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Math Help - Find the value of k

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    Find the value of k

    (page 350 #9)
    Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

    The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance
    Attached Thumbnails Attached Thumbnails Find the value of k-untitled.jpg  
    Last edited by fabxx; September 19th 2008 at 11:36 PM.
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    Quote Originally Posted by fabxx View Post
    (page 350 #9)
    Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

    The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance
    Are points A, B and C meant to coincide with the vertices of the triangle?
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    yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.
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    Okay
    Quote Originally Posted by fabxx View Post
    yes they are. sorry for the sloppy graph. A, C should be the points for both the parabola and the triangle.
    Note that k is positive.
    A and C are the x intercepts of the curve y=k-x.
    This means that y=0 for these points.
    Thus x=+ or - sqrt(k)
    The distance AB is then 2sqrt(k).

    Can you see what the altitude of the triangle from point B is ? It's the y-ordinate of B, which is the y intercept of the curve. This means that its x is 0. So what is its y ?

    Then remember that area of triangle = (1/2)*AB*altitude.

    Solve for k !
    Last edited by Moo; September 20th 2008 at 03:06 AM.
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    Quote Originally Posted by fabxx View Post
    (page 350 #9)
    Sorry for the sloppy graph. The outter line should be a curve concave down. And inside the curve is a triangle.

    The question: The figure above shows the graph of y=k-x^2, where k is constant. If the area of triangle ABC is 64, what is the value of k? I don't get why the correct answer is 16. Thanks in advance
    The x-intercepts of the parabola are x = \pm \sqrt{k} and the y-coordinate of the turning point is y = k. So you have a triangle of baselength 2\sqrt{k} and height k.

    Therefore solve k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64 for k.
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    Quote Originally Posted by Moo View Post
    Okay
    A and B are the x intercepts of the curve y=k-x.
    shouldn't the x intercepts of the curve be A and C?


    Also another question is why do you make y=0?

    Thanks in advance
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    Quote Originally Posted by mr fantastic View Post

    Therefore solve k \sqrt{k} = 64 \Rightarrow k^{3/2} = 64 for k.
    I don't get how you got from k \sqrt{k} = 64  to \Rightarrow k^{3/2} = 64 and to the final k=16.
    Can you please show me the steps?

    Thanks in advance
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    Moo
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    Quote Originally Posted by fabxx View Post
    shouldn't the x intercepts of the curve be A and C?


    Also another question is why do you make y=0?

    Thanks in advance
    Excuse me, I put B instead of C. Now I've edited.

    The x intercepts are the points intercept points of the curve with the x-axis. But any point on the x-axis has a null ordinate. This is why y=0 for these points.

    Quote Originally Posted by fabxx View Post
    I don't get how you got from k \sqrt{k} = 64  to \Rightarrow k^{3/2} = 64 and to the final k=16.
    Can you please show me the steps?

    Thanks in advance
    \sqrt{k}=k^{1/2}

    and remember a rule of exponents : a^b a^c=a^{b+c}
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    Quote Originally Posted by moo
     \Rightarrow k^{3/2} = 64 for k.
    how do you get from k^(3/2) to k=16?

    Thanks again!! sorry for the trouble.
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    Quote Originally Posted by fabxx View Post
    how do you get from k^(3/2) to k=16?

    Thanks again!! sorry for the trouble.
    k^{3/2} = (k^{1/2})^3 = (\sqrt{k})^3 = 64

    Cube root both sides: \sqrt{k} = 4.

    Square both sides: k = 16.
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  11. #11
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    How come you cannot use the 30, 60, 90 triangle method?
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