal line; some examples are shown on the graph: one intersects twice, the other two do not intersect at all. If a horizontal line touches the curve just once—and so is a tangent—it will do so at one of the turning points. For a single intersection, we again require a discriminant to be zero. A horizontal line has equation $y=c$: solving this simultaneously with the equation of the curve gives $$c = \frac{x^2+3}{x-1},$$ which rearranges to \[x^{2}-cx+(3+c)=0.\]

The discriminant of this quadratic is

\begin{align*}(-c)^{2}-4(3+c)&=c^{2}-4c-12\\&=(c-6)(c+2),\end{align*}

so the values of $c$ that give a discriminant of zero are $-2$ and $6$. These correspond to the $y$-coordinates of the turning points, and it’s a quick task to find the corresponding $x$ values, giving the turning points as $(-1,-2)$ and $(3,6)$:

## Cubics and beyond

So far, our problems have all boiled down to setting a quadratic’s discriminant to zero and working from there. Unfortunately, with many functions this approach won’t be possible. If we want to use our method above for finding a tangent to a cubic, we find it’s not so straightforward—when we substitute our straight line equation into our cubic we get another cubic rather than a quadratic, so there’s no way of setting the discriminant equal to zero. We can observe that when we reach this second cubic it must have a double factor at the point of tangency, so we can force it to be of the form $(x-a)^{2}(x-b)$ by equating coefficients.

There is, however, a lovely trick using polynomial division to find tangents to polynomial curves, which was brought to my attention by David T Williams on Twitter (and it was still *Twitter* then…), which is to use the fact that when a polynomial, $p(x)$, is divided by $(x-a)^2$, the remainder gives us the expression for the tangent at $x=a$.

To see why this is the case, write

\[p(x)=q(x)(x-a)^{2}+r(x).\]

When $x$ is close to $a$ we can see that $x-a$ is small and so $(x-a)^{2}$ is very small—in other words $p(x)\approx r(x)$. Since $(x-a)^{2}$ is a quadratic, the remainder will be a linear function, so we’re approximating $p(x)$ by a linear function $r(x)$ at $x=a$. But this is exactly what a tangent is! (For those who have studied Taylor series, our remainder $r(x)$ represents the first two terms of the Taylor series for $p(x)$ about $x=a$.) This method works nicely for polynomial functions only, where we can use long division to find the appropriate linear remainder.

As an example, let’s find the tangent to $y=x^{4}-7x^{3}+20x^{2}-37x+42$ at $x=3$.

All we need to do is divide the polynomial by $(x-3)^{2}=x^{2}-6x+9$:

And so the required tangent is $y=2x-3$:

So it turns out there’s a lot we can do with curves without resorting to calculus. In the example above we don’t even need to calculate the $y$-value at $x=3$. if you’re fluent with polynomial division it’s quicker than using differentiation.

As a final teaser, try the following problem using (a) a discriminant method and (b) calculus:

If $x^{2}-2xy-4x+4y^{2}=12$, what’s the maximum value of $y-2x$?