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    So I’ve got a question where f(x) = (x - 1) (x + 2) and I have to draw the graph y = f(x + 2). I’ve got the x axis intersections as -4 and -1, however I don’t understand how the y axis intersection is 4... I mean if x is 0, y = -1 x 2 = -2; and then add 2 makes 0? Not 4? Please help where I’ve went wrong!
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    if x =0 then y=f(2) f(2) =?
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    If x=0, f(x+2) = f(0+2) = f(2)

    (2-1)(2+2) = 4
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    (Original post by funkybinoculars)
    So I’ve got a question where f(x) = (x - 1) (x + 2) and I have to draw the graph y = f(x + 2). I’ve got the x axis intersections as -4 and -1, however I don’t understand how the y axis intersection is 4... I mean if x is 0, y = -1 x 2 = -2; and then add 2 makes 0? Not 4? Please help where I’ve went wrong!
    graph moves two steps to the left-the y-intercept should be 4 now. if not then im not sure
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    Be careful for the difference between "f(x+a)" and "f(x) + a"!
    (Where "a" is a constant).

    For x-y graph transformations:

    - f(x-a),
    Shifts f(x) by a-units on the X-AXIS to the right
    (Therefore negative values of "a" shift f(x) by "|a|"-units to the left on the x-axis).

    We can see this by your example, f(x) = (x-1)(x+2), there are x-intercepts at x=1 and x=-2.
    Therefore f(x+2) = (x+1)(x+4), has x-intercepts at x=-1 and x = -4, (shifted by -2 units on x-axis).
    -----------------------------------------------------------------------------------------------------------------------------

    -f(x) + a,
    Shifts f(x) up by "a" units on the Y-AXIS
    (Similarly, if "a" is a negative value, the f(x) is shifted DOWN by "|a|" units on the y-axis).

    For your example, f(x) = (x-1)(x+2), has y-intercept at y = (-1)(2) = -2.
    f(x) + 2 = (x-1)(x+2) + 2, has y-intercept y= (-1)(2) + 2 = 0.
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    (Original post by funkybinoculars)
    So I’ve got a question where f(x) = (x - 1) (x + 2) and I have to draw the graph y = f(x + 2). I’ve got the x axis intersections as -4 and -1, however I don’t understand how the y axis intersection is 4... I mean if x is 0, y = -1 x 2 = -2; and then add 2 makes 0? Not 4? Please help where I’ve went wrong!


    f(x) = (x - 1)(x + 2) becomes f(x + 2)

    This means that every value of x in the orignal equation becomes 'x + 2'.

     (x - 1)(x + 2) \to\ ( (x + 2) - 1)( (x +2) +2) \to\ (x + 1)(x +4)

    Your problem was that you were substituting x = 0 into f(x) and not f(x+2)
 
 
 
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