Visual walkthrough — Graphs of functions — plotting, reading key features

Before any symbol, let us agree on the stage.

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Step 1 — Draw the empty stage (the two axes)

WHAT. We start with a flat plane and draw two number lines crossing at a right angle. The horizontal one we call the ==$x$-axis== (the input line). The vertical one we call the ==$y$-axis== (the output line). Where they cross is the origin, the point $(0,0)$.

WHY. A function is a machine: you feed it a number (the input) and it hands back a number (the output). To see both numbers at once we need a place for each: input goes left–right, output goes up–down. Two perpendicular lines give every pair of numbers exactly one home.

PICTURE. The dot in red is the origin. A point is written $(x,y)$: walk right by $x$, then up by $y$.

Two words will recur, so let us pin them down now before using them.

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Step 2 — Turn the rule into a point

WHAT. Our rule is $g(x)=\underset{\text{square the input}}{\underbrace{x^2}}\underset{\text{subtract four inputs}}{\underbrace{-4x}}\underset{\text{add three}}{\underbrace{+3}}.$ Feed in one input, say $x=0$. Then $g(0)=0^2-4(0)+3=3$. That single fact becomes the single point $(0,3)$.

WHY. The whole graph is defined as the collection of all points $(x,g(x))$. So the smallest possible piece of a graph is one input paired with its one output. We must be able to make one dot before we can make a curve.

PICTURE. The red dot sits directly above $x=0$ at height $3$. Notice the vertical dashed line: it climbs from the input $0$ on the $x$-axis up to the output height $3$.

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Step 3 — Sample several inputs (make a scatter of dots)

WHAT. We repeat Step 2 for a spread of inputs. Every square below is one honest calculation:

input $x$	$x^2-4x+3$	output $g(x)$	dot
$0$	$0-0+3$	$3$	$(0,3)$
$1$	$1-4+3$	$0$	$(1,0)$
$2$	$4-8+3$	$-1$	$(2,-1)$
$3$	$9-12+3$	$0$	$(3,0)$
$4$	$16-16+3$	$3$	$(4,3)$

WHY. One dot tells you nothing about shape. Several dots, spread across the input line, start to reveal a trend. We deliberately include $x=1$ and $x=3$ because there the output is $0$ — those dots sit on the $x$-axis and are special (they are the roots, where the curve crosses the input line).

PICTURE. Five red dots. Read them left to right: the heights go $3,0,-1,0,3$ — down, then back up. The lowest dot is at $(2,-1)$.

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Step 4 — Find the exact bottom (the vertex) instead of guessing

WHAT. The dots suggest the lowest point is near $x=2$, but "near" is not good enough. We locate it exactly by completing the square — a rewrite that hides the whole expression inside one perfect square. Watch the middle step, do not skip it: $x^2-4x+3=\underset{\text{a perfect square: }(x-2{)}^2}{\underbrace{(x^2-4x+4)}}\underset{\text{take back the 4 we added}}{\underbrace{-4}}+3=\underset{\text{never negative}}{\underbrace{(x-2{)}^2}}\underset{\text{the floor}}{\underbrace{-1}}.$ Here is the trick term by term: to build $(x-2{)}^2=x^2-4x+4$ we needed a $+4$, but the original had no $+4$, so we add $4$ and immediately subtract $4$ (adding zero, so nothing changes), then combine $-4+3=-1$.

The piece $(x-2{)}^2$ is a square, so it can never be below $0$; its smallest possible value is $0$, and that happens exactly when $x-2=0$, i.e. $x=2$. There $g=0-1=-1$.

WHY THIS TOOL. Why completing the square and not just sampling more dots? Because a square tells us the guaranteed minimum algebraically: $(\text{anything}{)}^2\ge 0$ always. Sampling could miss the true bottom between two chosen inputs; the square proves the floor is exactly $-1$ at exactly $x=2$. This exact turning point is the vertex.

PICTURE. The red dot is the vertex $(2,-1)$. The dashed vertical line through it is the axis of symmetry: fold the plane along it and the left dots land on the right dots ($x=1\leftrightarrow x=3$, $x=0\leftrightarrow x=4$).

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Step 5 — Join the dots into the smooth curve

WHAT. Now we sweep a smooth, unbroken curve through all five dots and the vertex, respecting the symmetry from Step 4.

