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A question for C1 Watch

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    Hey guys this is a question for C1 edexcel











    I get parts a b and c. But d i don't get.

    I get where they got y=x^2 + 2x - 5 from . bit i dont get is (x+1)^2 - 6 . minimum at (-1,-6). how do they get that

    thanks guys
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    (Original post by jayseanfan)
    I get where they got y=x^2 + 2x - 5 from . bit i dont get is (x+1)^2 - 6 . minimum at (-1,-6). how do they get that

    thanks guys
    You need to learn how to complete the square.

    If a quadratic is in the form y = k(x-a)^2 + b, then the coordinates of the vertex are (a, b).
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    (Original post by Mr M)
    You need to learn how to complete the square.

    If a quadratic is in the form y = k(x-a)^2 + b, then the coordinates of the vertex are (a, b).
    Why can't you factorise?
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    (Original post by jayseanfan)
    Why can't you factorise?
    Try it. It doesn't work. (with integer values anyway)
    Completing the square in situations where factorising isn't possible is easier. Also you get the minimum point, where as if you factorised, you'd only get the roots of the equation.
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    (Original post by jayseanfan)
    ...
    you need to think about what value of x will make the (x+1)^2 as small as possible (the smaller that bracket the smaller the value of y, right?)
    Spoiler:
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    if something is squared it can't have a value less than 0. so essentially you want (x+1) to be equal to 0 and when x = -1, (x+1)^2 = 0.


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    the corresponding y value when x = -1 is -6, therefore the minimum point is (-1,-6)


    p.s. if a quadratic isn't factorisable use either complete the square or differentiation to find minimum points

    differentiation:
    y = x^2 + 2x - 5
    dy/dx = 2x - 2

    minimum point => dy/dx = 0
    2x + 2 = 0
    2x = -2
    x = -1
    plug into the formula to get y = -6

    therefore min point at (-1,-6)

    (i prefer differentiation but that's probably because i do a **** load of it in further maths. it makes a lot of sense when you understand the concept)
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    (Original post by F1Addict)
    Try it. It doesn't work. (with integer values anyway)
    Completing the square in situations where factorising isn't possible is easier. Also you get the minimum point, where as if you factorised, you'd only get the roots of the equation.
    -Thanks for your reply
    - Yes you are right is doesn't work.
    - Once you have completed the square, how do you use it to get the minimum point?
    - Also if I did factorise it I would have put x=0 and found the y intercept and put y=0 to find the x -axis intercepts . So basically I would have found were the curve is crossing the axes, so why in this case do you use the minimum point instead or using the intercepts to sketch the curve? And can you even use the mimum point to sketch the curve?
    - Why havent they labelled on the minimum point on their graph ?
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    (Original post by jayseanfan)
    Once you have completed the square, how do you use it to get the minimum point?
    Did you even read my previous post?
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    (Original post by jayseanfan)
    -Thanks for your reply
    - Yes you are right it doesn't work.
    - Once you have completed the square, how do you use it to get the minimum point?
    See here: http://www.thestudentroom.co.uk/show...1#post28857201
    - Also if I did factorise it I would have put x=0 and found the y intercept and put y=0 to find the x -axis intercepts . So basically I would have found were the curve is crossing the axes, so why in this case do you use the minimum point instead or using the intercepts to sketch the curve? And can you even use the minimum point to sketch the curve?
    Yes, the y-intercept is when x=0.
    Yes, if you factorised, you are essentially putting y=0 into the equation.
    In this case, you're using the minimum point instead as you don't know the x-intercepts. You can use the minimum point to sketch a curve, as you should know the general shape, where the minimum point lies, and the y-intercept, so you can sorta guage whereabouts the curve will cross the x-axis.

    - Why havent they labelled on the minimum point on their graph?
    The graph was drawn in part a, not part d. You worked out the minimum point in part d, not part a, so it wasn't labelled in the diagram. If you do know the minimum point, its common practice to label it.
    My post in bold.
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    (Original post by F1Addict)
    My post in bold.
    Thanks for the reply. very helpfull

    when i complete the square for y=x^2 + 2x - 5 i get = (x+1)^2 - 1 ?
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    (Original post by jayseanfan)
    Thanks for the reply. very helpfull

    when i complete the square for y=x^2 + 2x - 5 i get = (x+1)^2 - 1 ?
    Then you don't know how to complete the square:

    x^2+2x-5 = (x+1)^2 - 1^2 - 5 = (x+1)^2 - 6
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    (Original post by Mr M)
    You need to learn how to complete the square.

    If a quadratic is in the form y = k(x-a)^2 + b, then the coordinates of the vertex are (a, b).

    Right, so I completed the square now.

    = (x+1)^2 -6

    so the minmum point should be (1, -6) but its (-1,-6) , why is it -1 and not +1 after all inside the bracket it says +1
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    (Original post by jayseanfan)
    Right, so I completed the square now.

    = (x+1)^2 -6

    so the minmum point should be (1, -6) but its (-1,-6) , why is it -1 and not +1 after all inside the bracket it says +1
    READ my post carefully (the one you quoted) paying particular attention to the signs.
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    You should know that a tranformation of y = f(x) to y = f(x - a) + b is a translation of a units in the positive x direction and b units in the positive y direction. If f(x) = x^2 then the vertex of the parabola is translated from (0, 0) to (a, b).
 
 
 
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