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Determine when y = x^2 + bx + 17 has horizontal tangent

  1. Apr 18, 2005 #1
    am i doing this right?
    Determine a value of the constant b so that the graph of y=x^2+bx+17 has a horizontal tangent at (2, 21+2b)
    ans: y'(2)=0
    y'=2x+b
    0=2(2)+b
    b=-4
    what do u need the y-coordinate (21+2b)for?
    My teacher also requires me to draw a diagram...can anybody show me a diagram of this problem?
     
  2. jcsd
  3. Apr 18, 2005 #2

    Ouabache

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    Did you try plotting your original equation with the value you calculated for b?
    From that plot you will see why you need to use the point(2,21+2b).
    (Hint: it has to do with the graphical interpretation for taking the first derivative of your equation)

    If you are having a problem drawing the diagram.. Try calculating some ordered pairs using a range of x values (say from -10 to 10) and plug that into your original equation. You already have b, so you should be able to calculate all your y-values. Then plot these ordered pairs on some graph paper.
     
    Last edited: Apr 18, 2005
  4. Apr 18, 2005 #3
    my friend taught me that
    21+2(-4)=13
    (2)^2+(-4)(2)+17=13
    i am not too sure why he did that...
     
  5. Apr 18, 2005 #4

    Ouabache

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    What does that suggest to you about the point(2,13) and the two
    equations you are now working with?

    After you plot the ordered pairs as i suggested, from your initial equation y(x)
    Do the same thing for the second equation y'(x) and note what is happening.
     
  6. Apr 18, 2005 #5
    You could have also done it by finding the correct vertex of the parabola.
    Nothing.
    It's a standard parabola translated a little. Complete the square to find the x and y offsets.
     
  7. Apr 19, 2005 #6

    Ouabache

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    How did you make out plotting your curves?
    Did you see why need to use the point(2,21+2b)?
     
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