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Math Help - How do you graph this function?

  1. #1
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    How do you graph this function?

    f(x)=x^2+6x
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  2. #2
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    Quote Originally Posted by juliajank View Post
    f(x)=x^2+6x
    There are alot of ways to do this the most simple is to plot points.

    So pick some values of x and plug them in to get y values and plot the points...

    here is one let x=1 \implies f(1)=1^2+6(1)=7 \implies (1,7)
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  3. #3
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    Thanks a lot for your answer, but I can't use the table of values anymore

    Do you have to factor out the common x first to make it
    x(x+6) and then graph it as a normal line, six units left?
    or are you supposed to graph it as a quadratic function? :S
    When you factor the x out, does the make whatever is in the brackets the only important thing now?
    I am so confused with what I have to do if there is more than one x!
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  4. #4
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    Quote Originally Posted by juliajank View Post
    Thanks a lot for your answer, but I can't use the table of values anymore

    Do you have to factor out the common x first to make it
    x(x+6) and then graph it as a normal line, six units left?
    or are you supposed to graph it as a quadratic function? :S
    When you factor the x out, does the make whatever is in the brackets the only important thing now?
    I am so confused with what I have to do if there is more than one x!

    There are many methods. What method are you supposed to use.

    Here is one...

    We could first find the x-intercepts by setting y=0

    0=x^2+6x=x(x+6) \implies x=0 or x=-6

    So the ordered pairs of the x intercepts are (0,0) and (-6,0)

    Frome here we can find the vertex by using the fact that a parabola is symmetric about its vertex. That means the x coordinate of the vertex is at the mid point of the x intercepts. x=\frac{0+(-6)}{2}=-3

    Now we can complete the order pair by plugging this back into the equation
    y=(-3)^2+6(-3)=9-18=-9 so the vertex is at (-3,-9)

    So if we plot these three points we can sketch the parabola.

    Another methold we be to complete the square...

    Do any of these sound familar...
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  5. #5
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    Thumbs up

    Yes! Thank you. I would have to solve it by completing the square which is not very hard. It's just figuring out what to do that's trickiest for me.
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  6. #6
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    Talking

    Quote Originally Posted by juliajank View Post
    It's just figuring out what to do that's trickiest for me.
    For worked examples of the general methodology, try here.
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