An equation describes a set of points

Try

A marker sits at (3,4). Does it belong to the graph of x²+y²=25? Substitute the horizontal coordinate for x and the vertical coordinate for y: 9+16=25, so it does. The point (3,−4) also works. The two points share an x-coordinate but are different solutions.

Here a solution is an ordered pair that makes the equation true. The graph of an equation is the set of all such pairs. This definition does not require one y-value per x-value. A function of x has that extra requirement; a general equation need not. Write down two points that satisfy the equation and one that fails before looking at the picture.

Predict

Predict the points on the coordinate axes. When y=0, x²=25 gives x=5 or x=−5; when x=0, y=5 or y=−5. Now predict whether any point with x=6 can belong. Its x² term alone is 36, and y² cannot be negative, so no real y can repair it.

The equal-scale diagram below shows the whole circle, not just the upper branch. Its text equivalent is: center (0,0), radius 5, intercepts (±5,0) and (0,±5), with marked points (3,4) and (3,−4).

Experiment

Construct a small table, retaining both signs where necessary. At x=0, y is ±5; at x=3, y is ±4; at x=4, y is ±3; at x=5, y is 0. Sketch these points on equal scales, then connect them using the distance description rather than inventing straight edges between samples.

x	Possible y values	Why
0	−5, 5	y²=25
3	−4, 4	y²=16
5	0	y²=0
6	None	y²=−11 has no real solution

Your action is to explain why the table alone does not establish the full graph. A finite table checks selected points; it does not rule out unexamined points or prove the arcs between them belong.

Understand

The distance from (x,y) to the origin is √(x²+y²). Therefore every solution of x²+y²=25 has distance 5, and every point at distance 5 satisfies the equation after squaring. These two implications prove that the graph is exactly the circle. A useful graph justification always asks both questions: does every claimed point work, and does every solution appear?

For a line, reversible algebra can answer both. The equation 2x+y=7 gives y=7−2x by subtracting 2x. Conversely substituting that expression gives 2x+(7−2x)=7 for every real x. No inputs were discarded. By contrast, dividing by an expression that might be zero can lose solutions. For xy=0, dividing by x keeps the line y=0 only when x≠0 and misses the full vertical line x=0.

Domain restrictions also survive simplification. The graph y=(x²−4)/(x−2) agrees with y=x+2 wherever x≠2, but has no point at x=2. Factoring cancels a nonzero factor; it does not authorize division by zero.

Apply

Now work with a fresh circle in the practice cards. Draw before using a formula, label axes and scales, and verify the listed points. For a line or factored equation, provide a two-direction argument. The written drawing rubric checks mathematical features, not artistic smoothness.

An unreliable argument says, “These five points lie on my curve, so it must be the whole graph.” Repair it by using the distance equation, reversible operations or a complete case split. The point checks are valuable error detectors, but a set equality needs a statement covering every allowed input.

Continue

A system asks for the points common to two equation graphs. An inequality instead selects a region, often with a boundary that is included or excluded. Carry the same membership habit into both: substitute a candidate into the original statements.

For optional depth, compare x=y² with y=x². Both have perfectly valid graphs; only the second is a function assigning one y to each real x. Changing which variable is regarded as the input changes the function question, but it does not change the equation-graph definition.