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Math Help - Help with functons and line and circle formulas

  1. #1
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    Help with functons and line and circle formulas

    Few questions from my homework.

    Find the equation of a circle with diameter endpoints (8,10) and (-2,14). I know I need to find the distance and midpoint but am having trouble with putting togther the exact formula.

    2. Find the equation of a line in slope intercept form passing through (-2,-5) and (3,5)

    And then find equation of a line that is parallel to the line above passing through (-3,4)

    Functions:
    How would you graph f(x)= sqaure root of x+2 - 3 Note: the 3 is not under the square root.


    Explain what is being done to f(x) in this equation y=-2f(x+3)-1


    How would you show that f(x)=8x^3 -3x is odd.

    Thanks to anyone who can help me out.
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  2. #2
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    The last one...

    The def of an odd function is f(-x)=-f(x)

    So lets start with the above

    f(x)=8x^3-3x

    f(-x)=8(-x)^3-3(-x) = -8x^3+3x

    if we factor out a minus sign

    f(-x)=8(-x)^3-3(-x) = -8x^3+3x=-\underbrace{(8x^3-3x)}_{\mbox{This is f(x)}}=-f(x)

    so by the above definition it is an odd function.
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  3. #3
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    Thanks for the help with that one, any idea with the others?
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  4. #4
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    #2

    we use this formula to find the slope

    m=\frac{y_2-y_1}{x_2-x_1}=\frac{-5-5}{-2-3}=\frac{-10}{-5}=2

    now we can use the slope intercept from to find the equation of the line.

    y=mx+b we know m from our above work so

    y=2x+b now using either point (3,5)

    5=2(3)+b \iff -1 =b so the equation is

    y=2x-1

    For part b use the same slope and procedure with the new point (-3,4)
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