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Math Help - URGENT HELP WITH some problems. :x

  1. #1
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    Question URGENT HELP WITH some problems. :x

    Okay it starts with how to use algebraic tests to check for symmetry with respect to both axes and the origin.

    13. y = -4x+1
    14. y= 5x-6
    15. y=5-x^squared
    16.y=x^squared - 10
    17. y=x cubed + 3
    18. y= -6 -x cubed
    19. y = square root of x +5
    20. y=|x| +9

    And..
    I know theres this formula for this its (x-h) squared + (y-k) squared = r squared but
    How to find the center of a radius of a circle when they give u just this

    21. x squared + y squared = 9
    22. x squared + y squared = 4
    23. (x+2)squared + y squared = 16
    24.x squared + (y-8)squared = 81
    25. (x - 1/2) squared + (y + 1)squared = 36
    26. (x+4) squared + (y - 3/2)squared = 100


    Whoever answers them all first gets thanks IF they helped me understand how to do them correctly
    much love
    -THANKS
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  2. #2
    o_O
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    If a function is even (i.e. symmetric about the y-axis), then: f(x) = f(-x), i.e. if you plug in -x to your function, you'll get the same function if you plugged in x.

    If a function is odd (i.e. symmetric about the origin), then: f(-x) = -f(x), i.e. if you plug in -x to your function, then you'll get the negative of f(x).

    Show us your work and we'll point out any errors.

    As for the circle, given (x-a)^2 + (y-b)^2 = r^2, the centre is given by (a,b) (note the minus signs in front of them in the equation). For example, (x - 5)^2 + (y + 2)^2 = 9 has centre (5, {\color{red}-}2)
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  3. #3
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    Quote Originally Posted by o_O View Post
    If a function is even (i.e. symmetric about the y-axis), then: f(x) = f(-x), i.e. if you plug in -x to your function, you'll get the same function if you plugged in x.

    If a function is odd (i.e. symmetric about the origin), then: f(-x) = -f(x), i.e. if you plug in -x to your function, then you'll get the negative of f(x).

    Show us your work and we'll point out any errors.

    As for the circle, given (x-a)^2 + (y-b)^2 = r^2, the centre is given by (a,b) (note the minus signs in front of them in the equation). For example, (x - 5)^2 + (y + 2)^2 = 9 has centre (5, {\color{red}-}2)
    Im confused by your first explanation but here if i have say y=-4x + 1
    is this function negative or odd? i don't understand.. do i have to plug in (1) to x and see because that is what i understand from your explanations..
    or would the function y=5x - 6 be positive because its slope is 5x and its starting point or b is -6?

    I am confused in that aspect and if i had something like
    y=x cubed + 3 to me this would be positive or am i not correct? if it positive then for me to find out if its symmetric about the origin how would i know this? and if its odd its symmetric about the origin is that what your saying because i get that if its odd its symmetric by the origin and if its even it is symmetric about the y axis but what i also don't understand on something like y=-4x +1 is how i found out what number is odd do i plug in 1 to find out what just y equals and determine if its symmetric about the y axis or origin? So confuseed.
    Or do i determine if the slopes odd / or the starting point?

    For the 2nd part I understand that something like x squared + y squared = 9
    is (x-1)squared + (y - 1) squared = 9 or 3 squared so is that how that is graphed at (-1,-1) and goes 3 to the sides to form the circle on a graph?

    and for something like 23. (x+2)squared + y squared = 16 how would i take out (x+2) squared to make it work with the formula square root (x+2) and Square root 16 to get 4 and then subtract 2 and have x + y squared = 2 Sorry if i confused you there. but im completely confused ha.
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