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    how do i find the minimum/maximum point of this equation:

    x2-8x+15

    the 2 being the squared
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    (Original post by maccaism)
    how do i find the minimum/maximum point of this equation:

    x2-8x+15

    the 2 being the squared
    What do you think you would need to do to the equation?

    (Think about previous questions or examples you've seen).
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    -3 -5

    so x=+3 and +5

    Not sure tbh by what you mean.
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    (Original post by maccaism)
    how do i find the minimum/maximum point of this equation:x2-8x+15the 2 being the squared
    Differentiate the original equation (dy/dx) , then equate the new equation to 0 and solve.

    EDIT: to find the x values
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    (Original post by maccaism)
    how do i find the minimum/maximum point of this equation:

    x2-8x+15

    the 2 being the squared
    Complete the square is probably what you want.

    Differentiating works but I don't think that's the aim of the Q
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    you could find the roots... the turning point is always equidistant from the roots on a quadratic function
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    (Original post by 08Mercyf)
    -3 -5

    so x=+3 and +5

    Not sure tbh by what you mean.
    i got that............. so is that the minimum point?
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    isn't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the Min value is -1 ??? No point differentiating it. This type of question comes up in a lot of past papers and this is usually what they ask you to do when finding the min/max value.
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    There is only one point as it is a quadratic equation.
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    (Original post by ihatehannah)
    isen't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the min value is -1 ???
    yep i did completing the square and i got that answer too. i just forgot how to find the minimum value and the x and y co ordinates.

    so the minimum value is -1?
    and would the x and y co ordinates be 3,5?
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    (Original post by ihatehannah)
    isen't the minimum point completing the square so (x+ 4)^2 + 15 which becomes (x+4)^2 -1 , therefore the min value is -1 ???
    Yeah it's x=4 y= -1, you don't have to differentiate but it is a personal preference

    EDIT: apart from it would be x-4 inside the brackets as there was a negative in the original equation
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    (Original post by maccaism)
    so the minimum value is -1?
    and would the x and y co ordinates be 3,5?
    The minimum value is the y value, so the coordinates would be 4,-1
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    (Original post by maccaism)
    yep i did completing the square and i got that answer too. i just forgot how to find the minimum value and the x and y co ordinates.

    so the minimum value is -1?
    and would the x and y co ordinates be 3,5?


    edit: (x-4)* so x= 4, the answer is (4,-1) , not ( -4,-1)
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    (Original post by ihatehannah)
    x= -4 , y=-1 , you mean.
    No the original equation is x^2 -8x +15 therefore...
    dy/dx = 2x -8
    2x-8=0
    2x=8
    x=4
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    (Original post by Nirvana27)
    No the original equation is x^2 -8x +15 therefore...
    dy/dx = 2x -8
    2x-8=0
    2x=8
    x=4
    oh my bad, thought it was x^2 + 8x + 15, yes x= 4.
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    (Original post by ihatehannah)
    the bracket ( x+4) , for it to equal to zero, x must be -4, and we know y= -1, therefore the answer is ( -4, -1)
    x^2 -8x +15
    (x-4)^2 -16 +15
    (x-4)^2 -1
    Doing it the other way
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    (Original post by Nirvana27)
    x^2 -8x +15
    (x-4)^2 -16 +15
    (x-4)^2 -1
    Doing it the other way
    So the minimum value is 4,-1
    x=4 and y= -1
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    (Original post by maccaism)
    So the minimum value is 4,-1
    x=4 and y= -1
    Yeah, the minimum value is always the y value so -1, but the min coordinates are (4,-1).
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    (Original post by Nirvana27)
    Yeah, the minimum value is always the y value so -1, but the min coordinates are (4,-1).
    Thank you very much!
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    (Original post by maccaism)
    how do i find the minimum/maximum point of this equation:

    x2-8x+15

    the 2 being the squared
    Firstly, you will need to find gradient of the equation by calculating the derivative.
    (X^2)-8X+15.......
    dy/dx = 2x-8
    To calculate the minima or maxima, use the calculated gradient.
    set the equation equal to zero..
    2X-8 = 0.
    Making X as the subject, 2X-8=0
    2X=8
    X=8/2
    X=4
    as the gradient changes from -ve to +ve, for example the point reached is 4 and is positive, so you will have a minimum at x= 4.
    Using the X value, substituting X=4 into y= (x^2)-8X+15
    you will get y= -1.
    therefore the minimum point is (4, -1) .
    sorry I have plotted the graph wrong, but I hope it will give you an Idea.
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