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Completing the Square

  1. Feb 3, 2014 #1

    I am having trouble with completing the square in this question. I've looked at several online videos and I am still confused as to why I am not getting an answer that makes sense. I keep ending up with complex numbers and I am not sure what the next step to take is.

    1. The problem statement, all variables and given/known data
    Determine the equation of the tangent to the curve defined by f(x)=x^2-6x+14 at (1, 9) and then sketch it.

    2. Relevant equations

    3. The attempt at a solution

    [tex]f'(1) = lim_h→0 \frac{f(x+h)-f(x)}{h}[/tex]
    [tex]= lim_h→0 \frac{[(1+h)^2-6(1+h)+14]-[1^2-6(1)+14]}{h}[/tex]
    [tex]= lim_h→0 \frac{[1+h+h^2+h+14]-[1-6+14]}{h}[/tex]
    [tex]= lim_h→0 \frac{[1+2h+h^2+14]-[1-6+14]}{h}[/tex]
    [tex]= lim_h→0 \frac{2h+h^2-6}{h}[/tex]
    [tex]= lim_h→0 (2+h-6)[/tex]
    [tex]= 2+(0)-6[/tex]
    [tex]= -4[/tex]

    Determine equation of tangent line:

    [tex]y = mx+b[/tex]
    [tex]9 = (-4)(-1) + b[/tex]
    [tex]9 = -4 + b[/tex]
    [tex]4 + 9 = b[/tex]
    [tex]b = 13[/tex]

    Therefore, the equation of the tangent to the curve is y= -4x + 13.

    Completing the square:

    [tex]x^2 - 6x + 14 = 0[/tex]
    [tex]x^2 - 6x + 9 + 14 - 9 = 0[/tex]
    [tex](x-3)^2 + 14 - 9 = 0[/tex]
    [tex](x-3)^2 + 5 = 0[/tex]
    [tex](x-3)^2 = -5[/tex]

    \sqrt{(x-3)^2} = +- \sqrt{-5}

    [tex]x-3 = \sqrt{5}i[/tex]
    [tex]x = \sqrt{5}i + 3[/tex]


    [tex]x-3 = -\sqrt{5}i[/tex]
    [tex]x = -\sqrt{5}i + 3[/tex]
  2. jcsd
  3. Feb 3, 2014 #2
    Hi NoLimits!

    This doesn't look right. Can you show how do you get ##(1+h)^2-6(1+h)+14## equal to ##1+h+h^2+h+14##?
  4. Feb 3, 2014 #3
    Oops. I messed up there, however the end result of the equation (-4) is still the same. The line should've read:

    [tex]1 + 2h + h^2 - 6 - 6h + 14[/tex]
  5. Feb 3, 2014 #4


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    Homework Helper

    By sheer luck you arrived at the correct answer (even after making many mistakes).

    Can you break down what you did here? It's wrong, but it looks like it's wrong in quite a few places.

    This looks good.

    What does finding the roots of the parabola have to do with finding the tangent line to the parabola at a point? And yes, you won't have real roots because the parabola is entirely above the x-axis.
  6. Feb 3, 2014 #5
    I'm not sure the relevance (if any) of finding the roots. I saw it done in a video, and since the question wants a sketch of the graph, I assumed that I would be needing two values (x and y) for the vertex, hence the need for two roots. Sorry about the screw-up on the equation - I usually do a better job checking before I post.

    This is the corrected line where I think it all went wrong:
  7. Feb 3, 2014 #6
    So assuming I do not need the square roots, then the vertex must be (3, -5) and the y-intercept must be 9? Given the equations I have, it seems like they are the only possible values. However, when using a function graphing tool the y-intercept doesn't appear to be 9, nor does the vertex look like (3, -5). I can't seem to figure out how to go from the completed square to the values I need.
  8. Feb 3, 2014 #7


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    Staff: Mentor

    Have you read what the Mentallic wrote?

    Completing the square is a way of finding roots of the equation - these are handy when you want to sketch the plot, but this equation has NO real roots, and doesn't cross the x-axis.
  9. Feb 3, 2014 #8
    Yes I have read it, which is why I responded with what I did, though I did not completely understand what was meant by it. The question wants me to sketch it, so if the equation has no real roots and doesn't cross the x-axis, how is that possible? Just plug in x-values to the original equation?
    Last edited: Feb 3, 2014
  10. Feb 3, 2014 #9
    Guess so. Thanks.
  11. Feb 3, 2014 #10


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    Homework Helper

    Not exactly. As you've shown, after completing the square you end up with


    Let's analyse this result for a second. We know that [itex]n^2\geq 0[/itex] for any real value of n, and most importantly, it's equal to 0 when n=0. So if we extend this to having (something)2 then that's also always going to be greater than or equal to 0. It's again equal to 0 when something = 0.
    What we have is [itex](x-3)^2[/itex]. This means when x-3 = 0, hence x=3, we get [itex](x-3)^2=0[/itex] so then


    So our minimum point on the parabola which is the vertex is at (3,5) which is above the x-axis and since this parabola has a positive coefficient of x2, this means that it opens upwards, and this makes sense with why you were finding complex roots because it never cuts the x-axis.

    The y-intercept happens when you plug x=0 into the equation of the parabola. Similarly when you're finding roots of the equation, the x-intercept happens when you find y=0.
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