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Mathematics Dictionary
Dr. K. G. Shih
How to Use Graphic Calculator ?
Subjects


  • Q01 | - How to start the program ?
  • Q02 | - Describe the curve of y = sinh(x)
  • Q03 | - Intersection of y = x^2 - 6*x + 8 with its inverse
  • Q04 | - Find asymptote of y = ((x-1)^3)/(2*x)
  • Q05 | - Sketch y = (x^3 + 1)/x
  • Q06 | -
  • Q07 | -
  • Q08 | -
  • Q09 | -
  • Q10 | -
    • Answers


    Q01. How to Start the program ?

    Start the program
    • Start Program ABH on current location
    • Run at current location
      • Step 1 : We select run on curent location (no download)
      • Step 2 : We select yes to run.
      • Now the program is ready.
    Get the menu
    • Click the Start Command : The subjects are appeared in the upper box.
    • Click a subject : the programs appeared in the lower box.
    Run a sketch program
    • Step 1 : Start the program.
    • Step 2 : Click start command.
    • Step 3 : Select a subject in the upper box.
    • Step 4 : Select a program in the lower box.
    • Step 5 : Give the data on keyboard. (No data if it is demo only).
    Back to menu
    • Click Back Command in the program ABH
    • Select other subject or program.

    Go to Begin

    Q02. Describe the curve of y = sinh(x)

    Program
    • Start Program ABH Program GC 05 01
    • Sketch the diagram
      • Click start
      • Click subject 5 in upper box
      • Click program 1 in lower box
      • No data is required.
    Describe the curve of y = sinh(x)
    • When x LT 0
      • Range is from -infinite to zero
      • The curve is increasing (y' GT 0)
      • The curve is concave downward (y" LT 0)
    • When x EQ 0
      • The value of y is zero
      • AT point x = 0 and y = 0, the curve is a point of inflection
    • When x GT 0
      • Range is from 0 to infinite
      • The curve is increasing (y' GT 0)
      • The curve is concave upward (y" GT 0)
    Reference
    • Compare with sine curve
    • Compare with tangent curve
    • See PM 14

    Go to Begin

    Q03. Interections of y = x^2 - 6*x + 8 with its inverse

    Program
    • Start Program ABH Program GC 06 10
    • Sketch the diagram
      • Click start
      • Click subject 06 in upper box
      • Click program 10 in lower box
      • Input data : 1, -6, 8
    Study
  • 1. How many intesections are there ?
  • 2. Estimate the points of intersections
  • 3. How to find the intersections ?
    • Find intersection of y = x^2 - 6*x + 8 with y = x
    • That is x^2 - (6*x + x) + 8 = 0
    • Use discriminant formula : D = b^2 - 4*a*c
    • Hence D = (-7)^2 - 4*1*8 = 25
    • Since D GT than zero, it at least has two points of intersection
      • x1 = (-(-7) + Sqr(D))/2 = (7 + 5)/2 = 6 and y1 = 6
      • x1 = (-(-7) - Sqr(D))/2 = (7 - 5)/2 = 1 and y2 = 1
  • 4. How to find other two points of intersections
    • Solve y = x^2 - 6*x + 8 and x = y^2 - 6*y + 8
    • Hence x = (x^2 - 6*x + 8)^2 - 6*(x^2 - 6*x + 8) + 8
    • Solve this quartic equation by dividing (x - 1) and (x - 6)
    • Then the remainder is a quadratic equation
    • Then we can find other two points of intersection
    Reference
    • Use Inverse command
    • Four demo of point intersections of y = a*x^2+b*x+c with its inverse
    • Case 1 : No point of intersection
    • Case 2 : one point of intersection
    • Case 3 : two points of intersection
    • Case 4 : Four points of intersection
    • Write down the four functions as home work
    Home work
    • Get the quartic equation
    • Use synthetic division to find the remainder after divide by (x-1) and (x-6)
    • Find other two roots

    Go to Begin

    Q04. Find asymptote of y = ((x-1)^3)/(2*x)

