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Stage 4 - Stage 5

Polynomial curve sketching

Lesson

Certain polynomials of degree $n$n can be factorised into $n$n linear factors over the real number field.

For example the $4$4th degree polynomial $P\left(x\right)=2x^4-x^3-17x^2+16x+12$P(x)=2x4x317x2+16x+12 can be re-expressed as $P\left(x\right)=\left(x+3\right)\left(2x+1\right)\left(x-2\right)^2$P(x)=(x+3)(2x+1)(x2)2. Note that there are two distinct factors, the $\left(x+3\right)$(x+3) and $\left(2x+1\right)$(2x+1) and two equal factors, the $\left(x-2\right)$(x2) appears twice.

A polynomial function $y=P\left(x\right)$y=P(x) that can be completely factorised into its linear factors can be easily sketched. The function $y=\left(x+3\right)\left(2x+1\right)\left(x-2\right)^2$y=(x+3)(2x+1)(x2)2 has roots immediately identifiable as $x=-3,-\frac{1}{2}$x=3,12 and $x=2$x=2.

Multiplicity

A root is said to have multiplicity $k$k if a linear factor occurs $k$k times in the polynomial function. For example the function of degree $6$6 given by $y=\left(x-1\right)\left(x+1\right)^2\left(x-2\right)^3$y=(x1)(x+1)2(x2)3 is said to have a root $x=1$x=1 of multiplicity $1$1 (a single root), another root of $x=-1$x=1 of multiplicity $2$2 (a double root or two equal roots) and a root $x=2$x=2 of multiplicity $3$3 (a triple root or three equal roots). In effect that is $6$6 roots all together.

The key to understanding how a root's multiplicity effects the graph is given in the following statement:

The curve's shape near a root depends on the root's multiplicity.

Look closely at the sketch of $y=\left(x-1\right)\left(x+1\right)^2\left(x-2\right)^3$y=(x1)(x+1)2(x2)3

 

The curve approaching a root of multiplicity $1$1 will behave like a linear function across that root (see position B on graph). The curve approaching a root of multiplicity $2$2 will behave like a quadratic function across that root (see position A on graph). The curve approaching a root of multiplicity $3$3 will behave like a cubic function across that root (see position C on graph).

In general, the curve approaching a root of multiplicity $k$k will behave like the function $y=x^k$y=xk across that root.

There is also two other considerations that will help with the graphing of the function.

Firstly, note that this function is an even degree function, so at the left and right extremes it will move away from the $x$x - axis in the same direction. Odd degree functions move away in opposite directions.

Secondly, because the coefficient of the highest power of the function is positive, the curve moves away from the $x$x - axis positively.

Thirdly, always check the position of the $y$y - intercept. For example, in the above sketch, at $x=0$x=0, $y=\left(0-1\right)\left(0+1\right)^2\left(0-2\right)^3=8$y=(01)(0+1)2(02)3=8 and this confirms the direction of the curve downward toward the first positive root.

In all this we need more mathematical tools to determine the positions of local turning points, but the important principles here allow us to understand the functions basic shape.

Example 1

The function $y=x\left(x+3\right)\left(x-2\right)$y=x(x+3)(x2) has three roots $0,-3$0,3 and $2$2 all of multiplicity $1$1. It is of degree $3$3 with a $y$y-intercept of $\left(0\right)\left(-3\right)\left(2\right)=0$(0)(3)(2)=0. The shape is shown as graph $G2$G2 here.

Example 2

The function $y=-\left(x-1\right)\left(x+2\right)^2$y=(x1)(x+2)2 has a root of multiplicity $1$1 at $x=1$x=1 and a root of multiplicity $2$2 ($2$2 equal roots) at $x=-2$x=2. The $y$y - intercept is $-\left(-1\right)\left(2\right)^2=4$(1)(2)2=4. The function is of odd degree, so its ends move off in different directions. Note that for large positive values of $x$x, the curve becomes very negative (See graph $G3$G3 below).

Example 3

The function $y=\frac{1}{2}\left(x-3\right)^2\left(x+1\right)^3$y=12(x3)2(x+1)3 is of degree $5$5, with a double root at $x=3$x=3 and a triple root at $x=-1$x=1. The y intercept is $4.5$4.5. The coefficient $\frac{1}{2}$12 at the front simply halves the size of every $y$y value, but note that there is no change at all in the position of the roots because half of zero is still zero (see graph $G4$G4 below).

Worked Examples

Question 1

Consider the curve $y=\left(x+3\right)\left(x+2\right)\left(x-2\right)$y=(x+3)(x+2)(x2).

  1. Find the $x$x-value of the $x$x-intercept(s).

  2. Find the $y$y-value of the $y$y-intercept(s).

  3. Sketch a graph of the curve.

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Question 2

Consider the function $y=-\left(x-2\right)^2\left(x+1\right)$y=(x2)2(x+1).

  1. Find the $x$x-value(s) of the $x$x-intercept(s).

    Write each line of working as an equation. If there is more than one answer, write all solutions on the same line separated by commas.

  2. Find the $y$y-value of the $y$y-intercept.

    Write each line of working as an equation.

  3. Which of the following is the graph of $y=-\left(x-2\right)^2\left(x+1\right)$y=(x2)2(x+1)?

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    A

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    B

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    C

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    D

Question 3

The graph of $y=P\left(x\right)$y=P(x) is shown. Plot the graph of $y=P\left(x+5\right)$y=P(x+5).

  1. Loading Graph...

 

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