Intersection of a reciprocal function and a line

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

Solving

At an intersection the two $y$ values are equal, so $\dfrac{1}{x} = x$. Multiplying both sides by $x$ gives $x^2 = 1$, so $x = \pm 1$, and the intersection points are $(1, 1)$ and $(-1, -1)$.

Whenever an equation has a denominator, you must check whether the factor you multiplied by can vanish. Here that factor is $x$, and $x = 0$ is already excluded from the domain. Multiplying by a value that was never allowed loses no solution and introduces no false one.

What the two points are

Since $y = \dfrac{1}{x}$ is odd and $y = x$ is symmetric about the origin too, the intersections form an origin-symmetric pair. These two points are the vertices of the hyperbola $xy = 1$: the places where its two branches come closest together, with the line $y = x$ as the hyperbola's axis of symmetry.

The other axis never meets it

Running the same computation with $y = -x$ gives $x^2 = -1$, which has no real solution. The line $y = -x$ is the hyperbola's other axis of symmetry, yet it never meets the curve. That is only natural: $\dfrac{1}{x}$ lives in the first and third quadrants, while $y = -x$ runs through the second and fourth.

Sliding the line

A line of slope $1$ cuts the hyperbola twice no matter where it is placed. A line of negative slope, on the other hand, can be tangent to it.

Checking the tangency

Since the derivative of $y = \dfrac{1}{x}$ is $y' = -\dfrac{1}{x^2}$, the tangent slope at $x = 1$ is $-1$, exactly the slope of the line $y = -x + 2$. Intersections of a rational function and a line, too, come down to a quadratic equation and its discriminant.

The large dots on the graph are the intersection points.