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    Stretch and label each pair of graphs on the same set of axes showing the coordinates of any points where the graphs intersect. Write down the equation of any asymptotes.
    1.y=x^2 and y=x^3
    i have drawn the graphs but according to answers the point where both graphs intersect is (0,0) and (1,1). Can someone explain how to get these points please?
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    (Original post by Chelsea12345)
    Stretch and label each pair of graphs on the same set of axes showing the coordinates of any points where the graphs intersect. Write down the equation of any asymptotes.
    1.y=x^2 and y=x^3
    i have drawn the graphs but according to answers the point where both graphs intersect is (0,0) and (1,1). Can someone explain how to get these points please?
    y=x^2 and y=x^3

    When the lines intersect, it means that both the x and y values are the same in both functions.

    Therefore  y = x^3 = y = x^2 , hence x^3 = x^2

    Subtract x^2 from both sides, factorise and hence solve.

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    x^3 - x^2 = 0

    x^2(x-1) = 0

    x = 1 or x = 0


    Then substitute in for the y values.
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    (Original post by Chelsea12345)
    Stretch and label each pair of graphs on the same set of axes showing the coordinates of any points where the graphs intersect. Write down the equation of any asymptotes.
    1.y=x^2 and y=x^3
    i have drawn the graphs but according to answers the point where both graphs intersect is (0,0) and (1,1). Can someone explain how to get these points please?
    To make things easier for yourself draw out a table with values for each graph.

    So for this intance for x^2 the table will look like this

    X: 1, 2, 3, 4, 5.
    Y: 1, 4, 9, 16, 25.

    You get the y values since when x=1, y=(1)^2=1

    And for x^3 table

    X: 1, 2, 3, 4, 5.
    Y: 1, 8, 27, 64, 125.

    From this we can tell the graphs intersect at (1,1). And by memory we know the x^2 and x^3 graphs cross at the origin.
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    (Original post by K-Man_PhysCheM)
    y=x^2 and y=x^3

    When the lines intersect, it means that both the x and y values are the same in both functions.

    Therefore  y = x^3 = y = x^2 , hence x^3 = x^2

    There are two ways of solving this: either divide both sides by x^2, or subtract x^2 from both sides.

     \dfrac{x^3}{x^2} = \dfrac{x^2}{x^2}

     x = 1

    or:

    x^3 - x^2 = x^2 - x^2

     x = 0

    Then substitute in for the y values.
    The way you laid it out is not how it's done conventionally. You can easily get around the "OR" bit by simply factorising x^3-x^2=0 which gives you both solutions. Never divide when searching for roots as you risk losing solutions if you don't come back to pre division - by dividing you are assuming x\not= 0 too which cannot be assumed without that information being derived or given.

    Also don't post full solutions.

    (Original post by undercxver)
    To make things easier for yourself draw out a table with values for each graph.

    So for this intance for x^2 the table will look like this

    X: 1, 2, 3, 4, 5.
    Y: 1, 4, 9, 16, 25.

    You get the y values since when x=1, y=(1)^2=1

    And for x^3 table

    X: 1, 2, 3, 4, 5.
    Y: 1, 8, 27, 64, 125.

    From this we can tell the graphs intersect at (1,1). And by memory we know the x^2 and x^3 graphs cross at the origin.
    Not quite how it's expected to be done at A-Level.
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    (Original post by RDKGames)
    The way you laid it out is not how it's done conventionally. You can easily get around the "OR" bit by simply factorising x^3-x^2=0 which gives you both solutions. Never divide when searching for roots as you risk losing solutions if you don't come back to pre division.

    Also don't post full solutions.

    Ahh yes, thank you for the correction. And yeah, I forgot about the not posting full solutions thing, sorry.
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    (Original post by RDKGames)

    Not quite how it's expected to be done at A-Level.
    Didn't realise this was A level Maths. :ninja2:
 
 
 
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