maths

1)y=x²-3x+2
2)y=8x-16/x-3

A)Sketch both on axes
B)Calculate coordinate of points of intersection

I have drawn both on axes. I am having trouble with B). I know you have to do y=y.

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  1. Well, the first one is a parabola of course.
    I complete the square to graph

    x^2 -3 x = y - 2
    x^2 - 3 x + 9/4 = y+ 1/4
    (x-3/2)^2 = y+ 1/4
    so vertex at (3/2, -1/4)
    now zeros
    x = 3/2 +/- (1/2)sqrt (9-8)
    = 1 or 2 at y = 0

    However the second one if it is really

    y = 8 x -(16/x) - 3

    must be hard to sketch because it is undefined at the origin.
    It goes through (1, -11)
    It goes through (2,5)
    so it crosses the parabola somewhere between x = 1 and x = 2
    look for that point somewhere around 1.25
    also look for a point at larger x
    and check for a point at negative x
    I would do it numerically but you can try setting them equal
    x^2 - 3 x + 2 = 6 x - 16/x -3
    x^3 - 3 x^2 +2 x = 6 x^2 -16 - 3x
    x^3 - 9 x^2 + 5 x +18 = 0
    or
    x^3 - 11 x^2 + 5 x = -16
    try x around 1 to 2 and look for -16
    x = 1 --> -5
    x = 1.2 --> -8.1
    x = 1.3 ---> -9.89
    x = 1.4 ---> -11.8
    x = 1.5 ---> -13.9
    x = 1.6 ---> -16 Oh, oh score
    Now you could call (x-1.6) a factor and divide to get a quadratic or just search the same way or other roots.

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  2. I don't know if your second equation is y = (8x-16)/(x-3) or 8x - (16/x) -3 or 8x - 16/(x-3)
    Without parentheses, it is ambiguous.
    Whatever the second equation is, if you have drawn the graphs, just see where they intersect. Or, as you mentioned, solve the equation y1 = y2, where the y's are the two different formulas.

    If your second equation is
    y2 = (8x-16)/(x-3) = 8(x-2)/(x-3),
    Then use also
    y1 = (x-2)(x-1)
    Then (x-2)(x-1)= 8(x-2)/(x-3) at the intersections.
    x-1 = 8/(x-3)
    x^2 -4x -5 = 0
    (x-5)(x+1) = 0
    x = 5 is one solution. At that intersection point, y = 25 -15 +2 = 12
    I'll leave the other for you to solve.

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  3. Remember x = 2, both sides 0

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