Explainer: Graphs of Polynomial Functions

In this explainer, we will learn how to investigate the graph of polynomial functions and identify the equation of a polynomial function from its graph and vice versa.

We look at how factors correspond to π‘₯-intercepts of the graph, what happens when factors are repeated, and how the sign of the leading coefficient affects the graph.

Consider the polynomial function 𝑓(π‘₯)=βˆ’6π‘₯+27π‘₯+6π‘₯βˆ’117π‘₯+54.οŠͺ

What can we tell about the graph 𝑦=𝑓(π‘₯) before we can sketch it?

  1. Since 𝑓(0)=54, the constant term, we know that the graph passes through (0,54).
    There is not much else! We can add that this is the appearance of a typical degree 4 curve.
  2. Since the degree of the polynomial is 4, we hope that it will look like a β€œtypical” quartic curve. But which of these is typical?
    At least one detail is true of the three examples above, which we note below.
    The curves all go to either +∞ or βˆ’βˆž whether independent variable π‘₯ goes towards +∞ or βˆ’βˆž. This behaviour is determined by the leading term, in our case the βˆ’6π‘₯οŠͺ term, which β€œdominates” when π‘₯ is very large. In our case, whether π‘₯=10 or π‘₯=βˆ’10, the leading term contributes βˆ’6Γ—ο€Ή10=βˆ’6Γ—10=βˆ’6,000β‹―0ο‡Œο†²ο‡ο†²ο‡Ž,οŠͺοŠͺ24zeros which is very large and negative. The π‘₯ term only has 18 zeros. So the graph of 𝑦=𝑓(π‘₯) is β€œfacing downwards” as in example (c) above.

So far, we have this picture.

The best information comes from a factored form of the polynomial 𝑓(π‘₯), when it is available. Here 𝑓(π‘₯)=βˆ’6π‘₯+27π‘₯+6π‘₯βˆ’117π‘₯+54=βˆ’3(2π‘₯βˆ’1)(π‘₯+2)(π‘₯βˆ’3),οŠͺ telling us that 𝑓(π‘₯)=0 when one of the factors (2π‘₯βˆ’1), (π‘₯+2), or (π‘₯βˆ’3) is zero. In other words, when π‘₯=12,π‘₯=βˆ’2,π‘₯=3.or

Since the curve must pass smoothly through the three points and not cross the π‘₯-axis anywhere else, we can sketch the graph as shown below.

Not only do the π‘₯-intercepts correspond to the distinct factors, but the way the curve crosses the π‘₯-axis is also determined by the multiplicity of the linear factors.

The way the curve 𝑦=𝑓(π‘₯) cuts the π‘₯-axis near π‘₯=12 and π‘₯=βˆ’2 is β€œlinear” since we have (2π‘₯βˆ’1) and (π‘₯+2).

Near π‘₯=3, the factor (π‘₯βˆ’3) makes the graph look very much like the curve 𝑦=βˆ’70(π‘₯βˆ’3) shown as the dashed curve. Recall that the curve 𝑦=βˆ’70(π‘₯βˆ’3) is just the curve 𝑦=βˆ’70π‘₯ translated horizontally by 3 units.

Similarly, if (π‘₯+πœ‹) is a factor of the polynomial 𝑔(π‘₯), then 𝑦=𝑔(π‘₯) looks like a multiple of (π‘₯+πœ‹) near the intercept (βˆ’πœ‹,0). This is a vertically stretched form of 𝑦=π‘₯ or 𝑦=βˆ’π‘₯ shifted πœ‹ units to the left.

Example 1: Identifying the Graph of a Polynomial Given in Factored Form

Which of the following is the graph of 𝑓(π‘₯)=(π‘₯+3)(π‘₯βˆ’1)(π‘₯βˆ’4)?

Answer

The zeros of 𝑓(π‘₯) are βˆ’3, 1, and 4 from the factorization. So the choice is between (C) or (E). As the leading coefficient of 𝑓(π‘₯) is 1, which is positive, then the correct sketch is (C).

The choice could also have been settled by noting that the 𝑦-intercept is (0,𝑓(0))=(0,12) which is above the π‘₯-axis, unlike the 𝑦-intercept in (E).

Example 2: Identifying the Graph of a Quadratic Using Factoring

Solve π‘₯βˆ’4π‘₯+4=0 by factoring, and hence determine which of the following figures would be a sketch of 𝑦=π‘₯+4π‘₯+4.

Answer

In order to sketch 𝑦=π‘₯βˆ’4π‘₯+4, we begin by factoring π‘₯βˆ’4π‘₯+4: π‘₯βˆ’4π‘₯+4=(π‘₯βˆ’2)(π‘₯βˆ’2)=(π‘₯βˆ’2) so that π‘₯βˆ’4π‘₯+4=0 has the single solution π‘₯=2. This gives (2,0) as the only π‘₯-intercept of 𝑦=π‘₯+4π‘₯+4. The choice is between (C) and (D).

