Lesson Explainer: Graphs of Polynomial Functions | Nagwa Lesson Explainer: Graphs of Polynomial Functions | Nagwa

Lesson Explainer: Graphs of Polynomial Functions Mathematics

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 𝑥-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(𝑥)=6𝑥+27𝑥+6𝑥117𝑥+54.

What can we tell about the graph 𝑦=𝑓(𝑥) 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 “typical” 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 whether independent variable 𝑥 goes towards + or . This behaviour is determined by the leading term, in our case the 6𝑥 term, which “dominates” when 𝑥 is very large. In our case, whether 𝑥=10 or 𝑥=10, the leading term contributes 6×10=6×10=6,0000,24zeros which is very large and negative. The 𝑥 term only has 18 zeros. So the graph of 𝑦=𝑓(𝑥) is “facing downwards” as in example (c) above.

So far, we have this picture.

The best information comes from a factored form of the polynomial 𝑓(𝑥), when it is available. Here 𝑓(𝑥)=6𝑥+27𝑥+6𝑥117𝑥+54=3(2𝑥1)(𝑥+2)(𝑥3), telling us that 𝑓(𝑥)=0 when one of the factors (2𝑥1), (𝑥+2), or (𝑥3) is zero. In other words, when 𝑥=12,𝑥=2,𝑥=3.or

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

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

The way the curve 𝑦=𝑓(𝑥) cuts the 𝑥-axis near 𝑥=12 and 𝑥=2 is “linear” since we have (2𝑥1) and (𝑥+2).

Near 𝑥=3, the factor (𝑥3) makes the graph look very much like the curve 𝑦=70(𝑥3) shown as the dashed curve. Recall that the curve 𝑦=70(𝑥3) is just the curve 𝑦=70𝑥 translated horizontally by 3 units.

Similarly, if (𝑥+𝜋) is a factor of the polynomial 𝑔(𝑥), then 𝑦=𝑔(𝑥) looks like a multiple of (𝑥+𝜋) near the intercept (𝜋,0). This is a vertically stretched form of 𝑦=𝑥 or 𝑦=𝑥 shifted 𝜋 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(𝑥)=(𝑥+3)(𝑥1)(𝑥4)?


The zeros of 𝑓(𝑥) are 3, 1, and 4 from the factorization. So the choice is between (C) or (E). As the leading coefficient of 𝑓(𝑥) 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 𝑥-axis, unlike the 𝑦-intercept in (E).

Example 2: Identifying the Graph of a Quadratic Using Factoring

Solve 𝑥4𝑥+4=0 by factoring, and hence determine which of the following figures would be a sketch of 𝑦=𝑥+4𝑥+4.


In order to sketch 𝑦=𝑥4𝑥+4, we begin by factoring 𝑥4𝑥+4: 𝑥4𝑥+4=(𝑥2)(𝑥2)=(𝑥2) so that 𝑥4𝑥+4=0 has the single solution 𝑥=2. This gives (2,0) as the only 𝑥-intercept of 𝑦=𝑥+4𝑥+4. The choice is between (C) and (D).

The leading term here is 𝑥, with leading coefficient 1, which is positive. Therefore, as 𝑥 goes to +, we expect that 𝑥4𝑥+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

