College Algebra to Calculus and the TI-83 and TI-83 Plus
Lesson #12
Polynomial and Rational functions and their graphs
Polynomial function: a polynomial function of degree n, where n is a whole number, is a
function of the form y  a n x n  a n 1 x n 1  a n  2 x n 2  ...  a 2 x 2  a1 x  a0 .
a 0 is called the constant term and it gives the y-intercept of the function
a n is called the leading coefficient.
A polynomial function of degree zero is a constant function.
A polynomial function of degree one is a linear function.
Quadratic Function: a polynomial function of degree 2 is called a quadratic function.
The standard form of a quadratic function y= ax 2  bx  c where a, b, c are real numbers and
a  0.
Exercise 1. Find the zeros of the quadratic function: y  2 x 2  3x  5 . Find the vertex of the
parabola.
 Y=
Y1= 2x^2 - 3x - 5
 WINDOW Xmin=-10 Xmax=10 Ymin= -20 Ymax= 20 GRAPH
 2nd CALC 2 (trace to the left of the first zero or enter a value of x to the left)
 ENTER
 (trace to the right of the first zero or enter a value for x to the right) ENTER
 (trace to a point between the boundaries found or enter a value for x) ENTER
answer: Zero at x =-1
 2nd CALC 2 (trace to the left of the second zero) ENTER
 (trace to the right of the second zero) ENTER
 (trace to a point between the bounds ) ENTER answer: Zero at x=2.5
Graph the parabola, trace to find the vertex, use 2nd CALC to find the vertex.
 GRAPH 2nd CALC 3 (minimum)
 (trace to a point to the left of the minimum point) ENTER
 (trace to a point to the right of the minimum point) ENTER
 (trace to a point between the boundaries found) ENTER
answer: x=.75 y=-6.125 or (3/4, -49/8)
Exercise 2. Find the zeros of the quadratic function: y  4x 2  4x  37 . Graph the parabola,
trace to find the vertex, use 2nd CALC to find the vertex.
 Y=
Y1= 4x^2 - 4x +37
 WINDOW Xmin=-20 Xmax=20 Ymin=-5 Ymax= 200 GRAPH
answer: no x-intercepts or zeros.
nd
 2 CALC 3 (minimum)
 (trace to a point to the left of the minimum point) ENTER
 (trace to a point to the right of the minimum point) ENTER
 (trace to a point between the boundaries found) ENTER
answer: x=.49999 y=36 or (0.5, 36) OR (1/2, 36)
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Exercise 3. Find the zeros of the quadratic function y  2 x 2  12 x  16 . Graph the parabola; use
2nd CALC to find the vertex.
 Y=
Y1= -2x^2 +12x - 16
 WINDOW Xmin=-1 Xmax=5 Ymin=-3 Ymax= 8 GRAPH
 2nd CALC 2 (trace to the left of the first zero) ENTER
 (trace to the right of the first zero) ENTER
 (trace to a point between the boundaries found) ENTER answer: Zero at x=2
 2nd CALC 2 (trace to the left of the second zero) ENTER
 trace to the right of the second zero) ENTER
 (trace to a point between the boundaries found) ENTER answer: Zero at x = 4
 2nd CALC 4 (maximum)
 (trace to a point to the left of the maximum point) ENTER
 (trace to a point to the right of the maximum point) ENTER
 (trace to a point between the boundaries found) ENTER answer: x=3, y=2
Exercise 4. At the AP factory the cost in dollars of manufacturing q units is given by the
2
quadratic function C(q)  q  2q  1000. If q(t) = 20t units are manufactured during the
first t hours of a production run, the total cost as a function of t is given by
2
C(t)  400t  40t 1000 dollars.
a) Enter the cost functions as Y1 and Y2.
 Y= (deselect are plots and functions)
 Y1=X^2 - 2X +1000
 Y2=400X^2 - 40X +1000 2nd QUIT
b) What is the cost of manufacturing 10 units? What is the cost of manufacturing the 10th unit?
 VARS Y-VARS 1
1 to select Y1
 Y1(10)
ENTER
answer: $1080 is the cost of manufacturing 10 units.
