Intersection of a reciprocal function and a line

We find where the reciprocal graph $y = \dfrac{1}{x}$ meets the line $y = x$.

At an intersection the two $y$ values are equal, so $\dfrac{1}{x} = x$. Multiplying both sides by $x$ gives $1 = x^2$, that is $x^2 = 1$, so $x = 1, -1$. Substituting back into $y = x$ gives $y = 1, -1$, so the intersection points are $(1, 1)$ and $(-1, -1)$. Note $x = 0$ makes the denominator $0$, where $y = \dfrac{1}{x}$ is undefined, so multiplying through by $x$ loses no solution.

Since $y = \dfrac{1}{x}$ is an odd function with point symmetry about the origin, and $y = x$ is symmetric about the origin too, the intersections form the origin-symmetric pair $(1, 1)$ and $(-1, -1)$. Sliding the line to $y = x + k$ moves the intersections: $\dfrac{1}{x} = x + k$ becomes $x^2 + kx - 1 = 0$, whose discriminant $k^2 + 4$ is always positive, so $y = x + k$ meets the curve at 2 points for any $k$. The large dots on the graph are the intersection points $(1, 1)$ and $(-1, -1)$.