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    Using differentiation, find the turning point of the function y = 4x^2 – 2x + 6 and sketch its curve.

    I find it easy to differentiate, this would be: 8x-2.
    But how would you drave the curve, finding maximum, minimum and point of inflexion?
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    (Original post by Deannnn97)
    Using differentiation, find the turning point of the function y = 4x^2 – 2x + 6 and sketch its curve.

    I find it easy to differentiate, this would be: 8x-2.
    But how would you drave the curve, finding maximum, minimum and point of inflexion?
    Well, you should know the general shape of these types of curves. Positive quadratics are a U shape, and the minimum is where dy/dx = 0. There's no inflexion point or maximum.
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    (Original post by 13 1 20 8 42)
    Well, you should know the general shape of these types of curves. Positive quadratics are a U shape, and the minimum is where dy/dx = 0. There's no inflexion point or maximum.
    No i couldnt go the classes where this was taught, so as the x is posotive it would be a U shape? So for 8x-2 what effect does this have on the position?
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    (Original post by Deannnn97)
    No i couldnt go the classes where this was taught, so as the x is posotive it would be a U shape? So for 8x-2 what effect does this have on the position?
    Yeah.

    For a general function y = ax^2 + bx + c, a > 0, we have a U shape; if a < 0 then we have an inverted U (you can think of it as smiley face vs sad face if you want lol).

    Basically, where dy/dx = 0, you have a "turning point" of the graph. Remember that dy/dx gives the gradient at any point, right, i.e. the slope of the tangent line at that point (the straight line that just touches the graph there). So when dy/dx = 8x - 2 = 0, the tangent line is horizontal, right? And when 8x - 2 > 0, the graph has positive gradient, when 8x - 2 < 0, the graph has negative gradient. So 8x - 2 = 0 in fact gives the x point which is at the bottom of that "U" shape.
 
 
 
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