Rational Functions and Asymptotes Section 2.6

Objectives Find the domain of rational functions. Find horizontal and vertical asymptotes of graphs of rational functions. Use rational functions to model and solve real life problems.

Continuity A graph is continuous over a domain, if and only if it is continuous at every point of the domain. A continuous function can be drawn without lifting the pen from the paper over its entire domain (ie. There are no breaks). If a relation is not continuous at some point (ie. there is a break in the graph), it is said the graph is discontinuous at that point.

Discontinuities Jump discontinuity –Cause: Piecewise function Line (or asymptotic) discontinuity –Cause: Rational function Point discontinuity –Cause: Rational function

Jump Discontinuity

Line Discontinuity

Point Discontinuity

Rational Functions Rational function is the ratio of two polynomial functions Asymptote comes from combining 3 Greek words “an-sum-piptein” meaning does not fall together with. An asymptote is a curve that another curve approaches but does not ultimately cross.

Rational Functions A rational number is a number that can be written in the form of a fraction. So likewise, a rational function is function that is presented in the form of a fraction. Two examples of rational functions; –we can generate a graph of polynomials when we divide them with the Factor Theorem  i.e. f(x) = (x 3 - 2x + 1)/(x - 1). 1)If x-1 is a factor of p(x), then we observe a hole in the graph of f(x). 2)If x-1 is not a factor of p(x), then we observe an vertical asymptote in the graph of f(x).

Rational Functions

Several examples of rational functions are the reciprocal functions of linear functions, i.e. f(x) = 1/(x + 2) and quadratic functions i.e. g(x) = 1/(x 2 - 3x - 10) and the tangent function y = tan(x).

Domain, Range and Zeroes of Rational Functions Given the rational function r(x) = n(x)/d(x), The domain of rational functions involve the fact that we cannot divide by zero. Therefore, any value of x that creates a zero denominator is a domain restriction. Thus in r(x), d(x) cannot equal zero. For the zeroes of a rational function, we simply consider where the numerator is zero (i.e. 0/d(x) = 0). So we try to find out where n(x) = 0 To find the range, we must look at the various sections of a rational function graph and look for max/min values EXAMPLES: Graph and find the domain, range, zeroes of –f(x) = 7/(x + 2), –g(x) = x/(x 2 - 3x - 4), and –h(x) = (2x 2 + x - 3)/(x 2 - 4)

Domain, Range and Zeroes of Rational Functions EXAMPLES: Graph and find the domain, range, zeroes of –f(x) = 7/(x + 2), D f =(- ∞, ∞ )\2, R f =(- ∞, ∞ )\0, zeros: none –g(x) = x/(x 2 - 3x - 4), D f =(- ∞, ∞ )\-1, 4, R f =(- ∞, ∞ ), zeros: x=0 –h(x) = (2x 2 + x - 3)/(x 2 - 4), D f =(- ∞, ∞ )\±2, Rf=(- ∞, ∞ )\2, zeros: x=1, -3/2

Your Turn: Find the domain of and discuss the behavior of f near any excluded x -values. The domain of f is all real numbers x except x =1. As x approaches 1 from the left, f decreases without bound. As x approaches 1 from the right, f increases without bound.

Reciprocal Function

y = 1/x, x≠0 Has a shape known as a rectangular hyperbola Properties of y = 1/x –The sign of y is the same as the sign of x –As x increases, values of y decease. –The function is undefined at x = 0, ∴ at x = 0 there exists a vertical asymptote. –The graph never crosses the x-axis. The x- axis is a horizontal asymptote.

Reciprocal Function Its domain is {x E R| x  0} Its range likewise is {y E R| y  0} We have a vertical asymptote on the y-axis (x = 0) and horizontal asymptotes on the x-axis (y = 0) Two key points are (1,1) and (-1,-1)

What does look like? Translation Rules Combining transformations…. For any graphy = f(x) The translation y = f(x - a) moves it ‘a’ units to the right The translation y = f(x + a) moves it ‘a’ units to the left The translation y = f(x) + b moves it ‘b’ units up … this can be considered as y - b = f(x) --> Shift 3 units parallel to x-axis

What does look like? --> Shift 3 units parallel to x-axis f(x) = 1/x f(x) = 1/(x-3) x=3

What does look like? --> then stretch by factor 2 parallel to y-axis The transformation y = a f(x) stretches by a factor ‘a’ along the y-axis The translation y = f(x/b) stretches by a factor ‘b’ along the x-axis --> Shift 3 units parallel to x-axis

What does look like? f(x) = 1/x f(x) = 1/(x-3) x=3 f(x) = 2/(x-3) --> then stretch by factor 2 parallel to y-axis --> Shift 3 units parallel to x-axis

What does look like? --> Shift 3 units parallel to x-axis The transformation y = - f(x) reflects in the x-axis The translation y = f(-x) reflects in the y-axis --> then reflect in x-axis

What does look like? --> Shift 3 units parallel to x-axis f(x) = 1/x f(x) = 1/(x-3) x=3 --> then reflect in x-axis f(x) = 1/(3-x)

Transformations of f(x) = 1/x Reflection about The x-axis Shrink or Stretch along y-axis Translation in the x-direction Has vertical asymptote at x=b Translation in The y-direction Has horizontal Asymptote y=c

Y = 1/x 2, x≠0 Has a shape known as a Truncus. Properties –Function is undefined at x = 0. –Asymptotes: Vertical x = 0 Horizontal y = 0 –Graph is symmetrical about the y-axis.

