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Math Help - Problem Check (Sketch graph of Function)

  1. #1
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    Problem Check (Sketch graph of Function)

    I wanted to check the work I've done in these problems. This is a 7 step process to draw the graph of a function.

    The steps to follow are:
    1. Determine domain
    2. Determine the x and y intercepts
    3. Find the vertical and horizontal asymptotes
    4. Determine the interval of decreaing and increasing.
    5. Determine local maximum and minimum values
    6. Determine the interval of concave up and concave down. Also find the inflection points.
    7. Sketch the graph.
    So here are the problems that I've done so far, I haven't finished them all the way but I just want to make sure I'm doing them right so far:

    1. y = x/x^2-9

    1. Domain is all real numbers except 3 and -3.
    2. y intercept: x = 3, x = -3 x intercept: y = 0
    3. Vertical Aysmptote: x = -3 , x = 3 Horizontal Asymptote: y = 0
    4. Increasing on: (-(infinity), 3) Decreasing: (3, (infinity))
    5. No local max or min
    6. concave up: (3, (infinity)) concave down: (-(infinity), 3)
    2. y = x^2/x^2+9

    1. Domain is all real numbers
    2. y intercept: x=0 x intercept: y=0
    3. No V.A. H.A.: y=1
    4. Increasing on: (0, (infinity)) Decreasing on: (-(infinity),0)
    5. No local max, local min at x=0
    6. concave up: (-1/4, 1/4) concave down: (-(infinity),-1/4)U(1/4, (infinity))
    3. y= x/x^3-1

    1. Domain, all real numbers except 1.
    2. V.A. at x=1 H.A. at y=0
    3. _
    4. increasing: (-(infinity), 1) decreasing; (1,(infinity)
    5. local max at x=-1
    4. y=x/(x^2-1)^1/2

    1. Domain: (-(infinity),-1)U(1, (infinity))
    2. V.A.: x=-1 and x=1 H.A.: y=0
    3. _
    4. Couldn't find any intervals of increase or decrease
    5. No local max or min
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  2. #2
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    Quote Originally Posted by dm10 View Post
    I wanted to check the work I've done in these problems. This is a 7 step process to draw the graph of a function.





    The steps to follow are:
    1. Determine domain
    2. Determine the x and y intercepts
    3. Find the vertical and horizontal asymptotes
    4. Determine the interval of decreaing and increasing.
    5. Determine local maximum and minimum values
    6. Determine the interval of concave up and concave down. Also find the inflection points.
    7. Sketch the graph.
    So here are the problems that I've done so far, I haven't finished them all the way but I just want to make sure I'm doing them right so far:

    1. y = x/x^2-9

    1. Domain is all real numbers except 3 and -3.
    2. y intercept: x = 3, x = -3 Mr F says: No. The y-intercept is got by substituting x = 0. You get y = 0. So y-intercept at (0, 0).
    3. x intercept: y = 0 Mr F says: No. This is the equation of a line. It doesn't represent a point. Points have coordinates, not equations. The x-intercept is (0, 0), NOT y = 0.
    4. Vertical Aysmptote: x = -3 , x = 3 Horizontal Asymptote: y = 0
    5. Increasing on: (-(infinity), 3) Decreasing: (3, (infinity)) Mr F says: No. There are three intervals you need to consider: (-oo, -3), (-3 3) and (3, +oo).
    6. No local max or min
    7. concave up: (3, (infinity)) concave down: (-(infinity), -3) Mr F says: There are two other intervals you need to consider ..... Note that there's a stationary point of inflection at (0, 0).
    2. y = x^2/x^2+9

    1. Domain is all real numbers
    2. y intercept: x=0 x intercept: y=0 Mr F says: See above.
    3. No V.A. H.A.: y=1
    4. Increasing on: (0, (infinity)) Decreasing on: (-(infinity),0)
    5. No local max, local min at x=0 Mr F says: Coordinates are required! (0, 0).
    6. concave up: (-1/4, 1/4) concave down: (-(infinity),-1/4)U(1/4, (infinity)) Mr F says: No. The x-coordinates of the points of inflection are {\color{red}x = \pm \sqrt{3}}. You're using the y-coordinates of the points of inflection.
    3. y= x/x^3-1

    1. Domain, all real numbers except 1.
    2. V.A. at x=1 H.A. at y=0
    3. Mr F says: What about x- and y-intercepts?
    4. increasing: (-(infinity), 1) Mr F says: How can this be correct when there's a local maximum with x-coordinate {\color{red}x = -\sqrt[3]{\frac{1}{2}}}?? It is also logically inconsistent if there was a maximum at x = -1.
    5. decreasing; (1,(infinity)
    6. local max at x=-1 Mr F says: No. See above.
    4. y=x/(x^2-1)^1/2

    1. Domain: (-(infinity),-1)U(1, (infinity))
    2. V.A.: x=-1 and x=1 H.A.: y=0
    3. _
    4. Couldn't find any intervals of increase or decrease Mr F says: Look harder. There is. Decreasing for x < -1. Increasing for x > 1.
    5. No local max or min
    Where you have wrong values, it would help if you showed your working so your mistakes can be pointed out.
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