Functions · Lesson 5

Polynomial and rational graphs

Zeros come from the factors. For p(x)=(x−4)(x+1)2p(x) = (x - 4)(x + 1)^2, the zeros are the xx-values that make a factor 0: x=4x = 4 and x=−1x = -1. Watch the sign: the factor x−4x - 4 gives x=4x = 4, and x+1x + 1 gives x=−1x = -1. The exponent on a factor is the zero's multiplicity. At a zero of odd multiplicity (1, 3, ...) the graph crosses the xx-axis; at a zero of even multiplicity (2, 4, ...) it touches the axis and turns back.

End behavior comes from the leading term. Far from the origin, a polynomial acts like its highest-degree term. Even degree: both ends point the same way, up when the leading coefficient is positive and down when it is negative. Odd degree: the ends point opposite ways; with a positive leading coefficient the graph falls on the left and rises on the right. In factored form, multiply the leading parts: −(x+3)(x−1)2-(x + 3)(x - 1)^2 has leading term −x3-x^3. The constant term only gives the yy-intercept.

Rational functions: factor and cancel first. In x−1(x−1)(x+2)\frac{x - 1}{(x - 1)(x + 2)}, the factor x−1x - 1 cancels. If a factor cancels completely, with no copy of it left in the denominator, its zero is a hole, one missing point. A zero of the denominator that is left after cancelling is a vertical asymptote. So this graph has a hole at x=1x = 1 and a vertical asymptote at x=−2x = -2. The height of a hole comes from the simplified form: 11+2=13\frac{1}{1 + 2} = \frac{1}{3}.

Horizontal asymptotes compare degrees. If the numerator's degree is smaller, the asymptote is y=0y = 0. If the degrees are equal, it is the ratio of the leading coefficients, so 2x+4x−1\frac{2x + 4}{x - 1} levels off at y=2y = 2. If the numerator's degree is larger, there is no horizontal asymptote.

The horizontal asymptote of a rational function, from the degrees of the numerator (top) and denominator (bottom)
DegreesHorizontal asymptoteExample
Top less than bottomy=0y = 03xx2+1\frac{3x}{x^2 + 1}: y=0y = 0
Top equal to bottomRatio of the leading coefficients6x−12x+5\frac{6x - 1}{2x + 5}: y=3y = 3
Top greater than bottomNonex2x−4\frac{x^2}{x - 4}: none
The graph of y = \frac{x - 1}{(x - 1)(x + 2)}, with a hole and a vertical asymptote−7−6−5−4−3−2−112345−5−4−3−2−112345xyhole (1, 1/3)
The graph of y=x−1(x−1)(x+2)y = \frac{x - 1}{(x - 1)(x + 2)}, with a hole and a vertical asymptote Two branches separated by a dashed vertical asymptote at x = −2. The dashed line y = 0 (the x-axis) is the horizontal asymptote. The right branch comes down from very high values just right of x = −2 and levels off toward the x-axis; it has an open circle, a hole, at (1, 1/3). The left branch stays below the x-axis.

Worked example Easy

Which statement describes the graph of p(x)=(x−7)(x+4)2p(x) = (x - 7)(x + 4)^2 where it meets the xx-axis?

  1. It touches the xx-axis at x=−4x = -4 and crosses it at x=7x = 7.Answer
  2. It crosses the xx-axis at x=−4x = -4 and touches it at x=7x = 7.
  3. It touches the xx-axis at x=4x = 4 and crosses it at x=−7x = -7.
  4. It crosses the xx-axis at both x=−4x = -4 and x=7x = 7.

How to solve it

  1. Set each factor equal to 0: x−7=0x - 7 = 0 gives x=7x = 7, and x+4=0x + 4 = 0 gives x=−4x = -4.
  2. The factor x−7x - 7 has exponent 1, which is odd, so the graph crosses at x=7x = 7.
  3. The factor (x+4)2(x + 4)^2 has exponent 2, which is even, so the graph touches at x=−4x = -4 and turns back.