    Method 1 : Computer program
    • Start Program ABH Program 03 05
    • Sketch the curve
      • Click start
      • Click subject 3 in upper box
      • Click program 5 in lower box
      • Give power M : 3
    Method 2 : Use signs and asymptotes
    • Asymptotes
      • Vertical asymptote at x = 0
      • Parabola asymptote is y = (x^2)/2 - (3*x)/2 + 3/2
    • Signs of the curve
      • If x LT 0, y is positive
      • If x is between 0 and 1, y is negative
      • If x is GT 1, y is positive
    Method 3 : Use y' and y"
    • When x LT 0
      • The curve is between parabola and x = 0.
      • The curve from +infinite decreases to minimum
      • The curve from minimum increase to +infinite at x = 0
      • The curve is concave upword
    • When x GT 0 and LT 1
      • The curve increases from -infinite at x = 0 to point of inflection
      • The curve is concave downward
    • When x GT 1
      • The curve increases from point of inflection to +infinite
      • The curve is concave upward
      • The curve is between the parabola and x = 0
    Method 4 : Use (x,y) points
    • See next question : GC 13 05

    Go to Begin

    Q05. Sketch y = (x^3 + 2)/x

    Method 1 : Computer diagram
    • Start Program ABH Program 03 08
    • Sketch the curve
      • Click start
      • Click subject 3 in upper box
      • Click program 8 in lower box
      • Give coefficients : 1, 0, 0 2, 0, 0, 1, 0
      • Note 1 : first four number are coefficients of x^3 + 1
      • Note 2 : second four number are coefficients of x
    Method 2 : Use asymptote and signes as a guide
    • Draw asymptote y = x^2
    • Draw vertical asymptote x = 0
    • When y is large, it will be between the parabola asymptote (See GC 13 04)
    • Find the signs
      • When x LT -1, y is positive
      • When x EQ -1, y = 0
      • When x between -1 and 0, y is negative
      • When x just less than 0, y is -infinite
      • When x just greater than 0, y is +infinite
      • When x GT 0, y is positive. It decreases first then increases
    • Disadvantage
      • We do not know the extreme points
      • We have to guess the concavity from the asymptotes
    Question
    • Can you see the parabola asymptote y = x^2 from the graph ?
    Method 3 : Use y' and y"
    • y' = 2*x - 2*x^(-2) = 2*(x^3 - 1)/(x^2)
      • If y' = 0, then x^3 - 1 = 0
      • Hence (x - 1)*(x^2 + x + 1) = 0
      • Hence we have one extreme point at x = 1 and y = 3
      • If x LT 0, y' is negative, the curve is decreasing
      • If 0 GT x LT 1, y' is negative, the curve is decreasing
      • If x GT 1, y' is positive, the curve is increasing
    • y" = 2 + 2*x^(-3) = 2*(x^3 + 1)/(x^3)
      • Concavity
        • If x LT -1, y" is positive, Hence the curve is concave upward
        • If -1 GT x LT 0, Y" is negative, Hence the curve is concave downward
        • If x GT 0, y" is positive, Hence the curve is concave upward
      • Extreme points
        • If x = -1, y" EQ 0 and y' NE 0, it is a point of inflection
        • If x = +1, y" GT 0 and y' EQ 0, it is a minimum point
    • Find few points
      • x = -1 y = -1; x = -2 y = 3
      • x = +1 y = +3; x = +2 y = 5
      • x = +0 y = +infinite
      • x = -0 y = -infinite
      • If x is large, y = x^2
    Method 4 : Use (x,y) only
    • x = -06, -04, -2, -1, -0, +0, +1, +2, +04, +06
    • y = -33, +14, -3, -0, -v, +v, +3, +5, +18, +39
    • Note 1 : v stands for infinite
    • Note 2 : More points more accurate
    • Note 3 : It is straight fordward and simple but need lot of calculation

    Go to Begin

    Q06.
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    Q07.
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    Q08. Limit

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    Q09.

    Go to Begin

    Q10. Enter the program

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    Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
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