WHY. The dots are only anchors. The function has an output for every real input (the domain is all real numbers), not just our five. The smooth curve fills in the infinitely many dots we did not compute. Between anchors the curve cannot jump, because a polynomial is continuous — no breaks, no holes.

PICTURE. The red curve is the graph of $g$. Trace it left to right: it comes down steeply from the upper left, bottoms out at the red vertex, and climbs back up to the upper right. The black dots are our anchors sitting neatly on it.

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Step 6 — Read the personality off the finished picture

WHAT. With the curve drawn, every key feature can now be seen, not computed.

WHY. This is the payoff of plotting: the picture answers questions instantly that algebra answers slowly.

PICTURE. Each label points to where you literally look:

• $y$-intercept $(0,3)$ — where the curve crosses the vertical output line.
• $x$-intercepts / roots $(1,0)$ and $(3,0)$ — where it crosses the horizontal input line.
• Decreasing on the left of the vertex, increasing on the right (see Solving equations graphically for reading crossings).
• Range $[-1,\infty )$ — the outputs the curve actually reaches, from the floor $-1$ upward forever.

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Step 7 — The edge case: what if the bowl opened downward or flattened?

WHAT. We tested only an upward bowl. First, a word we now need: the leading coefficient of a polynomial is the number multiplying its highest power of $x$. For $g(x)=x^2-4x+3$ the highest power is $x^2$ and its multiplier is $1$, so the leading coefficient is $+1$. This one number controls the shape. Two boundary situations must be shown so no reader is surprised:

1. Downward parabola, e.g. $-(x-2{)}^2-1$: leading coefficient $-1$, so the same shape flipped, giving a maximum at the vertex. Notice this curve sits entirely below the $x$-axis — it never crosses it, so it has no real roots. (Algebraically, its discriminant $b^2-4ac$ is negative; a negative discriminant means "no real $x$-intercept.")
2. Degenerate case — a straight line, e.g. $2x-3$ (leading coefficient of $x^2$ is $0$). No vertex, no turning point at all; it only ever increases (see Transformations of functions).

WHY. The sign of the leading coefficient decides up vs down; when that coefficient vanishes the "parabola" collapses into a line. Covering both signs and zero means the reader has met every possibility.

PICTURE. Left: red upward bowl (minimum, two roots). Middle: red downward bowl (maximum, no real roots). Right: red straight line (no turning point).

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The one-picture summary

Everything above, compressed: the empty axes, the five sampled dots, the exact vertex, the smooth red curve, and every labelled feature — on one canvas.

Recall Feynman retelling — say it in plain words

I want to see what the machine $g(x)=x^2-4x+3$ does. First I draw two crossing rulers: one for the number I feed in, one for the number that comes out. I pick a few inputs — 0, 1, 2, 3, 4 — do the arithmetic, and drop a dot at each answer's height. The dots go 3, 0, −1, 0, 3: they dip down and come back up, so I suspect a bowl. To find the exact bottom I rewrite the rule as a perfect square minus one — I add and subtract a 4 to build $(x-2{)}^2$, leaving $-1$ behind — and since a square can't be negative, the smallest output is −1, right at input 2, on the mirror line $x=2$; that's the vertex. I join the dots into a smooth bowl (allowed, because polynomials never break), and now I just read: it hits the up-axis at 3, crosses the flat axis at 1 and 3, falls then rises, and never dips below −1. Finally I remember the bowl could open the other way (a peak instead of a valley, maybe with no roots at all), or flatten into a line if the squared term disappears — so I've covered every case.

Recall Quick self-test

What is the vertex of $g(x)=x^2-4x+3$? ::: $(2,-1)$, found from $(x-2{)}^2-1$. Why can we join the sampled dots with a smooth curve? ::: $g$ is a polynomial, hence continuous — no jumps or holes. What decides whether the vertex is a minimum or a maximum? ::: The sign of the leading coefficient (the $x^2$ multiplier): positive → opens up (minimum), negative → opens down (maximum). What happens to the vertex if the leading coefficient is $0$? ::: There is no vertex — the graph becomes a straight line with no turning point. When does a parabola have no real roots? ::: When it sits entirely above or below the $x$-axis (negative discriminant), so it never crosses.

Related: Symmetry (even and odd functions) · Asymptotes · Vertical line test · Calculus: derivatives and graphs.