1.DegreeNumber of intercepts with horizontal line
2.Odd/even degreeBehavior (1) as 𝑥±
3.Sign of leading coefficientBehavior (2) as 𝑥
4.Value of constant term𝑦-Intercept
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=𝑘 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 “arms” as 𝑥± 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 𝑥+. If the leading coefficient is positive, this is 1; otherwise, it is 4. The following figure shows the graphs of two degree 5 polynomials, 𝑃 and 𝑃, with leading coefficients of different signs.
  4. A polynomial expression 𝑃(𝑥) is a sum of multiples of powers of 𝑥 and a term of “degree 0”—the constant term. Setting 𝑥=0 says this constant is 𝑃(0), which gives the 𝑦-coordinate of the point on the curve 𝑦=𝑃(𝑥) that is also on the 𝑦-axis.
  5. The graph 𝑦=𝑃(𝑥) above cuts the 𝑥-axis at the number 4. This means that 𝑃(4)=0 and therefore (𝑥4) is a factor of 𝑃(𝑥). Likewise, 𝑃(𝑥)=(𝑥+6)(𝑥+3)(𝑥4)𝑄(𝑥) for some polynomial 𝑄(𝑥).
  6. Knowing that a number (such as 2) is the zero of a polynomial 𝑃(𝑥) tells us that the graph 𝑦=𝑃(𝑥) meets the 𝑥-axis at (2,0). But they can “meet” in essentially three different ways, as illustrated in the figure below.
    These are as follows:
    1. A sharp crossing like 𝑦=𝑥2: this is the same as 𝑦=𝑥(𝑥2) or indeed 𝑦=(𝑥2)𝑄(𝑥) for any 𝑄(𝑥) that is not zero at 𝑥=2.
    2. A tangency like 𝑦=(𝑥2) with the entire graph to the “same side” of the 𝑥-axis: this is the case for 𝑦=(𝑥2) or indeed 𝑦=(𝑥2)𝑄(𝑥), where 𝑘 is an even number, and 𝑄(2)0.
    3. A tangency like 𝑦=(𝑥2), or the 𝑦=(𝑥2) shown: with odd powers (𝑥2), 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=𝑥(𝑥+1)(𝑥+3)
  2. 𝑦=(𝑥1)(𝑥+3)(𝑥3)
  3. 𝑦=(𝑥+1)(𝑥+3)(𝑥3)
  4. 𝑦=(𝑥1)(𝑥+3)(𝑥3)
  5. 𝑦=(𝑥+1)(𝑥+3)(𝑥3)


From the intercepts 3,1, and 3, we know that (𝑥+3), (𝑥+1), and (𝑥3) must be factors of the polynomial used. This leaves the two choices (C) and (E).

Since the curve is in the 4th quadrant as 𝑥+, the leading coefficient must be negative. So the only possibility is (C), where the coefficient of 𝑥 is 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=(𝑥+1)(𝑥+7)
  2. 𝑦=(𝑥1)(𝑥7)
  3. 𝑦=(𝑥7)(𝑥1)
  4. 𝑦=(𝑥+7)(𝑥+1)
  5. 𝑦=(𝑥+7)(𝑥+1)


From the 𝑥-intercepts at 1 and 7, we know that (𝑥1) and (𝑥7) must be factors of the polynomial. The fact that the curve is tangent at 𝑥=7 in the “parabolic” way means that the factor (𝑥7) must appear an even number of times: as (𝑥7),(𝑥7),.

So this is either (𝑥1)(𝑥7) or (𝑥1)(𝑥7). The behavior as 𝑥 (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=(𝑥+5)(𝑥5)
  2. 𝑦=(𝑥5)(𝑥+5)
  3. 𝑦=(𝑥+5)(𝑥5)
  4. 𝑦=(𝑥+5)(𝑥5)
  5. 𝑦=(𝑥5)(𝑥+5)


The 𝑥-intercepts at 5 and 5 tell us that the polynomial must have factors (𝑥+5) and (𝑥5) respectively.

The crossing at (5,0) says the (𝑥5) factor is to the order 1—it is linear—while that at (5,0) says that we have an odd power of (𝑥+5) that is at least three.

From the given choices, we have either (𝑥+5)(𝑥5) or (𝑥+5)(𝑥5).

The behavior of 𝑦 as 𝑥+, in that 𝑦, tells us that the leading coefficient has to be negative. The solution must be 𝑦=(𝑥+5)(𝑥5) 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 → Graph

  1. The distinct factors give the 𝑥-intercepts, with (𝑥+𝑎) corresponding to 𝑎.
  2. The multiplicity 𝑘 tells us how the curve meets 𝑦=0. When 𝑘 is even, so that the factor is one of (𝑥+𝑎), (𝑥+𝑎),, then the curve is entirely to one side of the 𝑥-axis near (𝑎,0). Otherwise, for one of (𝑥+𝑎), (𝑥+𝑎),, the curve crosses the 𝑥-axis. It is tangent only when 𝑘>1.

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

  • 𝑓(0) determines the 𝑦-intercept (0,𝑓(0)).
  • The leading coefficient in the expanded form 𝑓(𝑥)=𝐴𝑥+ is the number 𝐴. If this is positive then the curve 𝑦=𝑓(𝑥) is in the 1st quadrant for large positive values of 𝑥. Otherwise, it is in the 4th quadrant.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.