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Thread: Applications of Polynomial Functions (Word Problem)

  1. #11
    Elite Member mmm4444bot's Avatar
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    This discussion is a duplicate thread.

    Galactus responded to this discussion at the other location, with an interpretation of the phrase "the arch must provide a minimum clearance of 4m over the highway" that is better than what I described in my post above.

    Galactus reads it as a 4-meter clearance at the lowest point directly above the outside edges of the road; I was reading it to mean a 4-meter clearance at the highest point (above the center of the road).

    I believe that Galactus is right, and I was wrong. Therefore, most of the information that I posted above is wrong, too.

    Please go to the other location HERE, and read Galactus' post.


    Oh, it looks like Elizabeth got it right, too. Credit goes to her, also.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  2. #12
    Elite Member stapel's Avatar
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    Duplicate threads merged.

  3. #13
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    Re: Applications of Polynomial Functions (Word Problem)

    OMG - I don't understand why I can't get this - I am wasting hours and getting no where

    How do you figure out what anything is with 4 = a(4-0)^2 + k ?

    How do I find a ?

    I understand the graph, but not the formulas and how to determine for a(x-h)^2 + k what the values of a, x, and k are, I know h=0 because that is the axis of symmetry and the x intercept value for the vertex point (0,k)

    I have the x-intercepts as (-8,0) (8,0)
    I have the points I can use as (-4,4) (4,4)

    You guys say to "just plug in points" but what points do I plug in where?!?!

    I don't even know what parts I'm not getting and what to look up on other help sites.......

  4. #14
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by plur222
    How do you figure out what anything is with 4 = a(4-0)^2 + k ? You don't. This is only one of two equations that you need to form.

    Did anybody teach you about solving systems of equations? That usually comes before assigning an exercise such as this one.

    (I posted the steps in forming of a system of equations when you still had this discussion in two different places. I showed you one method for solving that system. I do not know whether or not you read it or understand it.)


    You guys say to "just plug in points" but what points do I plug in where?!?!

    You substitute the x-coordinate for x, and you subsitute the y-coordinate for y.
    Galactus gave you a formula for y:

    y = a(x)^2 + k

    Do you see the symbol x? Replace it with the number 4.

    Do you see the symbol y? Replace it with the number 4.

    We can do this because we know that the point (4, 4) satisfies the parabola's equation, and we know that (4, 4) means x = 4 and y = 4. These substitutions give us ONE equation containing the symbols a and k.

    4 = a(4)^2 + k

    It simplifies.

    16a + k = 4

    We don't stop here.

    We need two equations with the symbols a and k, in order to find their values.

    If we were to now use the point (-4, 4) to write the second equation, we would end up with the same equation as we did with (4, 4). That's no good; we need two different equations.

    So, use the point (8, 0) instead. Substitute the number 8 for x and the number 0 for y.

    0 = a(8)^2 + k

    It simplifies.

    64a + k = 0

    Now we have a "system of two equations" containing the two symbols a and k.

    16a + k = 4
    64a + k = 0

    What do you know about solving systems of equations?

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #15
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    Re: Applications of Polynomial Functions (Word Problem)

    This is what I have come up with
    I'm feeling slightly less crazy
    Am I on the right track?
    I thought "a" would = -1/4 .... not -1/12

    I really appreciate everyone's help - this site is a godsent!

    I will be going back and studying further - but it would kill me to know I have let the two other people in my group down tonight.

  6. #16
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    Re: Applications of Polynomial Functions (Word Problem)

    Your worksheet indicates you got a = -1/12, which is correct; so what are you asking?

    By the way, you can solve such equations much quicker this way (and save on paper!):
    16a + k = 4 [1]
    64a + k = 0 [2]

    Subtract the equations; [1] - [2]:
    -48a = 4
    a = 4 / -48
    a = -1/12
    I'm just an imagination of your figment !

  7. #17
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    Re: Applications of Polynomial Functions (Word Problem)

    My final equation has to be in the format: a(x-h)^2 + k

    I was told the final equation would be y = - (x/4)^2 + 4 which would mean a = 1/4 yes?

    Is there a way I can check my work, assuming the following is correct:

    a = - 1/12

    k = - 16/3

    y = 0

    Which x value would I put into the equation? Or I suppose -8 and 8 would give me the same since it's being squared.

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    Re: Applications of Polynomial Functions (Word Problem)

    This is where I am at
    Attached Images Attached Images

  9. #19
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by plur222

    I was told the final equation would be y = - (x/4)^2 + 4 which would mean a = 1/4 yes?


    That equation models an arch whose vertex is 4 meters above the center of the highway.

    We changed the interpretation (i.e., the 4-meter height is no longer at the vertex), so the values of a and k changed.

    Not that it matters for this exercise, but -(x/4)^2 would not mean that a = 1/4 because the 4 is being squared and there is a negative sign in front. It would mean that a = -1/16.

    -
    Is there a way I can check my work, assuming the following is correct:

    a = - 1/12

    k = - 16/3


    You got the right values on your latest work shown above, but it looks like you made a typographical error here because k is not negative.

    a = -1/12

    k = 16/3

    The equation that models this arch is:

    y = (-1/12)x^2 + 16/3

    You can check this equation by making sure that you get 0 for y when you let x be 8 or -8. Make sure that you get 4 for y when you let x be 4 or -4. You could check to make sure that you get 16/3 for y when you let x be zero, but that's obvious, right?

    -
    I suppose -8 and 8 would give me the same since it's being squared.


    Yup, yup.

    Letting x be -4 and 4 leads to the same situation. This is all because of symmetry.

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    Re: Applications of Polynomial Functions (Word Problem)

    GOOD! DONE! Huge thanks to everyone that helped - I have studying to do, I'm pretty sure they won't give me 11 hours on the exam ... lol

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