 VARS Y-VARS 1
1 to select Y1
 Y1(10) - VARS Y-VARS 1 1
 Y1(10) - Y1(9)
ENTER
answer: $17 is the cost of manufacturing the 10th unit.
c) What is the production cost by the end of the fourth hour? What is the production cost for the
fourth hour?
 VARS Y-VARS 1
2 to select Y2
Y2(4)
Answer: $7240
 VARS Y-VARS 1 2 to select Y2
Y2(4) - VARS Y-VARS 1 2
 Y2(4) - Y2(3)
ENTER
answer: $2760
d) In how many hours will the total manufacturing cost reach $26280?
 MATH 0  CLEAR eqn: 0= VARS Y=VARS 1 2 to select Y2
 eqn: 0= Y2(X) -26280 ENTER
 X=5 (or, any other guess) ALPHA SOLVE
answer.: 8 hours 2nd QUIT
e) How many units can be produced at a minimum cost?
 WINDOW Xmin=-50 Xmax=50 Ymin=-500 Ymax= 3000 GRAPH

MATH
6
fMin( VARS Y-VARS 1 , X, 0-50, 50) ENTER
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answer: 1 unit
Exercise 5. a) Graph the polynomial function y  x 4  x3  10 x 2  4 x  24 and find the zeros,
local maximum and minimum values.
 Y=
Y1=x^4 -x^3 -10x^2 +4x +24
 WINDOW: Xmin= -5 Xmax=5 Ymin=-10 Ymax=30 GRAPH
b) Find the zeros of the polynomial function using 2nd CALC
 2nd CALC 2 (zero)
 trace to the left of the first zero, from left to right. ENTER
 trace to the right of the first zero ENTER
 trace to a point between the lower bound and the upper bound. ENTER
answer: x = -1.9999999 or x = -2
nd
 2 CALC 2 (zero)
 trace to the left of the second zero from left to right ENTER
 trace to the right of the second zero ENTER
 trace to a point between the lower bound and the upper bound. ENTER
answer: x = 2
 2nd CALC 2 (zero)
 trace to the left of the third zero from left to right. ENTER
 trace to the right of the third zero. ENTER
 trace to a point between the lower bound and the upper bound. ENTER
answer: x = 3
Note: the polynomial in factored form is y  (x  2) 2 (x  2)( x  3)
c) Find the local maximum of the function
 2nd CALC 4 (local maximum)
 trace to the left of the relative maximum. ENTER
 trace to the right of the relative maximum. ENTER
 trace to a point between the lower bound and the upper bound. ENTER
answer: x = 0.19575164 and y = 24.393787
or (another method)
 MATH 7 (fMax) (VARS Y-VARS 1 1 to select Y1 )
 fMax(Y1, X, -5, 5)
ENTER
answer .1957509203
nd
 VARS Y-VARS 1 1 Y1 ( 2 ANS ) ENTER answer: 24.39378689
d) Find the local minimum of the function
 2nd CALC 3 (local minimum)
 trace to the left of the first relative minimum ENTER
 trace to the right of the relative minimum ENTER
 trace to a point between the lower bound and the upper bound ENTER
answer: x=-2 and y=0, or
nd
 2 CALC 3 (local minimum)
 trace to the left of the second relative minimum ENTER
 trace to the right of the relative minimum ENTER
 trace to a point between the lower bound and the upper bound. ENTER
answer: x = 2.5542466 and y = -5.124256
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Exercise 6. Suppose the number of worker-hours required distributing the newspaper to x% of
100 x
the population of a certain city is given by the rational function f ( x) 
, graph the
150  x
function for 0 ≤ x ≤ 100
 Y= Y1=100x  (150 – x)
 WINDOW Xmin=0 Xmax=100 Ymin=0 Ymax=250 GRAPH
b) Find Y1(0)
Y1(10) Y1(25) Y1(50) Y1(75) Y1(100)
 TRACE
0 (to let x = 0)
ENTER
answer: When x=0, y=0
 10 (to let x = 10)
ENTER
answer: x=10
y=7.1428571
 25
ENTER
answer: x=25
y=20
 50
ENTER
answer: x=50
y=50
 75
ENTER
answer: x=75
y=100
 100
ENTER
answer: x=100 y=200
Exercise 7. Consider the rational function y 
(2 x  5)( x  4) 2
( x  6)( x  7) 2
a) Graph the function.
