4A - Sketching polynomial graphs

How much of a curve can you predict before plotting any points? Factors reveal where a polynomial meets an axis, while the highest power tells you where its ends go. You will combine these clues to make a labelled sketch without finding exact turning points.

Intercepts and factors

The equation $y=(x+2)(x-1)(x-3)$ looks like an instruction to multiply three numbers. Can those same brackets tell you where its graph crosses the axes?

On the $x$ -axis, $y=0$ . A product is zero when at least one factor is zero, so $x=-2$ , $x=1$ and $x=3$ are the three roots. On the $y$ -axis, $x=0$ , so the height is $2(-1)(-3)=6$ . These are exact points, even on a rough sketch.

Intercept

An intercept is a point where a graph meets a coordinate axis. An $x$ -intercept has coordinates $(r,0)$ , where $r$ is a real root of $f(x)=0$ . The $y$ -intercept is $(0,f(0))$ .

Set the output to zero for $x$ -intercepts; set the input to zero for the $y$ -intercept. Keep the coordinates in that order.

Worked example 1: reading factors

Question. Find the intercepts of $y=(x+2)(x-1)(x-3)$ .

Plan. Set each factor equal to zero, then substitute $x=0$ .

\begin{aligned} x+2=0&\implies x=-2 &&\text{first factor}\\ x-1=0&\implies x=1 &&\text{second factor}\\ x-3=0&\implies x=3 &&\text{third factor}\\ f(0)&=2(-1)(-3)=6 &&\text{height on the }y\text{-axis} \end{aligned}

Answer. $(-2,0)$ , $(1,0)$ , $(3,0)$ and $(0,6)$ .

Check. Substitution of each stated root makes one bracket zero. Expanding gives $f(x)=x^3-2x^2-5x+6$ , whose constant term independently confirms $f(0)=6$ .

Now change only the multiplier in front. This changes the heights, but a nonzero multiplier cannot change which inputs make the product zero.

Worked example 2: a negative multiplier

Question. Find the intercepts of $y=-2(x+2)(x-1)(x-3)$ .

Plan. Keep the roots of the three brackets and recalculate the height at $x=0$ .

\begin{aligned} -2(x+2)(x-1)(x-3)=0&\implies x=-2,\ 1,\ 3 &&\text{divide by nonzero }-2\\ f(0)&=-2\times2\times(-1)\times(-3)=-12 &&\text{substitute }0 \end{aligned}

Answer. $(-2,0)$ , $(1,0)$ , $(3,0)$ and $(0,-12)$ .

Check. Every height in example 1 is multiplied by $-2$ . In particular, $6$ becomes $-12$ , and zero heights remain zero.

A root is an input, while an intercept is a point. Writing the point helps you avoid swapping the axes.

Crossing or touching an axis

A root tells you where the graph reaches the $x$ -axis; its repeated factors tell you what happens there.

Compare $y=(x+1)(x-2)$ with $y=(x+1)(x-2)^2$ . Near $x=2$ , the first bracket is positive in both cases. The factor $x-2$ changes from negative to positive, but its square stays nonnegative. The first graph crosses the axis at $2$ ; the second touches it and returns to the same side.

[DIAGRAM: asset_name: A repeated factor changes crossing into touching; asset_slug: crossing-touching; caption: Squaring the factor x minus 2 makes the graph touch the axis at 2; description: Side-by-side coordinate graphs show y equals (x plus 1)(x minus 2), crossing at minus 1 and 2, and y equals (x plus 1)(x minus 2) squared, crossing at minus 1 but touching from above at 2. Intercepts are labelled and sign strips show plus minus plus for the first graph and minus plus plus for the second.]
Squaring the factor x minus 2 makes the graph touch the axis at 2

Multiplicity

If $f(x)=(x-r)^m g(x)$ , where $g(r)\ne0$ , the root $r$ has multiplicity $m$ . An odd multiplicity makes the graph cross the $x$ -axis; an even multiplicity makes it touch without crossing.

Why does this work? Close enough to $r$ , $g(x)$ keeps its sign. An odd power of $x-r$ changes sign across $r$ , whereas an even power does not. A root of multiplicity three still crosses, even though it can look flatter than a simple crossing.

Worked example 3: simple crossings

Question. Describe how $y=(x+1)(x-2)$ meets the $x$ -axis.

Plan. Use one test input in each interval separated by the roots.

\begin{array}{c|ccc} x&-2&0&3\\\hline (x+1)(x-2)&4&-2&4\\ \text{sign}&+&-&+ \end{array}

The sign changes at both $-1$ and $2$ , so the curve passes through the axis at each root.

Answer. It crosses at $(-1,0)$ and $(2,0)$ .

Check. Both factors occur once. Each therefore has odd multiplicity, agreeing with the two sign changes.

Change just one factor to a square. The test values now show exactly which crossing changes.

Worked example 4: a repeated root

Question. Describe how $y=(x+1)(x-2)^2$ meets the $x$ -axis.

Plan. Find multiplicities, then verify the signs on the three intervals.

\begin{array}{c|ccc} x&-2&0&3\\\hline (x+1)(x-2)^2&-16&4&4\\ \text{sign}&-&+&+ \end{array}

At $-1$ the sign changes; at $2$ it stays positive. The squared factor is nonnegative, and $x+1>0$ near $2$ , so the contact is from above.

Answer. It crosses at $(-1,0)$ and touches from above at $(2,0)$ .

Check. The multiplicities are $1$ and $2$ : odd then even, agreeing with the sign table.

Test values work because a polynomial is continuous and cannot change sign inside an interval without another root. A sketch must respect these signs.

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4A - Sketching polynomial graphs

Intercepts and factors

Crossing or touching an axis

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