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    1. When f(x) = x^2 - x - 3 draw the graph of y= f(x) for -3 ≤ x ≤ 4
    Would I use the coordinates from the table of values to plot the graph or would I graph the inequality -3 ≤ x ≤ 4 ?
    The coordinates are:
    (-3,9)(-2,3)(-1,-1)(0,-3)(1,-3)(2,-1)(3,3)(4,9)

    (I have plotted both of the graphs but I am not sure which is correct)
    Then, by drawing a suitable line, estimate the gradient of the graph at the point x = -1.

    2. When g(x) = 6 - x^3/3 plot the graph of y = g(x) -2 ≤ x ≤ 3 on the same grid as before.
    Again would I plot the inequality or use the coordinates from the table of values?
    The coordinates in this question are:
    (-2,8.67)(-1,6.33)(0,6)(1,5.67)(2,3.33)(3,-3)

    3. Show that the equation f(x) = g(x) simplifies to x^3 c 3x^2 - 3x - 27 =0.
    Use your graph to write down the solution of x^3 c 3x^2 - 3x - 27 =0.

    I would like some help with how to answer these questions, thank you.
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    Hey man, these should be the answers, but you was right about the coordinate points. Feel free to
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    Question 1 - Since f(x) = y, substitute each value in that range from -3 to 4 to find its y value, and use those coordinate points to plot the graph.



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    Question 2 - Since g(x) = y, do the same thing as question 1 and find the y value when you substitute each value of x in the range of -2 and 3, then plot the coordinates.



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    Question 3 - Equate both of the equations and bring everything to one side so that you get the answer given, but make sure you show every step. Then since it says use the graph, as long as you have drawn them accurately, you should be able to get an approximation of where they both intersect.

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    Name:  F252AB3F-D780-4010-AAAE-53C0F50AFF61.jpeg
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Size:  148.1 KBWould this be the first graph? How would I find the gradient of this graph? Would I plot a diagonal tangent at x = -1 and draw a triangle from this to find the gradient using increase in y / increase in x? If yes, I drew this tangent and drawing a triangle I found the gradient to be 6/2 = 3. Is this correct?

    Bradley00
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    Name:  600DB700-1E5A-4FF2-9586-20C57FF19998.jpeg
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Size:  153.7 KBAnd is this the correct second graph? Bradley00
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    Both graphs look great to me, as long as you have input your x values and got your corresponding y value, then you should be all good. As for your gradient question, usually the question will give you something along the lines of 'find the gradient at point (x,y) on this graph'. This is when you draw your tangent and find the gradient
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    For the third question f(x) = g(x)
    x^2-x-3 = 6 - x^3/3

    Move all of the terms to one side:

    x^2 - x - 3 - 6 + x^3/3 = 0

    Simplify -3 - 6 to -9:
    x^2 - x - 9 + x^3/3

    Multiply by 3:

    3x^2 - 3x - 27 + x^3 = 0

    Rearrange to the desired form:

    x^3 + 3x^2 - 3x - 27 = 0

    Is this correct?

    Bradley00
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    Absolutely correct for the rearranging on question 3
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    Thank you for all of your help. One last question in regard to finding the solution of x^3 - 3x^2 - 3x - 27 =0, you previously stated that I would find need to find the intersection of both graphs. Would this be the overall solution or do I have to do something with these coordinates?
    The point of intersection I found was (2.2,0.6), does this sound right?

    I really appreciate all of your help, thank you very much 😊

    Bradley00
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    No, you wouldn't have to do anything with the coordinate points, once you find them then that's it, and you are very welcome.

    Feel free to ask about any other questions in the future, and If I can't answer I'm 100% sure that someone else will
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    (Original post by Bradley00)
    No, you wouldn't have to do anything with the coordinate points, once you find them then that's it, and you are very welcome.

    Feel free to ask about any other questions in the future, and If I can't answer I'm 100% sure that someone else will
    Thank you very much you have helped massively 😊
 
 
 
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