The highest power of $x$ fixes the overall shape:

• Linear ($x^1$): straight line.
• Quadratic ($x^2$): parabola.
• Cubic ($x^3$): a continuous S-shaped curve.
• Reciprocal ($1/x$): two separate hyperbola branches.

This lets you match an unlabelled graph to its equation almost instantly — spot the highest power, then check the sign of the leading coefficient.

For $y=ax^3+bx^2+cx+d$: if $a>0$, the curve rises from bottom-left to top-right; if $a<0$, it falls from top-left to bottom-right. The $y$-intercept is always $(0,d)$ (substitute $x=0$).

Build a table of values across the given domain, then join the points with a single smooth curve — never straight segments, and never a sharp V-shape at a turning point. A cubic's gradient changes continuously, so the curve must flow through any maximum/minimum.

For $y=\dfrac{a}{x}$, the function is undefined at $x=0$ (division by zero), so the curve can never cross or touch the $y$-axis (the vertical asymptote) or the $x$-axis (the horizontal asymptote, since $a/x\to0$ as $x\to\pm\infty$). If $a>0$, the two branches sit in quadrants 1 and 3; if $a<0$, in quadrants 2 and 4.

To capture the sharp curvature near $x=0$, your table of values must include small fractional inputs like $x=\pm0.5, \pm0.25$ — plotting only integers makes the curve look wrong.

This shifts the whole graph vertically by $k$. The vertical asymptote stays at $x=0$ (division by zero is unaffected), but the horizontal asymptote moves to $y=k$. Unlike the basic reciprocal, this shifted graph can cross the $x$-axis — solve $0=\dfrac{a}{x}+k$ to find where.

Plot $y_1=x^3-4x$ and the horizontal line $y_2=2$ on the same axes; the $x$-coordinates of their intersection points are the solutions.

Worked Example

Complete a table for $y=x^3-6x+1$ at $x=-3,-2,-1,0,1,2,3$, then describe the shape.

Since $a=1>0$, the curve runs bottom-left to top-right with two turning points (a local max then a local min) — a standard positive cubic S-shape.

Work through a short checklist: identify the highest power present (fixing the family — linear, quadratic, cubic, or reciprocal); check the sign of the leading coefficient (fixing the overall direction); find the $y$-intercept by substituting $x=0$ (for polynomials this is just the constant term); and for reciprocal graphs, check whether the curve is shifted by looking for a non-zero horizontal asymptote. Running through these four checks in order is far faster and more reliable than trying to recognise a graph's shape from memory alone.