Graphing Techniques Including Asymptotes

Steps for Graphing Rational Functions

A comprehensive graph of a rational function will show the following characteristics.

  1. all x- and y-intercepts

  2. all asymptotes: vertical, horizontal, and/or oblique

  3. the point at which the graph intersects its nonvertical asymptote (if there is any such point)

  4. the behavior of the function on each domain interval determined by the vertical asymptotes and x-intercepts

Graphing a Rational Function

Let LaTeX: f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}\:f(x)=p(x)q(x)define a function where p(x) and q(x) are polynomials and the rational expression is written in lowest terms.

To sketch its graph, follow these steps.

  1. Find any vertical asymptotes.
  2. Find any horizontal asymptotes.
  3. Find the y-intercept by evaluating f(0).
  4. Plot the x-intercepts, if any, by solving f(x) = 0. (These will be the zeros of the numerator, p(x).)
  5. Determine whether the graph will intersect its nonvertical asymptote y = b or y = mx + b by solving
    f(x) = b or f(x) = mx + b.

  6. Plot selected points, as necessary. Choose an x-value in each domain interval determined by the vertical asymptotes and x-intercepts.

  7. Complete the sketch.

Example

Graph LaTeX: f\left(x\right)=\frac{x+1}{2x^2+5x-3}.f(x)=x+12x2+5x−3.

Solution

  1. Since LaTeX: 2x^2+5x-3=\left(2x-1\right)\left(x+3\right),\:2x2+5x−3=(2x−1)(x+3), the vertical asymptotes have equations LaTeX: x=\frac{1}{2}x=12 and LaTeX: x=-3.x=−3.
  2. By dividing the numerator and the denominator of f(x) by LaTeX: x^2x2 , we find that the horizontal asymptote is the x-axis.
  3. The y-intercept is LaTeX: \left(0,-\frac{1}{3}\right),\:(0,−13),since LaTeX: f\left(0\right)=\frac{0+1}{2\left(0^2\right)+5\left(0\right)-3}=-\frac{1}{3}.\:The\:y-intercept\:is\:the\:ratio\:of\:the\:constant\:terms.f(0)=0+12(02)+5(0)−3=−13.They−interceptistheratiooftheconstantterms.
  4. The x-intercept is found by solving f(x) = 0.  LaTeX: \frac{x+1}{2x^2+5x-3}=0\:\:If\:a\:fraction\:\left(or\:rational\:expression\right)\:is\:equal\:to\:zero\:then\:its\:numerator\:must\:be\:equal\:to\:zero.x+12x2+5x−3=0Ifafraction(orrationalexpression)isequaltozerothenitsnumeratormustbeequaltozero.
    LaTeX: x+1=0x+1=0
    LaTeX: x=-1x=−1
    The x-intercept is LaTeX: \left(-1,0\right).(−1,0).

  5. To determine whether the graph intersects its horizontal asymptote, solve LaTeX: f\left(0\right)=0\longleftarrow This\:is\:the\:y-value\:of\:the\:horizontal\:asymptotef(0)=0⟵Thisisthey−valueofthehorizontalasymptote

    Since the horizontal asymptote is the x-axis, the solution of this equation was found in Step 4. The graph intersects its horizontal asymptote at (−1, 0).

  6. Plot a point in each of the intervals determined by the x-intercepts and vertical asymptotes, to get an idea of how the graph behaves in each interval.

    Interval Test Point Value of f(x) Sign of f(x) Graph Above or Below the x-Axis
    LaTeX: \left(-\infty,-3\right)(−∞,−3) LaTeX: -4−4 LaTeX: -\frac{1}{3}−13 Negative Below
    LaTeX: \left(-3,-1\right)(−3,−1) LaTeX: -2−2 LaTeX: \frac{1}{5}15 Positive Above
    LaTeX: \left(-1,\frac{1}{2}\right)(−1,12) 0 LaTeX: -\frac{1}{3}−13 Negative Below
    LaTeX: \left(\frac{1}{2},\infty\right)(12,∞) 2 LaTeX: \frac{1}{5}15 Positive Above

  7. Complete the sketch. This function is decreasing on each interval of its domain—that is, on LaTeX: \left(-\infty,-3\right),\:\left(-3,\frac{1}{2}\right),\:(−∞,−3),(−3,12),and LaTeX: \left(\frac{1}{2},\infty\right).(12,∞).
    Graph incorporating all of the information from steps 1-7.PNG