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Math Help - Another Increasing/Decreasing Functions Question

  1. #1
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    Another Increasing/Decreasing Functions Question

    "Use calculus techniques to show that the graph of the quadratic function f(x) = ax^2 + bx + c, a > 0, is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)."

    I have no idea where to start really. Can someone lead me in the right direction?

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    "Use calculus techniques to show that the graph of the quadratic function f(x) = ax^2 + bx + c, a > 0, is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)."

    I have no idea where to start really. Can someone lead me in the right direction?

    Thanks!
    find the derivative and note that it is negative on the interval x < (-b/2a) and positive on the interval x > (-b/2a)
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  3. #3
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    f(x) = ax^2 + bx + c
    f'(x) = 2ax + b
    0 = 2ax + b
    -b = 2ax
    -b/2a = x

    How do I show that it is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    f(x) = ax^2 + bx + c
    f'(x) = 2ax + b
    0 = 2ax + b
    -b = 2ax
    -b/2a = x

    How do I show that it is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)?
    there is only one point where the derivative is zero, so we don't really have to worry about fluxuations. just pick a point to the right of -b/2a and one to the left. for example, you may consider the derivative at x = -b/2a + 1 and x = -b/2a - 1. see what happens
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    there is only one point where the derivative is zero, so we don't really have to worry about fluxuations. just pick a point to the right of -b/2a and one to the left. for example, you may consider the derivative at x = -b/2a + 1 and x = -b/2a - 1. see what happens
    I know what you're talking about, but I don't know how to represent it on paper. How do I show that it is increasing on the one interval, but decreasing on the other interval?

    (I know that the quadratic function looks like a parabola and I can think about how at the zero slope point it would change from decreasing to increasing, but I can't represent it)
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Here you go

    f'(x)=2ax+b which equals zero at x=\frac{-b}{2a}...now to test before \frac{-b}{2a} put \frac{-b}{2a}-1 in the derivative...you get f'\bigg(\frac{-b}{2a}-1\bigg)=2*a*\bigg(\frac{-b}{2a}-1\bigg)+b=-2a..assuming that a>0 if a<0 [tex]f(x)]/math] is decreasing on x\in\bigg(-\infty,\frac{-b}{2a}\bigg)

    QED
    Last edited by Mathstud28; April 6th 2008 at 12:28 PM.
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  7. #7
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    Ahh, okay. After plugging in (-b/2a - 1) and (-b/2a +1) into the derivative function, I get -2a and 2a respectively.

    Meaning that the function is decreasing on the interval x < -b/2a and the function is increasing on the interval x > -b/2a.

    Thank you.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Ahh, okay. After plugging in (-b/2a - 1) and (-b/2a +1) into the derivative function, I get -2a and 2a respectively.

    Meaning that the function is decreasing on the interval x < -b/2a and the function is increasing on the interval x > -b/2a.

    Thank you.
    good. and note that you know that -2a is negative and 2a is positive respectively because they told you a > 0. you have to watch out for that
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