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core 1 maths help?

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    I need to find the co-ordinate of the stationary point of the curve y=x^2-4x+5.
    so to find the x co-ordinates do i factorise or complete the square? i tried factorisng the equation but it doesnt work?
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    (Original post by Chelsea12345)
    I need to find the co-ordinate of the stationary point of the curve y=x^2-4x+5.
    so to find the x co-ordinates do i factorise or complete the square? i tried factorisng the equation but it doesnt work?
    Well you can't always factorise. With quadratics and their stationary points you have 2 options:

    1) Complete the square to find the x-coordinate and then by inspection (or substitution) find the y-coordinate

    2) Differentiate the function and set the differential equal to 0 and solve it. Setting it to 0 will ensure that the rate of change is 0 as to be expected at a stationary point.
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    (Original post by RDKGames)
    Well you can't always factorise. With quadratics and their stationary points you have 2 options:

    1) Complete the square to find the x-coordinate and then by inspection (or substitution) find the y-coordinate

    2) Differentiate the function and set the differential equal to 0 and solve it. Setting it to 0 will ensure that the rate of change is 0 as to be expected at a stationary point.
    if you complete the square you only get one x-cordinate,dont you need 2 of them??
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    (Original post by Chelsea12345)
    if you complete the square you only get one x-cordinate,dont you need 2 of them??
    Yeah, you get the same through differentiation. You find the x-coordinate first and then plug it back through the quadratic to see what the y-coordinate is.
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    (Original post by Chelsea12345)
    if you complete the square you only get one x-cordinate,dont you need 2 of them??
    Hi, as this is a quadratic, your only stationary point is your vertex. Complete the square to find this.
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    (Original post by RDKGames)
    Yeah, you get the same through differentiation. You find the x-coordinate first and then plug it back through the quadratic to see what the y-coordinate is.
    EDIT: Never mind, I was being an idiot LOLOL
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    (Original post by TiernanW)
    When you complete the square isn't it just the equation re-wrote in another form? So aren't you then finding the points where it crosses the x-axis or am I being an idiot?
    Completing the square is indeed a different form to write a quadratic, and no it does not necessarily mean you are finding the roots of the equation.

    If I put some arbitrary quadratic in the form of (x-a)^2+b then I can see how much the parabola x^2 has been shifted along the x-axis (in this case it has been shifted by vector [a,0] as far as the horizontal transformation is concerned). Since the x-coordinate of the stationary point of x^2 is at 0, the shift will be the same for it to get the stationary point of (x-a)^2+b.

    If you want to find the roots, you would have to solve (x-a)^2+b=0
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    (Original post by slowdive)
    Hi, as this is a quadratic, your only stationary point is your vertex. Complete the square to find this.
    okay thankyou!!
 
 
 
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