The leading term here is π‘₯, with leading coefficient 1, which is positive. Therefore, as π‘₯ goes to +∞, we expect that π‘₯βˆ’4π‘₯+4 will also go to +∞. This tells us that the correct choice is (C).

This is the reverse to what you should know about the characteristics of the graph of a polynomial function determined by the characteristics of the polynomial expression itself, which we summarize.

Relating Characteristics, Polynomial ⟢ Graph

ExpressionGraph
1.DegreeNumber of intercepts with horizontal line
2.Odd/even degreeBehavior (1) as π‘₯β†’Β±βˆž
3.Sign of leading coefficientBehavior (2) as π‘₯β†’βˆž
4.Value of constant term𝑦-Intercept
5.Zeros/factorsπ‘₯-Intercepts
6.Multiplicity of factorCurvature at the intercept

Let us list the dictionary above.

  1. Degree 0 and 1 polynomials give rise to straight lines: horizontal if the degree is 0, not if it is 1. Degree 2 polynomials have graphs that are all parabolas, with mirror symmetry along a vertical axis. In higher degree, the shapes of the graphs are more varied. What is true is that
    1. the number of intercepts with a horizontal line 𝑦=π‘˜ is never more than the degree;
    2. this number can be less than the degree.
    Of course, if the polynomial has degree 1 (linear), then every line meets the curve exactly once. If the degree is 2, a horizontal meets the curve in either 2, 1, or 0 places.
    For degree 3, we can get 1, 2, or 3, with no guarantees as for quadratic polynomials.
  2. Behavior (1) is whether the two β€œarms” as π‘₯β†’Β±βˆž on the sketched curve are both in the upper or lower half planes (𝑦>0 or 𝑦<0), which is true if and only if the polynomial has an even degree. Otherwise, whatever the degree, it must be odd.
  3. Behavior (2) is which of quadrants 1 or 4 the graph lies in as π‘₯β†’+∞. If the leading coefficient is positive, this is 1; otherwise, it is 4. The following figure shows the graphs of two degree 5 polynomials, π‘ƒοŠ§ and π‘ƒοŠ¨, with leading coefficients of different signs.
  4. A polynomial expression 𝑃(π‘₯) is a sum of multiples of powers of π‘₯ and a term of β€œdegree 0”—the constant term. Setting π‘₯=0 says this constant is 𝑃(0), which gives the 𝑦-coordinate of the point on the curve 𝑦=𝑃(π‘₯) that is also on the 𝑦-axis.
  5. The graph 𝑦=𝑃(π‘₯) above cuts the π‘₯-axis at the number 4. This means that 𝑃(4)=0 and therefore (π‘₯βˆ’4) is a factor of 𝑃(π‘₯). Likewise, 𝑃(π‘₯)=(π‘₯+6)(π‘₯+3)(π‘₯βˆ’4)𝑄(π‘₯) for some polynomial 𝑄(π‘₯).
  6. Knowing that a number (such as 2) is the zero of a polynomial 𝑃(π‘₯) tells us that the graph 𝑦=𝑃(π‘₯) meets the π‘₯-axis at (2,0). But they can β€œmeet” in essentially three different ways, as illustrated in the figure below.
    These are as follows:
    1. A sharp crossing like 𝑦=π‘₯βˆ’2: this is the same as 𝑦=π‘₯(π‘₯βˆ’2) or indeed 𝑦=(π‘₯βˆ’2)𝑄(π‘₯) for any 𝑄(π‘₯) that is not zero at π‘₯=2.
    2. A tangency like 𝑦=(π‘₯βˆ’2) with the entire graph to the β€œsame side” of the π‘₯-axis: this is the case for 𝑦=(π‘₯βˆ’2)οŠͺ or indeed 𝑦=(π‘₯βˆ’2)𝑄(π‘₯), where π‘˜ is an even number, and 𝑄(2)β‰ 0.
    3. A tangency like 𝑦=(π‘₯βˆ’2), or the 𝑦=(π‘₯βˆ’2) shown: with odd powers (π‘₯βˆ’2), the curve passes through (2,0) but lies on both sides of the axis near this point.

We can use these facts as in the following examples.

Example 3: Identifying the Equation of a Curve from a Sketch

Which of the following could be the equation of the given sketch?

