1.2 - Reciprocal and square-root graphs

Reciprocal graph

The graph $y=1/f(x)$ has a point $(x,1/b)$ for each point $(x,b)$ on $y=f(x)$ with $b\ne0$. Its domain contains exactly the inputs where $f$ is defined and nonzero.

Worked example 1: a zero becomes a pole

Question. Use $f(x)=x-1$ to describe the graph of $y=1/f(x)$.

Plan. Transfer the zero, signs and fixed heights from the line.

To the left of $1$, both $f$ and its reciprocal are negative; to the right, both are positive. Approaching $1$ gives $-\infty$ on the left and $+\infty$ on the right. At infinity, the reciprocal tends to zero. The points with original heights $-1,1$, namely $(0,-1)$ and $(2,1)$, remain fixed.

Answer. A reciprocal curve with asymptotes $x=1$, $y=0$, passing through $(0,-1)$ and $(2,1)$, with no zero.

Check. Directly, $1/f(3)=1/2$, whereas $1/f(-1)=-1/2$. These confirm the side and relative height of each branch.

Worked example 2: an old pole becomes a hole

Question. Describe $y=1/f(x)$ for $f(x)=(x-1)/(x+2)$.

Plan. Record where $f$ is undefined or zero before reversing the fraction.

The original zero at $1$ becomes the vertical asymptote $x=1$. At the original pole $-2$, the reciprocal tends to zero but is still undefined: there is a hole at $(-2,0)$. At infinity, $f\to1$, so the new horizontal asymptote is also $y=1$. The $y$-axis intersection is $(0,-2)$.

Answer. Asymptotes $x=1$, $y=1$; hole $(-2,0)$; no $x$-axis intersection; $y$-axis intersection $(0,-2)$.

Check. At $x=4$, the original output is $1/2$, and the new output is $2$. At $x=-2$, the displayed reversed fraction alone would give zero, but $f(-2)$ does not exist, so neither does $1/f(-2)$.

From a squared output to two branches

For the real relation $y^2=f(x)$, retain only original-domain inputs with $f(x)\ge0$, and draw

At $f(x)=0$, the two branches meet at $y=0$. These graph rules are not in the formula booklet: learn them.

Worked example 3: two halves of a sideways parabola

Question. Describe the relation $y^2=x-1$.

Plan. Restrict the input using nonnegativity, then include both square roots.

We require $x-1\ge0$, so $x\ge1$. Hence $y=\pm\sqrt{x-1}$. Equivalently,

The branches meet at $(1,0)$, and pass through $(2,1)$, $(2,-1)$, $(5,2)$ and $(5,-2)$.

Answer. A sideways parabola opening to the right, symmetric about the $x$-axis, with leftmost point $(1,0)$. Its tangent there is vertical.

Check. In the parametrisation $(x,y)=(t^2+1,t)$, $dx/dt=2t$ and $dy/dt=1$. At $t=0$, the direction is vertical, confirming that the two branches join smoothly rather than forming a corner.

Worked example 4: selecting intervals before taking roots

Question. Describe $y^2=(x-1)/(x+2)$, including its asymptotes.

Plan. Make a sign chart at $-2$ and $1$, then transfer the limiting heights.

For these inputs, $y=\pm\sqrt{(x-1)/(x+2)}$. The two right-hand branches meet at $(1,0)$. On the left of $-2$, the fraction tends to $+\infty$, so the branches tend to $+\infty$ and $-\infty$. At both infinite ends, the fraction tends to $1$, giving horizontal asymptotes $y=1$ and $y=-1$. There are no points for $-2<x<1$.

Answer. Domain in $x$: $( -\infty,-2)\cup[1,\infty)$. Vertical asymptote $x=-2$; horizontal asymptotes $y=\pm1$; sole axis intersection $(1,0)$; symmetry about the $x$-axis.

Check. At $x=-3$, $y^2=4$, giving $y=\pm2$. At $x=2$, $y^2=1/4$, giving $y=\pm1/2$. At $x=0$, the right side is negative, so no real point exists.

Worked example 5: an inequality on a reciprocal graph

Question. For $f(x)=(x-1)/(x+2)$, find the inputs where $1/f(x)\ge2$.

Plan. Find the crossing with height $2$, then compare the branch with that line on its domain.

The reciprocal is $h(x)=(x+2)/(x-1)$, with $x\ne-2,1$. Now

The critical inputs are $1$, where the expression is undefined, and $4$, where it is zero. The ratio is negative for $x<1$, positive for $1<x<4$, and negative for $x>4$. The inherited exclusion $-2$ lies outside the solution interval.

Answer. $1<x\le4$.

Check. At $x=2$, $h=4\ge2$; at $x=4$, $h=2$; at $x=5$, $h=7/4<2$. A point to the left, $x=0$, gives $-2<2$.

Worked example 6: intersections on both square-root branches

Question. Find all points of $y^2=(x-1)/(x+2)$ on the lines $y=2$ and $y=-2$.

Plan. Square either stated height and solve the same original-height equation.

For both heights, $y^2=4$, so

This input is allowed and lies in the positive interval $x<-2$.

Answer. The points are $(-3,2)$ and $(-3,-2)$.

Check. The original right side at $-3$ is $(-4)/(-1)=4$. Both stated outputs square to $4$, verifying both points rather than just the upper one.

Watch out

From $(x+2)/(x-1)\ge2$, multiplying by $x-1$ without a sign check can reverse or lose part of the inequality. Subtract the line height and use its critical inputs instead.