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Cubic Graphs

I think I understand cubic graphs except for one thing: How do I know whether the graph crosses a point or touches it? I know how to get the points that the graph will cross or touch, but I do not know how to determine whether it will cross or touch the point in question. Could anyone elaborate?
Original post by Illidan2
I think I understand cubic graphs except for one thing: How do I know whether the graph crosses a point or touches it? I know how to get the points that the graph will cross or touch, but I do not know how to determine whether it will cross or touch the point in question. Could anyone elaborate?


Huh? How can a graph 'touch' a point...? It either goes through it or it doesn't.
Reply 2
I'm only going on what my textbook tells me. Sometimes, the graph will dip under the axis after crossing the point, and sometimes, in some of the graphs, the graph will 'touch' the point, not cross over the axis, and continue going up or down without crossing over.
Ignore previous post. The graph will touch the x-axis if 2 or 3 of the roots are the same. For example, the cubic graph (x+1)(x+2)^2 will cross the x-axis at -1 and touch the point -2, not go through it.
Reply 4
Oh, I see! This makes perfect sense now. Thank you so much! :smile:
Original post by lmaofolks
Ignore previous post. The graph will touch the x-axis if 2 or 3 of the roots are the same. For example, the cubic graph (x+1)(x+2)^2 will cross the x-axis at -1 and touch the point -2, not go through it.


It still goes THROUGH (2,0)(-2,0)...

Original post by Illidan2
I'm only going on what my textbook tells me. Sometimes, the graph will dip under the axis after crossing the point, and sometimes, in some of the graphs, the graph will 'touch' the point, not cross over the axis, and continue going up or down without crossing over.


I think what you mean is 'How do I determine whether the x-axis will be tangent to the cubic?' in which case, you should know that a cubic will ALWAYS have 1 real root, ie it will always cross the x-axis at some point. If you know that point, you can divide your cubic through by (xa)(x-a) where a is that point, and see how many roots the leftover quadratic has. If 0, then the cubic does not touch the x-axis (excluding the point aa), if 1 then the cubic touches the x-axis once, and if 2 then the cubic intersects the x-axis two more times.

Also if you can factorise your cubic such that (xa)(xb)(xc)=0(x-a)(x-b)(x-c)=0 then if any two pairs of a,b,ca,b,c are equal to each other, and the other is distinct, then you'll have the situation where it does touch the x-axis.
(edited 6 years ago)
Reply 6
Original post by RDKGames
It still goes THROUGH (2,0)(-2,0)...



I think what you mean is 'How do I determine whether the x-axis will be tangent to the cubic?' in which case, you should know that a cubic will ALWAYS have 1 real root, ie it will always cross the x-axis at some point. If you know that point, you can divide your cubic through by (xa)(x-a) where a is that point, and see how many roots the leftover quadratic has. If 0, then the cubic does not touch the x-axis (excluding the point aa), if 1 then the cubic touches the x-axis once, and if 2 then the cubic intersects the x-axis two more times.

Also if you can factorise your cubic such that (xa)(xb)(xc)=0(x-a)(x-b)(x-c)=0 then if any two pairs of a,b,ca,b,c are equal to each other, and the other is distinct, then you'll have the situation where it does touch the x-axis.


I'll try to remember that. Thank you :smile:
Reply 7
I was asked to sketch the graph y=x3+4x2+4x y=x^3+4x^2+4x .

I obtained the coordinates thus:

x(x2+4x+4)=0[br]x(x+1)(x+4)=0[br][br]x=0,x=1,x=4.[br][br]Sothegraphcrossesthexaxisat(0,0),(1,0)and(4,0). x(x^2+4x+4)=0[br]x(x+1)(x+4)=0[br][br]x=0, x=-1, x=-4.[br][br]So the graph crosses the x-axis at (0,0), (-1,0) and (-4,0).

I have attached a screenshot of the way my textbook derives the solution.

Why is the way I derived the coordinates incorrect? I have derived coordinates of cubic graphs from equations in the form y=x(x+a)(x+b) y=x(x+a) (x+b) before.
Reply 8
Ignore this. I made a silly factorisation error. Had I not done this, I would have derived the solution the same way.

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