 Y=
Y1 = ((2x-5)(x+4)^2)  ((x-6)(x+7)^2)
 WINDOW Xmin=-10 Xmax=10 Ymin=-30 Ymax=30 GRAPH
 to obtain a closer picture at the zeros, use Ymin=-1 Ymax=1
b) Find the maximum value of the function between x=-4 and x=2.5
 GRAPH 2nd CALC 4
 trace to a point to the left of the maximum, ENTER
 trace to a point to the right of the maximum, ENTER
 trace to a point between the lower and upper boundaries, ENTER
 Maximum at x=-.089165, y=.2723390507
Another method:
 MATH 7 fMax( VARS Y-VARS 1 1 fMAX(Y1, x, -4, 2.4) ENTER
X= -0891632319
nd
 VARS Y-VARS 1 1 Y1 2 ANS
ENTER
Y=.2723390507
c) Graph the rational function and find the y-intercept.
 GRAPH
 TRACE
0 ENTER y=.27210884 is the y-intercept
d) Find the horizontal asymptote by finding the values of Y1(1000000) and Y1(-1000000)
 VARS Y-VARS 1 1 Y1(2nd EE 6) ENTER 1.999995
 VARS Y-VARS 1 1 Y1(-EE6) ENTER 2.000005
answer: the horizontal asymptote is y=2
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Exercise 8. Consider the rational function y 
x3  2 x 2  4 x  8
x 4  5 x 3  3x 2  5 x  4
a) Find the zeros of the function (the zeros of the numerator)
 Y=
y1=x^3 – 2x^2 – 4x +8
 WINDOW Xmin=-5 Xmax=5 Ymin=-10 Ymax=15 GRAPH
 2nd CALC 2 Left Bound? Trace to a point to the left of the zero ENTER
 Right Bound? Trace to a point to the right of the zero ENTER
 Trace to a point between the bounds ENTER
Answer Zero x = -2, y=0
nd
 2 CALC 2 Left Bound? Trace to a point to the left of the next zero ENTER
 Right Bound? Trace to a point to the right of the zero ENTER
 Trace to a point between the bounds ENTER Answer: Zero x = 2, y=0.
Note: 2 is a zero of multiplicity two because the graph bounces at x = 2.
Note: the factored form of the numerator is (x  2) 2 ( x  2) (Check by Algebra)
b) Find the equations of the vertical asymptotes (the poles or zeros of the denominator)
 Y=
y2=x^4+5x^3 +3x^2 – 5x -4 (deselect y1)
 WINDOW Xmin=-7 Xmax=5 Ymin=-20 Ymax=10 GRAPH
 2nd CALC 2 Left Bound? Trace to a point to the left of the first zero ENTER
 Right Bound? Trace to a point to the right of the zero ENTER
 Trace to a point between the bounds ENTER
Answer: Zero x = -4 y=0
nd
 2 CALC 2 Left Bound? Trace to a point to the left of the next zero ENTER
 Right Bound? Trace to a point to the right of the zero ENTER
 Trace to a point between the bounds ENTER
Answer: Zero x = -1 y=0.
nd
 2 CALC 2 Left Bound? Trace to a point to the left of the next zero ENTER
 Right Bound? Trace to a point to the right of the zero ENTER
 Trace to a point between the bounds ENTER
Answer: Zero x = 1 y=0
 Answer: The vertical asymptotes are the lines x = - 4, x = -1 (multiplicity 2) and x = 1
The factor form of the denominator is ( x  4)( x  1) 2 ( x  1)
c) Write the function in factored form.
( x  2)( x  2) 2
answer: y 
( x  4)( x  1) 2 ( x  1)
d) Graph the rational function and find the y-intercept
 Y=
Y3= VAR Y-VAR 1 1  VAR Y-VAR 1 2
 WINDOW Xmin=-7 Xmax=3 Ymin=-5 Ymax=5 GRAPH
 TRACE 0 ENTER
answer: y = -2 is the y-intercept
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