# At what points do the functions y = x^2-x and y = sin pix intersect?

Jan 24, 2017

These two equations intersect at the points $\left(0 , 0\right)$ and $\left(1 , 0\right)$

#### Explanation:

First, let us take a look at:

$y = {x}^{2} - x$

We can factor this as:

$y = x \left(x - 1\right)$

so this quadratic has $x$ intercepts at $\left(0 , 0\right)$ and $\left(1 , 0\right)$

It has minimum value at the midpoint of these two $x$ coordinates, where $x = \frac{1}{2}$:

$y = \textcolor{b l u e}{\frac{1}{2}} \left(\textcolor{b l u e}{\frac{1}{2}} - 1\right) = - \frac{1}{4}$

So note that $y = {x}^{2} - x < 0$ for all $x \in \left(0 , 1\right)$

The intersections of $y = {x}^{2} - x$ with the horizontal line $y = 1$ are at the points given by solving:

$0 = {x}^{2} - x - 1$

$\textcolor{w h i t e}{0} = {x}^{2} - x + \frac{1}{4} - \frac{5}{4}$

$\textcolor{w h i t e}{0} = {\left(x - \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(x - \frac{1}{2} - \frac{\sqrt{5}}{2}\right) \left(x - \frac{1}{2} + \frac{\sqrt{5}}{2}\right)$

That is:

$x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$

Note that $\frac{1}{2} + \frac{\sqrt{5}}{2} \approx 1.618 \in \left(1 , 2\right)$

Note that $\frac{1}{2} - \frac{\sqrt{5}}{2} \approx - 0.618 \in \left(- 1 , 0\right)$

So ${x}^{2} - x \in \left(0 , 1\right]$ when $x \in \left[\frac{1}{2} - \frac{\sqrt{5}}{2} , 0\right)$ or $x \in \left(1 , \frac{1}{2} + \frac{\sqrt{5}}{2}\right]$

Outside these intervals, ${x}^{2} - x > 1$ so cannot be equal to $\sin \left(\pi x\right)$

Now consider $y = \sin \left(\pi x\right)$ in each of these intervals:

• If $x \in \left[\frac{1}{2} - \sqrt{2} , 0\right)$ then $\sin \left(\pi x\right) < 0$

• If $x = 0$ then $\sin \left(\pi x\right) = 0 = {x}^{2} - x$

• If $x \in \left(0 , 1\right)$ then $\sin \left(\pi x\right) > 0$

• If $x = 1$ then $\sin \left(\pi x\right) = 0 = {x}^{2} - x$

• If $x \in \left(1 , \frac{1}{2} + \frac{\sqrt{5}}{2}\right)$ then $\sin \left(\pi x\right) < 0$

So in each of the intervals $\left[\frac{1}{2} - \frac{\sqrt{5}}{2} , 0\right)$, $\left(0 , 1\right)$ and $\left(1 , \frac{1}{2} + \frac{\sqrt{5}}{2}\right]$ the two functions have opposite signs and cannot intersect.

So the only two points of intersection are:

$\left(x , y\right) = \left(0 , 0\right)$

$\left(x , y\right) = \left(1 , 0\right)$

graph{(y-x^2+x)(y - sin(pix)) = 0 [-2.105, 2.895, -1.19, 1.31]}