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Math Help - Need Help with a Few Basic Problems Please

  1. #1
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    Need Help with a Few Basic Problems Please

    1. Determine if the graph of the equation y = 16 - x^2 is symmetric with respect to:

    a) the x-axis (yes or no, w/ brief explanation)
    b) the y-axis (yes or no, w/ brief explanation)
    c) the origin (yes or no, w/ brief explanation)
    d) Find the x-intercept(s).
    e) Find the y-intercepts(s).

    2. Given the equation of a circle, find the center and the radius.

    x^2 + y^2 + 5x - 2y + 6 = 0

    3. Write the equation in general form of the line that contains the point (7,3) and is parallel to the line passing through the points (1,1) and (-1,5).

    Any help is appreciated, thanks!
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Need Help with a Few Basic Problems Please

    definitions

    A function is symmetric about the x-axis if and only if replacing y with -y will yield the same function (explain why)
    A function is symmetric about the y-axis if and only if replacing x with -x will yield the same function (explain why)
    A function is symmetric about the origin if and only if it is both symmetric about the x and y axis

    x-intercepts are points which lie on the line y = 0. find where y= 16-x^2 and y = 0 intersect
    y-intercepts are points which lie onthe line x = 0. find where y = 16-x^2 and x = 0 intersect

    2. You can re-write the equation by rearranging and grouping terms, and then completing the square for both x and y

    x^2 + 5x + y^2 - 2y + 6 = x^2 + 5x + (5/2)^2 -(5/2)^2 + y^2 - 2y + 1 - 1 + 6 = (x+(5/2))^2 + (y+1)^2 + -(5/4) = 0

    Hence the equation can be re-written as (x+(5/2))^2 + (y+1)^2 = (5/4) In this form, can you tell me what the center and radius is? If you have a problem with completing the square let me know.

    3.
    To be parallel with a line means having the same slope. calculate the slope of the line through the points (1,1) and (-1,5) using m = \frac{y_1 - y_0}{x_1 - x_0} (rise over run)
    Using m and the point (7,3) plug them into the point slope form y-y_1= m(x-x_1) which is just rewriting the above slope formula to get the equation for your desired line.
    Thanks from StephenSkinner
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  3. #3
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    Re: Need Help with a Few Basic Problems Please

    Hi, thanks for the reply.

    For #1, I got:

    a) no
    b) yes
    c) no
    d) x-int = 4, -4
    e) y-int = 16

    #2:

    center = (-5/2 , -1)
    radius = 5/4

    #3:

    0 = -2x + y + 17

    Can you confirm that these are right?
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  4. #4
    Senior Member MacstersUndead's Avatar
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    Re: Need Help with a Few Basic Problems Please

    For #2, recall that the formula for a circle of center (a,b) with radius r is (x-a)^2 + (y-b)^2 = r^2. so what's incorrect is that r = sqrt(5/4). As long as you know how to complete the square, the process of finding the points of interest is completing squares and writing the general form into standard form.

    The rest is correct
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