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    loveia's Avatar
    loveia Posts: 3, Reputation: 1
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    #1

    Nov 29, 2008, 08:58 PM
    I'm confused
    f(x)=xsquared + 6x -4 Find the vertex, the y-intercept, and the equation of the axis of symmetry. Then graph the function. :confused:
    pimp_mah_alpaka's Avatar
    pimp_mah_alpaka Posts: 103, Reputation: 1
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    #2

    Nov 30, 2008, 03:22 AM

    Ask the teacher. If you want to learn, learn by going to the teacher so he/she can take you through step by step instructions. This'll be more easier when you get another question like this one
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Dec 1, 2008, 08:42 AM

    First - the y-intercept is the point where x = 0. So simply substitute x=0 and see what you get for f(0).

    The next thing is to restate the equation into a standard form that is like this:

    y-b = (x-a)^2

    The reason is that if you can find the values of 'a' and 'b', then the vertex is at (a,b). You need to emorize this basic form. So, starting with:

    y = x^2 +6x -4

    First - complete the square - take half the value of the coefficient of the x term, square it, and then add and subtract that value on the right hand side. This is a trick that is very useful to have in your arsenal - it comes up a lot! Here the coefficient for the x term is 6, so half of that squared is 9:

    y= x^2 + 6x + 9 -9 - 4 = x^2 - 6x + 9 -13

    Bring the 13 to the left hand side:

    y +13 = x^2 +6x + 9

    Now you can factor the right hand side:

    y+13 = (x+3)^2

    Now you can see why completing the square was a good idea - it gets you to the form you wanted of (x-a)^2.

    Note that the basic eqation has a negative sign in front of both the 'a' and 'b' terms, so we need to restate this as:

    y - (-13) = (x - (-3))^2

    So now we have a = -3 and b = -13. This tells you the vertex is at the point (-3,-13).
    Finally, the axis of symmetry goes through the vertex, and is parallel to the y axis. Can you take it from here?

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