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Thread: helpp

  1. #1
    Junior Member
    Sep 2008


    help with curve sketching, i have a question how do u sketch this,
    y= (x-2)^3
    i know u have to find the roots, but if u get (x-2), isn't the root 2, and it goes through a curve because of its to the power of 3?

    also one question consider this exmaple. y= -5 (x+5) (x-5), would the -5 consider as a root or 0?, and also if a questions like this
    y = -(x^2-25)(x^-1), would the negative sign be consider the graph as negative parabola? or would the negative sign be ignored?
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  2. #2
    MHF Contributor
    Apr 2008
    1. As far as sketching $\displaystyle y = (x - 2)^3$, this is simply a right shift of $\displaystyle y = x^3$. You'll need a few points on the graph to help you sketch it. The root is indeed 2. What is y when x = 0?

    2. Zero is not a root of $\displaystyle y = -5(x + 5)(x - 5)$, because x = 0 gives y = 125. The roots are dependent on the polynomial factors of x + 5 and x - 5, not the multiple of -5.

    3. In your equation $\displaystyle y = \frac{-(x^2 - 25)}{x}$ the negative sign is not ignored. You can distribute it across to yield the equation $\displaystyle y = \frac{25 - x^2}{x}$. Additionally, this graph is not a parabola. When I write it in this form, can you see why?
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