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Math Help - Im so stuck point on graph

  1. #1
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    Im so stuck point on graph

    I think this is the right forum, probably wrong but oh well

    f(x)=ax3-3x. The point (-2,0) lies on the graph of y=f(x). Find the value of a.

    Thanks for the future help ppl.

    xxJoJoxx
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jojo View Post
    I think this is the right forum, probably wrong but oh well

    f(x)=ax3-3x. The point (-2,0) lies on the graph of y=f(x). Find the value of a.

    Thanks for the future help ppl.

    xxJoJoxx
    The point (-2,0) lies on y-f(x), so f(-2)=0, so:

    a(-2)^3-3(-2)=-8a+6=0,

    so rearranging:

    8a=6,

    or a=3/4.

    RonL
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  3. #3
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    Hello again soz to ask but theres a b bit I swear I'm actually good at maths I've been off for 2 weeks tho

    "Find the coordinates of the point where the line 8y-6x=48 cuts the X-axis."

    I think its that x=0 thing but cant remember. Thanks again
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jojo View Post
    Hello again soz to ask but theres a b bit I swear I'm actually good at maths I've been off for 2 weeks tho

    "Find the coordinates of the point where the line 8y-6x=48 cuts the X-axis."

    I think its that x=0 thing but cant remember. Thanks again
    No it cuts the x-axis when y=0. So you need to solve:

    -6x=48,

    which gives x=-8.

    RonL
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