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Completing the Square + Differentiation - check watch

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    Hi

    I need feedback on the following pretty straightforward question - are my answers right?
    I also need to know if the layout of my answer is suitable for examination purposes.

    Thanks!


    My Question:

    Write  x^2 + 10x + 38 in the form (x + b)^2 + c where the values of
    b and c are to be found.

    a) State the minimum value of  x^2 + 10x + 38 and the value of x for which this occurs.

    b) Determine the values of x for which  x^2 + 10x + 38 \ge 22.


    My Answer:

     x^2 + 10x + 38 in the form  (x + b)^2 + c:

     x^2 + 10x + 38 = (x + 5)^2 + 13 (Completed the square)

    Therefore b = 5 and c = 13.

    a) Let f(x) =  x^2 + 10x + 38

    The function is defined for all real numbers.

    f '(x) = 2x + 10

    f '(x) = 0 when x = -5

    f '(-5) = (-5)^2 + 10(-5) +38 = 13

    The minimum point is (-5, 13).

    Therefore the minimum value of  x^2 + 10x + 38 is 13 when x = -5.


    b)  x^2 + 10x + 38 \ge 22

     x^2 + 10x + 38 - 22 \ge 0

     x^2 + 10x + 16 \ge 0

     (x + 8)(x + 2)\ge 0

     x \ge -8 and x \le -2

    Therefore  -8 \le x \le  -2

    Furthermore, the values of x for which (x + 8)(x + 2) = 0 are x = -8
    and x = -2 (which are the critical values of the inequality).
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    it's all right except part c - draw the graph of y = (x+8)(x+2) and see where it's >0 - that is, above the x-axis
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    (Original post by AsAllTheWay)
    Hi

    I need feedback on the following pretty straightforward question - are my answers right?
    I also need to know if the layout of my answer is suitable for examination purposes.

    Thanks!


    My Question:

    Write  x^2 + 10x + 38 in the form (x + b)^2 + c where the values of
    b and c are to be found.

    a) State the minimum value of  x^2 + 10x + 38 and the value of x for which this occurs.

    b) Determine the values of x for which  x^2 + 10x + 38 \ge 22.


    My Answer:

     x^2 + 10x + 38 in the form  (x + b)^2 + c:

     x^2 + 10x + 38 = (x + 5)^2 + 13 (Completed the square)

    Therefore b = 5 and c = 13.

    a) Let f(x) =  x^2 + 10x + 38

    The function is defined for all real numbers.

    f '(x) = 2x + 10

    f '(x) = 0 when x = -5

    f '(-5) = (-5)^2 + 10(-5) +38 = 13

    The minimum point is (-5, 13).

    Therefore the minimum value of  x^2 + 10x + 38 is 13 when x = -5.


    b)  x^2 + 10x + 38 \ge 22

     x^2 + 10x + 38 - 22 \ge 0

     x^2 + 10x + 16 \ge 0

     (x + 8)(x + 2)\ge 0

     x \ge -8 and x \le -2

    Therefore  -8 \le x \le  -2

    Furthermore, the values of x for which (x + 8)(x + 2) = 0 are x = -8
    and x = -2 (which are the critical values of the inequality).
    Correct, except for these two bits  -8 \le x \le  -2

    Edit: I was too quick.

    These are also wrong  x \ge -8 and x \le -2

    It should be  x \le -8 and x \ge -2
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    (Original post by Pheylan)
    it's all right except part c - draw the graph of y = (x+8)(x+2) and see where it's >0 - that is, above the x-axis
    Sorry, the answer should be x< =-8 or x>=-2 the two inequalities can't be put together
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    (Original post by steve2005)
    Part (c) is correct.
    no it isn't



    >0 when x<-8 or x>-2
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    (Original post by Pheylan)
    no it isn't



    >0 when x<-8 or x>-2
    Yes , I was wrong.
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    In part a where they ask you to state the minimum value of the function and its x coordinate, they don't want you to differentiate, they want you to use your answer to the previous part to deduce the answer. The fact that they specifically asked you to state the answer means you shouldn't have to calculate anything; the answer is there already, you just have to write it down.

    The constants b and c give the coordinates of the vertex of the curve, which are (-b,c)
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    Oh right. Thanks
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    The coordinates of the turning-point of a quadratic equation are \boxed{\displaystyle\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)}

    So for x^2+10x+38, we have:

    Spoiler:
    Show
    \displaystyle\left(\frac{-10}{2}, -\frac{\left(10^2-4(38)\right)}{4}\right) = \left(-5, -\frac{\left(100-152\right)}{4}\right) = \left(-5, -\frac{\left(-52\right)}{4}\right) = \left(-5, \frac{52}{4}\right) = \left(-5, 13\right).


    The form that you were asked to put on was (x+b)^2+c, which is as same as \displaystyle\left(x+\frac{b}{2a  }\right)^2 -\frac{\left(b^2-4ac\right)}{4a}. That means when you have a quadratic equation in the form (x+b)^2+c, the coordinates of the turning-point are \boxed{\left(-b, c\right)}.
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    (Original post by Quadrifolium)
    The coordinates of the turning-point of a quadratic equation are \boxed{\displaystyle\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)}

    So for x^2+10x+38, we have:

    Spoiler:
    Show
    \displaystyle\left(\frac{-10}{2}, -\frac{\left(10^2-4(38)\right)}{4}\right) = \left(-5, -\frac{\left(100-152\right)}{4}\right) = \left(-5, -\frac{\left(-52\right)}{4}\right) = \left(-5, \frac{52}{4}\right) = \left(-5, 13\right).


    The form that you were asked to put on was (x+b)^2+c, which is as same as \displaystyle\left(x+\frac{b}{2a  }\right)^2 -\frac{\left(b^2-4ac\right)}{4a}. That means when you have a quadratic equation in the form (x+b)^2+c, the coordinates of the turning-point are \boxed{\left(-b, c\right)}.
    Why go through all that hassle when differentiating gives the same result...with much less effort
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    (Original post by AsAllTheWay)
    f '(-5) = (-5)^2 + 10(-5) +38 = 13

    .
    probably it's just a typo, since it's pretty clear you mean f and not f'
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    (Original post by antonfigo)
    Why go through all that hassle when differentiating gives the same result...with much less effort
    Why do you think they have asked to put the equation into the form (x+b)^2+c?
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    (Original post by Quadrifolium)
    Why do you think they have asked to put the equation into the form (x+b)^2+c?
    Ah I see...but still theres no hence!
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    (Original post by antonfigo)
    Ah I see...but still theres no hence!
    If you see, then why do you think that putting the equation into the form (x+b)^2+c and then simply realising that the turning-point is at (-b, c) is "hassling"?
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    (Original post by antonfigo)
    Ah I see...but still theres no hence!
    You might want to re-read my post (below)... I linked to a wikipedia page that explains how, once you've completed the square, the coordinates of the vertex can simply be stated.

    (Original post by TheNack)
    In part a where they ask you to state the minimum value of the function and its x coordinate, they don't want you to differentiate, they want you to use your answer to the previous part to deduce the answer. The fact that they specifically asked you to state the answer means you shouldn't have to calculate anything; the answer is there already, you just have to write it down.

    The constants b and c give the coordinates of the vertex of the curve, which are (-b,c)
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    (Original post by TheNack)
    You might want to re-read my post (below)... I linked to a wikipedia page that explains how, once you've completed the square, the coordinates of the vertex can simply be stated.
    Oh yea. Thanks
 
 
 
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Updated: January 29, 2010
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