Why each choice is right or wrong

  • A. Correct. The squared factor (x+4)2(x + 4)^2 gives an even-multiplicity zero at x=−4x = -4, where the graph touches; x−7x - 7 appears once, so the graph crosses at x=7x = 7.
  • B. Incorrect. This swaps the multiplicities. The exponent 2 is on x+4x + 4, so the graph touches at x=−4x = -4; the single factor x−7x - 7 makes it cross at x=7x = 7.
  • C. Incorrect. This flips the signs of the zeros. The factor x+4x + 4 is 0 at x=−4x = -4, not 4, and x−7x - 7 is 0 at x=7x = 7, not −7-7.
  • D. Incorrect. Saying it crosses at both ignores the exponent. A zero of even multiplicity, like x=−4x = -4 from (x+4)2(x + 4)^2, is a touch-and-turn point.

Worked example Medium

The graph of the polynomial y = p(x)−5−4−3−2−11234−12−10−8−6−4−2246810120xy(−3, 0)(0, −3)(1, 0)
The graph of the polynomial y=p(x)y = p(x) A smooth curve that comes down from the upper left, crosses the x-axis at (−3, 0), keeps falling to a low point near (−1.7, −9.5), rises through the y-intercept (0, −3), touches the x-axis at (1, 0) without crossing it, then turns and falls toward the lower right.

The graph of a polynomial function y=p(x)y = p(x) is shown in the standard (x,y)(x, y) coordinate plane. Its only xx-intercepts are (−3,0)(-3, 0) and (1,0)(1, 0). Which of the following could be p(x)p(x)?

  1. (x+3)(x−1)2(x + 3)(x - 1)^2
  2. −(x−3)(x+1)2-(x - 3)(x + 1)^2
  3. −(x+3)2(x−1)-(x + 3)^2(x - 1)
  4. −(x+3)(x−1)2-(x + 3)(x - 1)^2Answer

How to solve it

  1. Zeros: (−3,0)(-3, 0) needs the factor x+3x + 3, and (1,0)(1, 0) needs x−1x - 1.
  2. The graph touches at x=1x = 1, so x−1x - 1 has an even power; it crosses at x=−3x = -3, so x+3x + 3 has an odd power.
  3. Ends: up on the left, down on the right means odd degree with a negative leading coefficient, −x3-x^3.
  4. Check the yy-intercept: −(0+3)(0−1)2=−3-(0 + 3)(0 - 1)^2 = -3, which matches (0,−3)(0, -3).

Why each choice is right or wrong

  • A. Incorrect. (x+3)(x−1)2(x + 3)(x - 1)^2 has the right zeros and powers, but its leading term is +x3+x^3, so it falls on the left and rises on the right, the opposite of this graph. Its yy-intercept is 3, not −3-3.
  • B. Incorrect. −(x−3)(x+1)2-(x - 3)(x + 1)^2 flips the signs of the zeros: its factors are 0 at x=3x = 3 and x=−1x = -1. A zero at x=−3x = -3 comes from the factor x+3x + 3.
  • C. Incorrect. −(x+3)2(x−1)-(x + 3)^2(x - 1) puts the even power on the wrong factor. It would touch the axis at x=−3x = -3 and cross at x=1x = 1; this graph crosses at −3-3 and touches at 1.
  • D. Correct. The factors give the zeros −3-3 and 1, the square makes the graph touch at 1, the leading term −x3-x^3 gives the ends, and p(0)=−3p(0) = -3 matches the graph.

Trap Calling a hole an asymptote

Every zero of the denominator breaks the graph, but not in the same way. If the factor cancels with one in the numerator, the graph has a hole there, not an asymptote. Listing every denominator zero as a vertical asymptote is a common slip, so factor and cancel before you decide.

Trap Flipped signs and the constant term

The factor x−3x - 3 gives the zero x=3x = 3, not −3-3; flipping every sign is a common slip. And the ends of a polynomial's graph depend only on the leading term. The constant term tells you where the graph crosses the yy-axis, nothing about where the ends go.

Calculator When a calculator is faster

Graphing shows zeros, turning points and asymptotes at once. If you test online, type y = x^2 - x - 2/x^2 - 1 into the built-in Desmos calculator and you see the vertical asymptote at x=1x = 1. The hole at x=−1x = -1 doesn't show, which is why you factor by hand. To match a graph with factored-form choices, graph the choices one at a time and compare the intercepts and the ends.

A hole is easy to miss on any calculator: the curve is drawn straight through it. Factor and cancel by hand to find holes. With a handheld calculator, testing a choice at one or two xx-values is often the quickest check.