  1. 𝑦=βˆ’π‘₯(π‘₯+1)(π‘₯+3)
  2. 𝑦=βˆ’(π‘₯βˆ’1)(π‘₯+3)(π‘₯βˆ’3)
  3. 𝑦=βˆ’(π‘₯+1)(π‘₯+3)(π‘₯βˆ’3)
  4. 𝑦=(π‘₯βˆ’1)(π‘₯+3)(π‘₯βˆ’3)
  5. 𝑦=(π‘₯+1)(π‘₯+3)(π‘₯βˆ’3)

Answer

From the intercepts βˆ’3,βˆ’1, and 3, we know that (π‘₯+3), (π‘₯+1), and (π‘₯βˆ’3) must be factors of the polynomial used. This leaves the two choices (C) and (E).

Since the curve is in the 4th quadrant as π‘₯β†’+∞, the leading coefficient must be negative. So the only possibility is (C), where the coefficient of π‘₯ is βˆ’1, while the leading coefficient in the other option is 1.

Example 4: Identifying the Equation of a Curve from a Sketch

Which of the following could be the equation of the given sketch?

  1. 𝑦=βˆ’(π‘₯+1)(π‘₯+7)
  2. 𝑦=(π‘₯βˆ’1)(π‘₯βˆ’7)
  3. 𝑦=(π‘₯βˆ’7)(π‘₯βˆ’1)
  4. 𝑦=βˆ’(π‘₯+7)(π‘₯+1)
  5. 𝑦=(π‘₯+7)(π‘₯+1)

Answer

From the π‘₯-intercepts at 1 and 7, we know that (π‘₯βˆ’1) and (π‘₯βˆ’7) must be factors of the polynomial. The fact that the curve is tangent at π‘₯=7 in the β€œparabolic” way means that the factor (π‘₯βˆ’7) must appear an even number of times: as (π‘₯βˆ’7),(π‘₯βˆ’7),β€¦οŠ¨οŠͺ.

So this is either (π‘₯βˆ’1)(π‘₯βˆ’7) or βˆ’(π‘₯βˆ’1)(π‘₯βˆ’7). The behavior as π‘₯β†’βˆž (the curve in quadrant 1) tells us that it must have a positive leading coefficient. So the answer is (C).

Example 5: Identifying the Equation of a Curve from a Sketch

Which of the following could be the equation of the given sketch?

  1. 𝑦=(π‘₯+5)(π‘₯βˆ’5)
  2. 𝑦=(π‘₯βˆ’5)(π‘₯+5)
  3. 𝑦=βˆ’(π‘₯+5)(π‘₯βˆ’5)
  4. 𝑦=(π‘₯+5)(π‘₯βˆ’5)
  5. 𝑦=βˆ’(π‘₯βˆ’5)(π‘₯+5)

Answer

The π‘₯-intercepts at βˆ’5 and 5 tell us that the polynomial must have factors (π‘₯+5) and (π‘₯βˆ’5) respectively.

The crossing at (5,0) says the (π‘₯βˆ’5) factor is to the order 1β€”it is linearβ€”while that at (βˆ’5,0) says that we have an odd power of (π‘₯+5) that is at least three.

From the given choices, we have either (π‘₯+5)(π‘₯βˆ’5) or βˆ’(π‘₯+5)(π‘₯βˆ’5).

The behavior of 𝑦 as π‘₯β†’+∞, in that π‘¦β†’βˆ’βˆž, tells us that the leading coefficient has to be negative. The solution must be 𝑦=βˆ’(π‘₯+5)(π‘₯βˆ’5) which is option (C).

In summary, this is the information we can gather from a polynomial expression in factored form, and how we can use it in identifying the graph.

Factored Polynomial β†’ Graph

  1. The distinct factors give the π‘₯-intercepts, with (π‘₯+π‘Ž) corresponding to βˆ’π‘Ž.
  2. The multiplicity π‘˜ tells us how the curve meets 𝑦=0. When π‘˜ is even, so that the factor is one of (π‘₯+π‘Ž), (π‘₯+π‘Ž),…οŠͺ, then the curve is entirely to one side of the π‘₯-axis near (βˆ’π‘Ž,0). Otherwise, for one of (π‘₯+π‘Ž), (π‘₯+π‘Ž),β€¦οŠ©, the curve crosses the π‘₯-axis. It is tangent only when π‘˜>1.

The follwoing is additional information about the polynomial expression 𝑓(π‘₯) that does not depend on the factoring but that informs about the shape of the curve 𝑦=𝑓(π‘₯):

  • 𝑓(0) determines the 𝑦-intercept (0,𝑓(0)).
  • The leading coefficient in the expanded form 𝑓(π‘₯)=𝐴π‘₯+β‹―οŠ is the number 𝐴. If this is positive then the curve 𝑦=𝑓(π‘₯) is in the 1st quadrant for large positive values of π‘₯. Otherwise, it is in the 4th quadrant.

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