What does 1/x 2 look like?

Reciprocal of Quadratic Functions Now we apply the same idea of a reciprocal to quadratic functions We have seen that the key points on a function are the x- intercepts (as these form the vertical asymptotes of the reciprocal function) and we know that quadratic functions have either 0,1, or 2 x-intercepts Therefore, we expect the reciprocal function to have either 0,1, or 2 vertical asymptotes The horizontal asymptote will remain y = 0, as our reciprocal function equation is 1/f(x) so as f(x) gets larger, the value of the reciprocal gets smaller

Graphs of Reciprocal Quadratic Functions Here are some graphs of quadratic functions that have 2 x-intercepts and their reciprocal function:

Graphs of Reciprocal Quadratic Functions Here are some graphs of quadratic functions that have 1 x-intercept and their reciprocal function:

Graphs of Reciprocal Quadratic Functions Here are some graphs of some quadratic functions that have no x- intercepts and their the reciprocal function:

Analysis of the Reciprocal of Quadratic Functions The domain is restricted to wherever the x-intercept(s) of the original quadratic function. A key point on the original was the vertex  in the reciprocal, the vertex relates to a range restriction  in that the reciprocal of the y-value of the vertex is a key point on the reciprocal The vertical asymptote(s) occur where the roots of the original were. The horizontal asymptote is y = 0 or the x-axis.

Vertical and Horizontal Asymptotes A vertical asymptote occurs when the value of the function increases or decreases without bound as the value of x approaches a from the right and from the left. We symbolically present this as f(x)  + ∞ as x  a + or x  a - We re-express this idea in limit notation  lim x  a+ f(x) = + ∞ A horizontal asymptotes occurs when a value of the function approaches a number, a, as x increases or decreases without bound. We symbolically present this as as f(x)  + a as x  + ∞ or x  - ∞ We can re-express this idea in limit notation  lim x  ∞ f(x) = a

Vertical and Horizontal Asymptotes

Finding the Equations of the Asymptotes Simplify (reduce to lowest terms – relatively prime) the rational function by dividing out common factors between the numerator and the denominator. Any factor eliminated will cause a hole discontinuity where it is equal to zero. To find the equation of the vertical asymptotes, we simply find the restrictions in the denominator (set the factor in the denominator equal to zero) and there is our equation of the asymptote i.e. x = a. To find the equation of the horizontal asymptotes, we can work through it in two manners. First is; 1)f(x)=0, if the degree of the numerator is less than the degree of the denominator. 2)f(x)=a/b, if the degree of the numerator equals the degree of the denominator, where a & b are the leading coefficients of the numerator & denominator respectively. 3)No horizontal asymptote if the degree of the numerator is greater than the degree of the denominator. The second approach, is to rearrange the equation to make it more obvious as to what happens when x gets infinitely positively and negatively large.

Vertical and Horizontal Asymptotes

Finding the Equations of the Asymptotes ex. Find the asymptotes of y = (x+2)/(3x-2) No factors can be eliminated between the numerator and the denominator  thus there are no point discontinuities. Set 3x-2 = 0 and solve  thus we have an vertical asymptote at x = 2/3. The degree of the numerator equals the degree of the denominator  thus we have horizontal asymptote at f(x)=a/b or f(x)=1/3.

Finding the Equations of the Asymptotes Alternatively, we can find the horizontal asymptotes of y = (x+2)/(3x-2) using algebraic methods  mutiple the numerator and denominator by 1/(x to the highest degree of x in the function). as x  + ∞, then 2/x  0

Vertical 4x - 2 =0 4x = 2 x = 0.5 is an asymptote Horizontal needs a little trick --> Divide numerator and denominator by x … as x gets very very big Then (gets near to 3/4) y = 3 / 4 is an asymptote Example:

Examples Further examples to do  Find vertical and horizontal asymptotes for: y = (4x)/(x 2 +1) y = (2-3x 2 )/(1-x 2 ) y = (x 2 - 3)/(x+5)

Solutions y = (4x)/(x 2 +1) No vertical, Horizontal f(x)=0 y = (2-3x 2 )/(1-x 2 ) Vertical x=1& x=-1, Horizontal f(x)=3 y = (x 2 - 3)/(x+5) Vertical x=-5, No Horizontal

Your Turn: Find the horizontal and vertical asymptotes of. Horizontal asymptote: y = 5 Vertical asymptotes: x = 1, x = -1

Your Turn: Find the horizontal and vertical asymptotes and holes in the graph of Vertical asymptote: x = -1 Horizontal asymptote: y = 1 Hole; (-3, 0)

Your Turn: For the function f, find the domain, the vertical asymptote of f, and the horizontal asymptote of f. Domain; all real numbers except Vertical asymptote: Horizontal asymptote: y = -2/3

Assignment Pg 152 – 155: odd, 31 – 37 odd